Category: Class 11

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 4 Bivariate Frequency Distribution and Chi Square Statistic Miscellaneous Exercise 4

Question 1.
Following data gives the coded price (X) and demand (Y) of a commodity.

Classify the data by taking classes 0 – 4, 5 – 9, etc. for X and 5 – 8, 9 – 12, etc. for Y.
Also find
(i) marginal frequency distribution of X and Y.
(ii) conditional frequency distribution of Y when X is less than 10.
Solution:
Given, X = coded price
Y = demand
Bivariate frequency table can be prepared by taking class intervals 0 – 4, 5 – 9,… etc for X and 5 – 8, 9 – 12,… etc for Y.
Bivariate frequency distribution is as follows.

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X < 10:

Question 2.
Following data gives the age in years and marks obtained by 30 students in an intelligence test.


Prepare a bivariate frequency distribution by taking class intervals 16 – 18, 18 – 20,…,etc. for age and 10 – 20, 20 – 30,…, etc. for marks.
Find
(i) marginal frequency distributions.
(ii) conditional frequency distribution of marks obtained when age of students is between 20 – 22.
Solution:
Let X = Age in years
Y = Marks
Bivariate frequency table can be prepared by taking class intervals 16 – 18, 18 – 20,…, etc for X and 10 – 20, 20 – 30,…, etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of Y when X is between 20 – 22:

Question 3.
Following data gives Sales (in Lakh ?) and Advertisement Expenditure (in Thousand ₹) of 20 firms.
(115, 61) (120, 60) (128, 61) (121, 63) (137, 62) (139, 62) (143, 63) (117, 65) (126, 64) (141, 65) (140, 65) (153, 64) (129, 67) (130, 66) (150, 67) (148, 66) (130, 69) (138, 68) (155, 69) (172, 68)
(i) Construct a bivariate frequency distribution table for the above data by taking classes 115 – 125, 125 – 135, ….etc. for sales and 60 – 62, 62 – 64, …etc. for advertisement expenditure.
(ii) Find marginal frequency distributions
(iii) Conditional frequency distribution of Sales when the advertisement expenditure is between 64 – 66 (Thousand ₹)
(iv) Conditional frequency distribution of advertisement expenditure when the sales are between 125 – 135 (lakh ₹)
Solution:
(i) Let X = Sales (in lakh ₹)
Y = Advertisement Expenditure (in Thousand ₹)
Bivariate frequency table can be prepared by taking class intervals 115 – 125, 125 – 135, …. etc for X and 60 – 62, 62 – 64, ….etc for Y.
Bivariate frequency distribution is as follows:

(ii) Marginal frequency distribution of X:

Marginal frequency distribution of Y:

(ii) Conditional frequency distribution of X when Y is between 64 – 66:

(iii) Conditional frequency distribution of Y when X is between 125 – 135:

Question 4.
Prepare a bivariate frequency distribution for the following data, taking class intervals for X as 35 – 45, 45 – 55, …. etc and for Y as 115 – 130, 130 – 145, … etc where, X denotes the age in years and Y denotes blood pressure for a group of 24 persons.
(55, 151) (36, 140) (72, 160) (38, 124) (65, 148) (46, 130) (58, 152) (50, 149) (38, 115) (42, 145) (41, 163) (47, 161) (69, 159) (60, 161) (58, 131) (57, 136) (43, 141) (52, 164) (59, 161) (44, 128) (35, 118) (62, 142) (67, 157) (70, 162)
Also find
(i) Marginal frequency distribution of X.
(ii) Conditional frequency distribution of Y when X < 45.
Solution:
Given X = Age in years
Y = Blood pressure
Bivariate frequency table can be prepared by taking class intervals 35 – 45, 45 – 55, …, etc for X and 115 – 130, 130 – 145, ….., etc for Y.
Bivariate frequency distribution is as follows:

(i) Marginal frequency distribution of X:

(ii) Conditional frequency distribution of Y when X < 45:

Question 5.
Thirty pairs of values of two variables X and Y are given below. Form a bivariate frequency table. Also find marginal frequency distributions of X and Y.

Solution:
Bivariate frequency table can be prepared by taking class intervals 80 – 90, 90 – 100, etc for X and 500 – 600, 600 – 700, …., etc for Y.
Bivariate frequency distribution is as follows:

Marginal frequency distribution of X:

Marginal frequency distribution of Y

Question 6.
The following table shows how the samples of Mathematics and Economics scores of 25 students are distributed:

Find the value of ϰ² statistic.
Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{35 \times 25}{50}\) = 17.5
E₁₂ = \(\frac{35 \times 25}{50}\) = 17.5
E₂₁ = \(\frac{15 \times 25}{50}\) = 7.5
E₂₂ = \(\frac{15 \times 25}{50}\) = 7.5
Table of expected frequencies.

Question 7.
Compute ϰ² statistic from the following data:

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{50 \times 60}{100}\) = 30
E₁₂ = \(\frac{50 \times 40}{100}\) = 20
E₂₁ = \(\frac{50 \times 60}{100}\) = 30
E₂₂ = \(\frac{50 \times 40}{100}\) = 20
Table of expected frequencies.

Question 8.
The attitude of 250 employees towards a proposed policy of the company is as observed in the following table. Calculate ϰ² statistic.

Solution:
Table of observed frequencies

Expected frequencies are given by
E_ij = \(\frac{\mathrm{R}_{\mathrm{i}} \times \mathrm{C}_{\mathrm{j}}}{\mathrm{N}}\)
E₁₁ = \(\frac{150 \times 95}{250}\) = 57
E₁₂ = \(\frac{150 \times 95}{250}\) = 57
E₁₃ = \(\frac{150 \times 60}{250}\) = 36
E₂₁ = \(\frac{100 \times 95}{250}\) = 38
E₂₂ = \(\frac{100 \times 95}{250}\) = 38
E₂₃ = \(\frac{100 \times 60}{250}\) = 24
Table of observed frequencies.

Question 9.
In a certain sample of 1000 families, 450 families are consumers of tea. Out of 600 Hindu families, 286 families consume tea. Calculate ϰ² statistic.
Solution:
The given data can be arranged in the following table.

Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{600 \times 450}{1000}\) = 270
E₁₂ = \(\frac{600 \times 550}{1000}\) = 330
E₂₁ = \(\frac{400 \times 450}{1000}\) = 180
E₂₂ = \(\frac{400 \times 550}{1000}\) = 220
Table of expected frequencies.

Question 10.
A sample of boys and girls were asked to choose their favourite sport, with the following results. Find the value of ϰ² statistic.

Solution:
Table of observed frequencies.

Expected frequencies are given by
E_ij = \(\frac{R_{i} \times C_{j}}{N}\)
E₁₁ = \(\frac{200 \times 126}{300}\) = 84
E₁₂ = \(\frac{200 \times 90}{300}\) = 60
E₁₃ = \(\frac{200 \times 69}{300}\) = 46
E₁₄ = \(\frac{200 \times 15}{300}\) = 10
E₂₁ = \(\frac{100 \times 126}{300}\) = 42
E₂₂ = \(\frac{100 \times 90}{300}\) = 30
E₂₃ = \(\frac{100 \times 69}{300}\) = 23
E₂₄ = \(\frac{100 \times 15}{300}\) = 5
Table of expected frequencies.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Miscellaneous Exercise 1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Miscellaneous Exercise 1

Question 1.
The data gives the number of accidents per day on a railway track. Compute Q₂, P₁₇, and D₇.
4, 2, 3, 5, 6, 3, 4, 1, 2, 3, 2, 3, 4, 3, 2
Solution:
The given data can be arranged in ascending order as follows:
1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6
Here, n = 15
Q₂ = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation
= value of (2 × 4)th observation
= value of 8th observation
∴ Q₂ = 3
P₁₇ = value of 17\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 17\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (17 × 0.16)th observation
= value of (2.72)th observation
= value of 2nd observation + 0.72 (value of 3rd observation – value of 2nd observation)
= 2 + 0.72 (2 – 2)
∴ P₁₇ = 2
D₇ = value of 7\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 7\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (7 × 1.6)th observation
= value of (11.2)th observation
= value of 11th observation + 0.2(value of 12th observation – value of 11th observation)
= 4 + 0. 2(4 – 4)
∴ D₇ = 4

Question 2.
The distribution of daily sales of shoes (size-wise) for 100 days from a certain shop is as follows:

Compute Q₁, D₂, and P₉₅.
Solution:
By arranging the given data in ascending order, we construct the less than cumulative frequency table as given below:

Here, n = 100
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation
= value of (25.25)th observation
Cumulative frequency which is just greater than (or equal) to 25.25 is 27.
∴ Q₁ = 3
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{100+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 10.1)th observation
= value of (20.2)th observation
Cumulative frequency which is just greater than (or equal) to 20.2 is 27.
∴ D₂ = 3
P₉₅ = value of 95\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 95\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (95 × 1.01)th observation
= value of (95.95)th observation
The cumulative frequency which is just greater than (or equal) to 95.95 is 100.
∴ P₉₅ = 8

Question 3.
Ten students appeared for a test in Mathematics and Statistics and they obtained the marks as follows:

If the median will be the criteria, in which subject, the level of knowledge of the students is higher?
Solution:
Marks in Mathematics can be arranged in ascending order as follows:
23, 23, 25, 25, 32, 35, 36, 37, 38, 42
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
Median = value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 32 + 0.5 (35 – 32)
= 32 + 0.5(3)
= 32 + 1.5
= 33.5
Marks in Statistics can be arranged in ascending order as follows:
22, 23, 26, 28, 29, 32, 34, 36, 45, 50
Here, n = 10
∴ Median = value of \(\left(\frac{n+1}{2}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{2}\right)^{\text {th }}\) observation
= value of (5.5)th observation
= value of 5th observation + 0.5(value of 6th observation – value of 5th observation)
= 29 + 0.5(32 – 29)
= 29 + 0.5(3)
= 29 + 1.5
= 30.5
∴ Median marks for Mathematics = 33.5 and
Median marks for Statistics = 30.5
∴ The level of knowledge in Mathematics is higher than that of Statistics.

Question 4.
In the frequency distribution of families given below, the number of families corresponding to expenditure group 2000 – 4000 is missing from the table. However, the value of the 25th percentile is 2880. Find the missing frequency.

Solution:
Let x be the missing frequency of expenditure group 2000 – 4000.
We construct the less than cumulative frequency table as given below:

Here, N = 75 + x
Given, P₂₅ = 2880
∴ P₂₅ lies in the class 2000 – 4000.
∴ L = 2000, h = 2000, f = x, c.f. = 14
∴ P₂₅ = L + \(\frac{h}{f}\left(\frac{25 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 2880 = 2000 + \(\frac{2000}{x}\left(\frac{75+x}{4}-14\right)\)
∴ 2880 – 2000 = \(\frac{2000}{x}\left(\frac{75+x-56}{4}\right)\)
∴ 880x = 500(x + 19)
∴ 880x = 500x + 9500
∴ 880x – 500x = 9500
∴ 380x = 9500
∴ x = 25
∴ 25 is the missing frequency of the expenditure group 2000 – 4000.

Question 5.
Calculate Q₁, D₆, and P₁₅ for the following data:

Solution:
Since the difference between any two consecutive mid values is 50, the width of each class interval is 50.
∴ the class intervals will be 0 – 50, 50 – 100, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
Cumulative frequency which is just greater than (or equal) to 125 is 160.
Q₁ lies in the class 100 – 150.
∴ L = 100, h = 50, f = 80, c.f. = 80
∴ Q₁ = L + \(\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 100 + \(\frac{50}{80}\)(125 – 80)
= 100 + \(\frac{5}{8}\)(45)
= 100 + 28.125
= 128.125
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 500}{10}\) = 300
Cumulative frequency which is just greater than (or equal) to 300 is 410.
∴ D₆ lies in the class 200 – 250.
∴ L = 200, h = 50, f = 150, c.f. = 260
∴ D₆ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{6 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 200 + \(\frac{50}{150}\)(300 – 260)
= 200 + \(\frac{1}{3}\)(40)
= 200 + 13.33
= 213.33
P₁₅ class = class containing \(\left(\frac{15 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{15 \mathrm{~N}}{100}=\frac{15 \times 500}{100}\) = 75
Cumulative frequency which is just greater than (or equal) to 75 is 80.
∴ P15 lies in the class 50 – 100.
∴ L = 50, h = 50, f = 70, c.f. = 10
∴ P₁₅ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{15 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 50 + \(\frac{50}{70}\) (75 – 10)
= 50 + \(\frac{5}{7}\) (65)
= 50 + \(\frac{325}{7}\)
= 50 + 46.4286
= 96.4286
∴ Q₁ = 128.125, D₆ = 213.33, P₁₅ = 96.4286

Question 6.
Daily income for a group of 100 workers are given below:

P₃₀ for this group is ₹ 110. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 50 – 100 and class 200 – 250 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 ……(i)
Given, P30 = 110
∴ P30 lies in the class 100 – 150.
∴ L = 100, h = 50, f = 25, c.f. = 7 + a
\(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
∴ P₃₀ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{30 \mathrm{~N}}{100}-\text { c.f. }\right)\)
∴ 110 = 100 + \(\frac{50}{25}\) [30 – (7 + a)]
∴ 110 – 100 = 2(30 – 7 – a)
∴ 10 = 2(23 – a)
∴ 5 = 23 – a
∴ a = 23 – 5
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18
∴ b = 20
∴ 18 and 20 are the missing frequencies of the class 50 – 100 and class 200 – 250 respectively.

Question 7.
The distribution of a sample of students appearing for a C.A. examination is:

Help C.A. institute to decide cut-off marks for qualifying for an examination when 3% of students pass the examination.
Solution:
To decide cut-off marks for qualifying for an examination when 3% of students pass, we have to find P97.
We construct the less than cumulative frequency table as given below:

Here, N = 1100
P97 class = class containing \(\left(\frac{97 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{97 \mathrm{~N}}{100}=\frac{97 \times 1100}{100}\) = 1067
Cumulative frequency which is just greater than (or equal) to 1067 is 1100.
∴ P₉₇ lies in the class 500 – 600.
∴ L = 500, h = 100, f = 130, c.f. = 970
∴ P₉₇ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{97 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 500 + \(\frac{100}{130}\)(1067 – 970)
= 500 + \(\frac{10}{13}\) (97)
= 500 + 74.62
= 574.62 ~ 575
∴ the cut off marks for qualifying an examination is 575.

Question 8.
Determine graphically the value of median, D₃, and P₃₅ for the data given below:

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (15, 8), (20, 22), (25, 30), (30, 55), (35, 70), (40, 84), (45, 90).

N = 90
For median, consider \(\frac{\mathrm{N}}{2}=\frac{90}{2}\) = 45
For D3, consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 90}{10}\) = 27
For P35, consider \(\frac{35 \mathrm{~N}}{100}=\frac{35 \times 90}{100}\) = 31.5
∴ We take the values 45, 27 and 31.5 on the Y-axis and draw lines from these points parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendicular on the X-axis.
Foot of the perpendicular represent the values of median, D3 and P35 respectively.
∴ Median ~ 29, D₃ ~ 23.5, P₃₅ ~ 26

Question 9.
The I.Q. test of 500 students of a college is as follows:

Find graphically the number of students whose I.Q. is more than 55 graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (30, 41), (40, 93), (50, 157), (60, 337), (70, 404), (80, 449), (90, 489), (100, 500)

To find the number of students whose I.Q. is more than 55, we consider the value 55 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point this line intersects the less than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of students whose I.Q. is less than 55.
∴ The foot of perpendicular ~ 244
∴ No. of students whose I.Q. is less than 55 ~ 244
∴ No. of Students whose I.Q. is more than 55 = 500 – 244 = 256

Question 10.
Draw an ogive for the following distribution. Determine the median graphically and verify your result by a mathematical formula.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (150, 2), (155, 7), (160, 16), (165, 31), (170, 47), (175, 54), (180, 59) and (185, 60).

N = 60
∴ \(\frac{\mathrm{N}}{2}=\frac{60}{2}\) = 30
∴ We take the value 30 on the Y-axis and from this point, we draw a line parallel to X-axis.
From the point where this line intersects the less than ogive, we draw a perpendicular on X-axis.
The foot perpendicular gives the value of the median.
∴ Median ~ 164.67
Now, let us calculate the median from the mathematical formula.
∴ \(\frac{\mathrm{N}}{2}\) = 30
The median lies in the class interval 160 – 165.
∴ L = 160, h = 5, f = 15, c.f. = 16
Median = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{2}-\mathrm{c} . \mathrm{f} .\right)\)
= 160 + \(\frac{5}{15}\) (30 – 16)
= 160+ \(\frac{1}{3}\) × 14
= 160 + 4.67
= 164.67

Question 11.
In a group of 25 students, 7 students failed and 6 students got distinction and the marks of the remaining 12 students are 61, 36, 44, 59, 52, 56, 41, 37, 39, 38, 41, 64. Find the median marks of the whole group.
Solution:
n = 25
Median = \(\frac{\mathrm{n}+1}{2}=\frac{25+1}{2}\) = 13th observation
We have been stated that 7 students failed (assuming passing marks on 35) and 6 students got distinction (assuming distinction as 70+), and the marks of the remaining 12 students (who will be situated between the two groups mentioned above, if arranged in ascending order), we have,
F, F, F, F, F, F, F, 36, 37, 38, 39, 41, 41, 44, 52, 56, 59, 61, 64, D, D, D, D, D, D
∴ median = 13th observation = 41.

Question 12.
The median weight of a group of 79 students is found to be 55 kg. 6 more students are added to this group whose weights are 50, 51, 52, 59.5, 60, 61 kg. What will be the value of the median of the combined group if the lowest and the highest weights were 53 kg and 59 kg respectively?
Solution:
n = 79
Median = 55kg
Lowest observation = 53 kg
Flighest observation = 59 kg
6 new students are added to the group having weights in Kg as follows:
50, 51, 52, 59.5, 60, 61
From the above, we see that of the 6 new students, 3 have weights which are below the lowest weight of the earlier group and 3 have weights which are above the highest weight of the earlier group.
∴ the median remains the same
∴ median = 55 kg.

Question 13.
The median of the following incomplete table is 92. Find the missing frequencies:

Solution:
Let a and b be the missing frequencies of class 50 – 70 and class 110 – 130 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 54 + a + b
Since, N = 80
∴ 54 + a + b = 80
∴ a + b = 26 …..(i)
Given, Median = Q₂ = 92
∴ Q₂ lies in the class 90 – 110.
∴ L = 90, h = 20, f = 20, c.f. = 24 + a
\(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 80}{4}\) = 40
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
∴ 92 = 90 + \(\frac{20}{20}\) [40 – (24 + a)
∴ 92 – 90 = 40 – 24 – a
∴ 2 = 16 – a
∴ a = 14
Substituting the value of a in equation (i), we get
14 + b = 26
∴ b = 26 – 14 = 12
∴ 14 and 12 are the missing frequencies of the class 50 – 70 and class 110 – 130 respectively.

Question 14.
A company produces tables which are packed in batches of 100. An analysis of the defective tubes in different batches has received the following information:

estimate the number of defective tubes in the central batch.
Solution:
To find the number of defective tubes in the central batch, we have to find Q₂.
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 4.5, 4.5 – 9.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 251
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 251}{4}\) = 125.5
Cumulative frequency which is just greater than (or equal to) 125.5 is 180.
∴ Q₂ lies in the class 9.5 – 14.5.
∴ L = 9.5, h = 5, f = 84, c.f. = 96
∴ Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 9.5 + \(\frac{5}{84}\) (125.5 – 96)
= 9.5 + \(\frac{5}{84}\) × 29.5
= 9.5 + \(\frac{147.5}{84}\)
= 9.5 + 1.76
= 11.26

Question 15.
In a college, there are 500 students in junior college, 5% score less than 25 marks, 68 scores from 26 to 30 marks, 30% score from 31 to 35 marks, 70 scores from 36 to 40 marks, 20% score from 41 to 45 marks and the rest score 46 and above marks. What are the median marks?
Solution:
Given data can be written in tabulated form as follows:

Since the given data is not continuous, we have to convert it into the continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be Less than 25.5, 25.5 – 30.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 500
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 500}{4}\) = 250
Cumulative frequency which is just greater than (or equal to) 250 is 313.
∴ Q₂ lies in the class 35.5 – 40.5.
∴ L = 35.5, h = 5, f = 70, c.f. = 243
∴ Median = Q₂ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)
= 35.5 + \(\frac{5}{70}\) (250 – 243)
= 35.5 + \(\frac{1}{14}\) (7)
= 35.5 + 0.5
= 36

Question 16.
Draw a cumulative frequency curve more than typical for the following data and hence locate Q₁ and Q₃. Also, find the number of workers with daily wages
(i) Between ₹ 170 and ₹ 260
(ii) less than ₹ 260

Solution:
For more than ogive points to be plotted are (100, 200), (150, 188), (200, 160), (250, 124), (300, 74), (350, 49), (400, 31), (450, 15), (500, 5)

Here, N = 200
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{200}{4}\) = 4
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 200}{4}\) = 150
We take the points having Y co-ordinates 50 and 150 on Y-axis.
From these points, we draw lines which are parallel to X-axis.
From the points of intersection of these lines with the curve, we draw perpendicular on X-axis.
X-Co-ordinates of these points gives the values of Q₁ and Q₃.
Since X-axis has daily wages more than and not less than the given amounts.
∴ Q₁ = Q₃ and Q₃ = Q₁
∴ Q₂ ~ 215 , Q₃ ~ 348

(i) To find the number of workers with daily wages between ₹ 170 and ₹ 260,
Take the values 170 and 260 on X-axis. From these points, we draw lines parallel to Y-axis.
From the point where they intersect the more than ogive, we draw perpendiculars on Y-axis.
The points where they intersect the Y-axis gives the values 178 and 114.
∴ Number of workers having daily wages between ₹ 170 and ₹ 260 = 178 – 114 = 64

(ii) To find the number of workers having daily wages less than ₹ 260, we consider the value 260 on the X-axis.
From this point, we draw a line that is parallel to Y-axis.
From the point where the line intersects the more than ogive, we draw a perpendicular on the Y-axis.
The foot of perpendicular gives the number of workers having daily wages of more than 260.
The foot of perpendicular ~ 114
∴ No. of workers whose daily wages are more than ₹ 260 ~ 114
∴ No. of workers whose daily wages are less than ₹ 260 = 200 – 114 = 86

Question 17.
Draw ogive of both the types for the following frequency distribution and hence find the median.

Solution:

For less than given points to be plotted are (10, 5), (20, 10), (30, 18), (40, 30), (50, 46), (60, 61), (70, 71), (80, 79), (90, 84), (100, 86)
For more than given points to be plotted are (0, 86), (10, 81), (20, 76), (30, 68), (40, 56), (50, 40), (60, 25), (70, 15), (80, 7), (90, 2)

From the point of intersection of two ogives. We draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.

Question 18.
Find Q1, D6 and P78 for the following data:

Solution:
Since the given data is not in the form of a continuous frequency distribution, we have to convert it into that form by subtracting 0.025 from the lower limit and adding 0.025 to the upper limit of each class interval.
∴ the class intervals will be 7.975 – 8.975, 8.975 – 9.975, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 50
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{50}{4}\) = 12.5
Cumulative frequency which is just greater than (or equal) to 12.5 is 15.
∴ Q₁ lies in the class 8.975 – 9.975.
∴ L = 8.975, h = 1, f = 10, c.f. = 5
Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 8.975 + \(\frac{1}{10}\) (12.5 – 5)
= 8.975 + 0.1(7.5)
= 8.975 + 0.75
= 9.725
D₆ class = class containing \(\left(\frac{6 \mathrm{~N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 50}{10}\) = 30
Cumulative frequency which is just greater than (or equal) to 30 is 35.
∴ D6 lies in the class 9.975 – 10.975.
∴ L = 9.975, h = 1, f = 20, c.f. = 15
D₆ = L + \(\frac{h}{f}\left(\frac{6 N}{10}-\text { c.f. }\right)\)
= 9.975 + \(\frac{1}{20}\) (30 – 15)
= 9.975 + 0.05(15)
= 9.975 + 0.75
= 10.725
P₇₈ class = class containing \(\left(\frac{78 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
\(\frac{78 \mathrm{~N}}{100}=\frac{78 \times 50}{100}\) = 39
Cumulative frequency which is just greater than (or equal) to 39 is 45.
∴ P₇₈ lies in the class 10.975 – 11.975.
∴ L = 10.975, h = 1, f = 10, c.f. = 35
∴ P₇₈ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{78 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10.975 + \(\frac{1}{10}\) (39 – 35)
= 10.975 + 0.1(4)
= 10.975 + 0.4
= 11.375

Question 19.

For the above data, find all quartiles and number of persons weighing between 57 kg and 72 kg.
Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 111
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation
∴ \(\frac{N}{4}=\frac{111}{4}\) = 27.75
Cumulative frequency which is just greater than (or equal) to 27.75 is 39.
∴ Q₁ lies in the class 50 – 55.
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\text { c.f. }\right)\)
= 50 + \(\frac{5}{20}\) (27.75 – 19)
= 50 + \(\frac{1}{4}\) × 8.75
= 50 + 2.1875
= 52.1875
Q₂ class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{2 N}{4}=\frac{2 \times 111}{4}\) = 55.5
Cumulative frequency which is just greater than (or equal) to 55.5 is 69.
∴ Q₂ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 30, c.f. = 39
∴ Q₂ = L + \(\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)
= 55 + \(\frac{5}{30}\) (55.5 – 39)
= 55 + \(\frac{1}{6}\) × 16.5
= 55 + 2.75
= 57.75
Q₃ class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 111}{4}\) = 83.25
Cumulative frequency which is just greater than (or equal) to 83.25 is 89.
∴ Q₃ lies in the class 60 – 65.
∴ L = 60, h = 5, f = 20, c.f. = 69
∴ Q₃ = L + \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)
= 60 + \(\frac{5}{20}\) (83.25 – 69)
= 60 + \(\frac{1}{4}\) × 14.25
= 60 + 3.5625
= 63.5625
In order to find the number of persons between 57 kg and 72 kg,
We need to find x in P_x, where P_x = 57 kg and y in P_y, where P_y = 72 kg
Then (y – x) would be the % of persons weighing between 57 kg and 72 kg
P_x = 57
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{x \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 57
∴ 55 + \(\frac{5}{30}\) (1.11x – 39) = 57
∴ \(\frac{1}{6}\) (1.11x – 39) = 2
∴ 1.11x – 39 = 12
∴ 1.11x = 51
∴ x = 45.95
∴ P_y = 72
∴ L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{y \times \mathrm{N}}{100}-\mathrm{c} . \mathrm{f} .\right)\) = 72
∴ 70 + \(\frac{5}{8}\) (1.11y – 99) = 72
∴ 0.625(1.11y – 99) = 2
∴ 1.11y – 99 = 3.2
∴ 1.11y = 102.2
∴ y = 92.07
∴ % of people weighing between 57 kg and 72 kg = 92.07 – 45.95 = 46.12 %
∴ No. of people weighing between 57 kg and 72 kg = 111 × 46.12% = 51.1932 ~ 51

Question 20.
For the following data showing weights of 100 employees, find the maximum weight of the lightest 25% of employees.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 100
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
Cumulative frequency which is just greater than (or equal) to 25 is 29.
∴ Q₁ lies in the class 55 – 60.
∴ L = 55, h = 5, f = 15, c.f. = 14
∴ Q₁ = L + \(\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{\mathrm{N}}{4}-\mathrm{c} . \mathrm{f} .\right)\)
= 55 + \(\frac{5}{15}\) (25 – 14)
= 55 + \(\frac{1}{3}\) × 11
= 55 + 3.67
= 58.67
∴ Maximum weight of the lightest 25% of employees is 58.67 kg.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.3

Question 1.
The following table gives the frequency distribution of marks of 100 students in an examination.

Determine D₆, Q₁, and P₈₅ graphically.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (20, 9), (25, 21), (30, 44), (35, 75), (40, 85), (45, 93), (50, 100).

Here, N = 100
For D₆, \(\frac{6 \mathrm{~N}}{10}=\frac{6 \times 100}{10}\) = 60
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₈₅, \(\frac{85 \mathrm{~N}}{100}=\frac{85 \times 100}{100}\) = 85
∴ We take the points having Y co-ordinates 60, 25 and 85 on Y-axis.
From these points, we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X co-ordinates of these points give the values of D₆, Q₁ and P₈₅.
∴ D₆ = 32.5, Q₁ = 26, P₈₅ = 40

Question 2.
The following table gives the distribution of daily wages of 500 families in a certain city.

Draw a ‘less than’ ogive for the above data. Determine the median income and obtain the limits of income of central 50% of the families.
Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

The points to be plotted for less than ogive are (100, 50), (200, 200), (300, 380), (400, 430), (500, 470), (600, 490) and (700, 500).

Here, N = 500
For Q₁, \(\frac{\mathrm{N}}{4}=\frac{500}{4}\) = 125
For Q₂, \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
For Q₃, \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 500}{4}\) = 375
∴ We take the points having Y co-ordinates 125, 250 and 375 on Y-axis.
From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of Q₁, Q₂ and Q₃.
∴ Q₁ ~ 150, Q₂ ~ 228, Q₃ ~ 297
∴ Median = 228
50% families lie between Q₁ and Q₃
∴ Limits of income of central 50% families are from ₹ 150 to ₹ 297

Question 3.
From the following distribution, determine the median graphically.

Solution:
To draw an ogive curve, we construct the less than and more than cumulative frequency table as given below:

The points to be plotted for less than ogive are (400, 50), (500, 121), (600, 310), (700, 415), (800, 475), (900, 513) and (1000, 520) and that for more than ogive are (300, 520), (400, 470), (500, 399), (600, 210), (700, 105), (800, 45), (900, 7).

From the point of intersection of two ogives, we draw a perpendicular on X-axis.
The point where it meets the X-axis gives the value of the median.
∴ Median ~ 574

Question 4.
The following frequency distribution shows the profit (in ₹) of shops in a particular area of the city.

Find graphically
(i) the Unfits of middle 40% shops.
(ii) the number of shops having a profit of fewer than 35,000 rupees.
Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (10, 12), (20, 30), (30, 57), (40, 77), (50, 94), (60, 100).

The Middle 40% value lies in between P₃₀ and P₇₀.
N = 100
For P₃₀ = \(\frac{30 \mathrm{~N}}{100}=\frac{30 \times 100}{100}\) = 30
For P₇₀ = \(\frac{70 \mathrm{~N}}{100}=\frac{70 \times 100}{100}\) = 70
∴ We take the points having Y co-ordinates 30 and 70 on Y-axis. From these points we draw lines parallel to X-axis.
From the points where these lines intersect the curve, we draw perpendiculars on X-axis.
X-Co-ordinates of these points give the values of P₃₀ and P₇₀.
∴ P₃₀ ~ 20, P₇₀ ~ 36
Limits of middle 40% shops lie between ₹ 20,000 to ₹ 36,000
To find the number of shops having a profit of less than ₹ 35,000, we take the value 35 on the X-axis.
From this point, we draw a line parallel to Y-axis, and from the point where it intersects the less than ogive we draw a perpendicular on Y-axis. It intersects the Y-axis at approximately 67.
∴ No. of shops having profit less than ₹ 35,000 is 67.

Question 5.
The following is the frequency distribution of overtime (per week) performed by various workers from a certain company. Determine the values of D₂, Q₂, and P₆₁ graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (8, 4), (12, 12), (16, 28), (20, 46), (24, 66) and (28, 80)
Here, N = 80

For D₂, we have to consider \(\frac{2 \mathrm{~N}}{10}=\frac{2 \times 80}{10}\) = 16
For Q₂, we have to consider \(\frac{\mathrm{N}}{2}=\frac{80}{2}\) = 40
and for P₆₁, we have to consider \(\frac{61 \mathrm{~N}}{100}=\frac{61 \times 80}{100}\) = 48.8
∴ We consider the values 16, 40 and 48.8 on the Y-axis.
From these points, we draw the lines which are parallel to the X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars to X-axis.
The values at the foot of perpendiculars represent the values of D₂, Q₂, and P₆₁ respectively.
∴ D₂ ~ 13, Q₂ ~ 19, P₆₁ ~ 20.5

Question 6.
Draw ogive for the following data and hence find the values of D₁, Q₁, and P₄₀.

Solution:
N = 100
To draw the less than ogive we have to plot the points (10, 4), (20, 6), (30, 24), (40, 46), (50, 67), (60, 86), (70, 96), (80, 99), (90, 100).

For D₁, we have to consider \(\frac{\mathrm{N}}{10}=\frac{100}{10}\) = 10
For Q₁, we have to consider \(\frac{\mathrm{N}}{4}=\frac{100}{4}\) = 25
For P₄₀, we have to consider \(\frac{40 \mathrm{~N}}{100}=\frac{40 \times 100}{100}\) = 40
∴ We consider the values 10, 25 and 40 on the Y-axis. From these points we draw lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The values at the foot of perpendicular represent the values of D₁, Q₁ and P₄₀ respectively.
∴ D₁ ~ 22, Q₁ ~ 30.5, P₄₀ ~ 37

Question 7.
The following table shows the age distribution of heads of the families in a certain country. Determine the third, fifth, and eighth decile of the distribution graphically.

Solution:
To draw an ogive curve, we construct a less than cumulative frequency table as given below:

Points to be plotted are (35, 46), (45, 131), (55, 195), (65, 270), (75, 360), (85, 400).

N = 400
For D₃, we have to consider \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 400}{10}\) = 120
For D₅, we have to consider \(\frac{5 \mathrm{~N}}{10}=\frac{5 \times 400}{10}\) = 200
For D₈, we have to consider \(\frac{8 \mathrm{~N}}{10}=\frac{8 \times 400}{10}\) = 320
∴ We consider the values 120, 200 and 320 on Y-axis. From these points we draw the lines parallel to X-axis.
From the points where they intersect the less than ogive, we draw perpendiculars on the X-axis.
The foot of perpendicular represent the values of D₃, D₅ and D₈.
∴ D₃ ~ 44, D₅ ~ 55.5 and D₈ ~ 70

Question 8.
The following table gives the distribution of females in an Indian village. Determine the median age graphically.

Solution:
To draw an ogive curve, we construct the less than cumulative frequency table as given below:


Points to be plotted are (10, 175), (20, 275), (30, 343), (40, 391), (50, 416), (60, 466), (70, 489), (80, 497), (90, 499), (100, 500).

N = 500
For median we have to consider \(\frac{\mathrm{N}}{2}=\frac{500}{2}\) = 250
∴ We consider the value 250 on Y-axis. From this point, we draw a line parallel to X-axis.
From the point it intersects the less than ogive, we draw a perpendicular to X-axis.
The foot perpendicular represents the value of the median.
∴ Median ~ 17.5

Question 9.
Draw ogive for the following distribution and hence find graphically the limits of the weight of middle 50% fishes.

Solution:
Since the given data is not continuous, we have to convert it into the continuous form by subtracting 5 from the lower limit and adding 5 to the upper limit of every class interval.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (895, 8), (995, 24), (1095, 44), (1195, 69), (1295, 109), (1395, 115), (1495, 120).

N = 120
For Q₁ and Q₃ we have to consider
\(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
\(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 120}{4}\) = 90
For finding Q₁ and Q₃ we consider the values 30 and 90 on the Y-axis.
From these points, we draw the lines which are parallel to X-axis.
From the points where these lines intersect the less than ogive, we draw perpendicular on X-axis.
The feet of perpendiculars represent the values Q₁ and Q₂.
∴ Q₁ ~ 1025 and Q₃ ~ 1248
∴ the limits of the weight of the middle 50% of fishes lie between 1025 to 1248.

Question 10.
Find graphically the values of D₃ and P₆₅ for the data given below:

Solution:
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
To draw an ogive curve, we construct the less than cumulative frequency table as given below:

Points to be plotted are (69.5, 20), (79.5, 60), (89.5, 110), (99.5, 160), (109.5, 180), (119.5, 190), (129.5, 200).

N = 200
For D₃, \(\frac{3 \mathrm{N}}{10}=\frac{3 \times 200}{10}\) = 60
For P₆₅, \(\frac{65 \mathrm{N}}{100}=\frac{65 \times 200}{100}\) = 130
∴ We take the values 60 and 130 on the Y-axis.
From these points we draw lines parallel to X-axis and from the points where these lines intersect less than ogive, we draw perpendiculars on X-axis.
The foot of perpendiculars represents the median of the values, D₃ and P₆₅.
∴ D₃ = 79.5, P₆₅ = 93.5

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.2

Question 1.
Calculate D₆ and P₈₅ for the following data:
79, 82, 36, 38, 51, 72, 68, 70, 64, 63
Solution:
The given data can be arranged in ascending order as follows:
36, 38, 51, 63, 64, 68, 70, 72, 79, 82
Here, n = 10
D₆ = value of 6\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 6\(\left(\frac{10+1}{10}\right)^{\text {th }}\) observation
= value of (6 × 1.1)th observation
= value of (6.6)th observation
= value of 6th observation + 0.6(value of 7th observation – value of 6th observation)
= 68 + 0.6(70 – 68)
= 68 + 0.6(2)
= 68 + 1.2
∴ D₆ = 69.2
P₈₅ = value of \(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of \(\left(\frac{10+1}{100}\right)^{\text {th }}\) observation
= value of (85 × 0. 11)th observation
= value of (9.35)th observation
= value of 9th observation + 0.35(value of 10th observation – value of 9th observation)
= 19 + 0.35(82 – 79)
= 79 + 0.35(3)
= 79 + 1.05
∴ P₈₅ = 80.05

Question 2.
The daily wages (in ₹) of 15 labourers are as follows:
230, 400, 350, 200, 250, 380, 210, 225, 375, 180, 375, 450, 300, 350, 250
Calculate D₈ and P₉₀.
Solution:
The given data can be arranged in ascending order as follows:
180, 200, 210, 225, 230, 250, 250, 300, 350, 350, 375, 375, 380, 400, 450
Here, n = 15
D₈ = value of 8\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 8\(\left(\frac{15+1}{10}\right)^{\text {th }}\) observation
= value of (8 × 1.6)th observation
= value of (12.8)th observation
= value of 12th observation – 0.8(value of 13th observation – value of 12th observation)
= 375 + 0.8(380 – 375)
= 375 + 0.8(5)
= 375 + 4
∴ D₈ = 379
P₉₀ = value of 90\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 90\(\left(\frac{15+1}{100}\right)^{\text {th }}\) observation
= value of (90 × 0.16)th observation
= value of (14.4)th observation
= value of 14th observation + 0.4 (value of 15th observation – value of 14th observation)
= 400 + 0.4(450 – 400)
= 400 + 0.4(50)
= 400 + 20
∴ P₉₀ = 420

Question 3.
Calculate 2nd decile and 65th percentile for the following:

Solution:
We construct the less than cumulative frequency table as given below:

Here, n = 200
D₂ = value of 2\(\left(\frac{n+1}{10}\right)^{\text {th }}\) observation
= value of 2\(\left(\frac{200+1}{10}\right)^{\text {th }}\) observation
= value of (2 × 20.1)th observation
= value of (40.2)th observation
Cumulative frequency which is just greater than (or equal to) 40.2 is 58.
∴ D₂ = 120
P₆₅ = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 65\(\left(\frac{200+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 2.01)th observation
= value of (130.65)th observation
The cumulative frequency which is just greater than (or equal to) 130.65 is 150.
∴ P₆₅ = 280

Question 4.
From the following data calculate the rent of the 15th, 65th, and 92nd house.

Solution:
Arranging the given data in ascending order.

Here, n = 100
P₁₅ = value of 15
= value of 15\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 15\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (15 × 1.01 )th observation
= value of (15.15)th observation
Cumulative frequency which is just greater than (or equal to) 15.15 is 25.
∴ P₁₅ = 11000
P₆₅ = value of 65\(\left(\frac{n+1}{100}\right)^{\text {th }}\)observation
= value of 65\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (65 × 1.01)th observation
= value of (65.65)th observation
Cumulative frequency which is just greater than (or equal to) 65.65 is 70.
∴ P₆₅ = 14000
P₉₂ = value of 92\(\left(\frac{n+1}{100}\right)^{\text {th }}\) observation
= value of 92\(\left(\frac{100+1}{100}\right)^{\text {th }}\) observation
= value of (92 × 1.01)th observation
= value of (92.92)th observation
Cumulative frequency which is just greater than (or equal to) 92.92 is 98.
∴ P₉₂ = 17000

Question 5.
The following frequency distribution shows the weight of students in a class.

(a) Find the percentage of students whose weight is more than 50 kg.
(b) If the weight column provided is of mid values then find the percentage of students whose weight is more than 50 kg.
Solution:
(a) Let the percentage of students weighing less than 50 kg be x.
∴ Px = 50

From the table, out of 20 students, 84 students have their weight less than 50 kg.
∴ Number of students weighing more than 50 kg = 120 – 84 = 36
∴ Percentage of students having there weight more than 50 kg = \(\frac{36}{120}\) × 100 = 30%

(b) The difference between any two consecutive mid values of weight is 5 kg.
The class intervals must of width 5, with 40, 45,….. as their mid values.
∴ The class intervals will be 37.5 – 42.5, 42.5 – 47.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 120
Let P_x = 50
The value 50 lies in the class 47.5 – 52.5
∴ L = 47.5, h = 5, f = 29, c.f. = 55

∴ x = 58 (approximately)
∴ 58% of students are having weight below 50 kg.
∴ Percentage of students having weight above 50 kg is 100 – 58 = 42
∴ 42% of students are having weight above 50 kg.

Question 6.
Calculate D₄ and P₄₈ from the following data:

Solution:
The difference between any two consecutive mid values is 5, the width of class interval = 5
∴ Class interval with mid-value 2.5 is 0 – 5
Class interval with mid value 7.5 is 5 – 10, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 100
D₄ class = class containing \(\left(\frac{4 \mathrm{N}}{10}\right)^{\text {th }}\) observation
∴ \(\frac{4 \mathrm{N}}{10}=\frac{4 \times 100}{10}\) = 40
Cumulative frequency which is just greater than (or equal to) 40 is 50.
∴ D₄ lies in the class 10 – 15.
∴ L = 10,h = 5, f = 25, c.f. = 25
∴ D₄ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{4 \mathrm{~N}}{10}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (40 – 25)
= 10 + \(\frac{1}{5}\) (15)
= 10 + 3
∴ D₄ = 13
P₄₈ class = class containing \(\left(\frac{48 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{48 \mathrm{~N}}{100}=\frac{48 \times 100}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P₄₈ lies in the class 10 – 15.
∴ L = 10, h = 5, f = 25, c.f. = 25
∴ P₄₈ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{48 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 10 + \(\frac{5}{25}\) (48 – 25)
= 10 + \(\frac{1}{5}\) (23)
= 10 + 4.6
∴ P₄₈ = 14.6

Question 7.
Calculate D₉ and P₂₀ of the following distribution.

Solution:
We construct the less than cumulative frequency table as given below:

Here, N = 240
D₉ class = class containing \(\left(\frac{9 \mathrm{~N}}{10}\right)^{\mathrm{th}}\) observation
∴ \(\frac{9 \mathrm{~N}}{10}=\frac{9 \times 240}{10}\) = 216
Cumulative frequency which is just greater than (or equal to) 216 is 225.
∴ D₉ lies in the class 80 – 100.
∴ L = 80, h = 20, f = 90, c.f. = 135
∴ D₉ = \(L+\frac{h}{f}\left(\frac{9 N}{10}-c . f .\right)\)
= 80 + \(\frac{20}{90}\)(216 – 135)
= 80 + \(\frac{2}{9}\)(81)
= 80 + 18
∴ D₉ = 98
P₂₀ class = class containing \(\left(\frac{20 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{~N}}{100}=\frac{20 \times 240}{100}\) = 48
Cumulative frequency which is just greater than (or equal to) 48 is 50.
∴ P₂₀ lies in the class 40 – 60.
∴ L = 40, h = 20, f = 35, c.f. = 15

∴ P₂₀ = 58.86

Question 8.
Weekly wages for a group of 100 persons are given below:

D₃ for this group is ₹ 1100. Calculate the missing frequencies.
Solution:
Let a and b be the missing frequencies of class 500 – 1000 and class 2000 – 2500 respectively.
We construct the less than cumulative frequency table as given below:

Here, N = 62 + a + b
Since, N = 100
∴ 62 + a + b = 100
∴ a + b = 38 …..(i)
Given, D₃ = 1100
∴ D₃ lies in the class 1000 – 1500.
∴ L = 1000, h = 500, f = 25, c.f. = 7 + a
∴ \(\frac{3 \mathrm{~N}}{10}=\frac{3 \times 100}{10}=30\)
∴ D₃ = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{3 \mathrm{~N}}{10}-\mathrm{c} . \mathrm{f} .\right)\)
∴ 1100 = 1000 + \(\frac{500}{25}\) [30 – (7 + a)]
∴ 1100 – 1000 = 20(30 – 7 – a)
∴ 100 = 20(23 – a)
∴ 100 = 460 – 20a
∴ 20a = 460 – 100
∴ 20a = 360
∴ a = 18
Substituting the value of a in equation (i), we get
18 + b = 38
∴ b = 38 – 18 = 20
∴ 18 and 20 are the missing frequencies of the class 500 – 1000 and class 2000 – 2500 respectively.

Question 9.
The weekly profit (in rupees) of 100 shops are distributed as follows:

Find the limits of the profit of middle 60% of the shops.
Solution:
To find the limits of the profit of the middle 60% of the shops, we have to find P₂₀ and P₈₀.
We construct the less than cumulative frequency table as given below:

Here, N = 100
P₂₀ class = class containing \(\left(\frac{20 \mathrm{N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{20 \mathrm{N}}{100}=\frac{20 \times 100}{100}=20\)
Cumulative frequency which is just greater than (or equal to) 20 is 26.
∴ P₂₀ lies in the class 1000 – 2000.
∴ L = 1000, h = 1000, f = 16, c.f. = 10
∴ P₂₀ = \(L+\frac{h}{f}\left(\frac{20 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 1000 + \(\frac{1000}{16}\) (20 – 10)
= 1000 + \(\frac{125}{2}\) (10)
= 1000 + 625
∴ P₂₀ = 1625
P₈₀ class = class containing \(\left(\frac{80 \mathrm{~N}}{100}\right)^{\text {th }}\) observation
∴ \(\frac{80 \mathrm{~N}}{100}=\frac{80 \times 100}{100}=80\)
Cumulative frequency which is just greater than (or equal to) 80 is 92.
∴ P₈₀ lies in the class 4000 – 5000.
∴ L = 4000, h = 1000, f = 20, c.f. = 72
∴ P₈₀ = \(L+\frac{h}{f}\left(\frac{80 \mathrm{~N}}{100}-\text { c.f. }\right)\)
= 4000 + \(\frac{1000}{20}\)(80 – 72)
= 4000 + 50(8)
= 4000 + 400
∴ P₈₀ = 4400
∴ the profit of middle 60% of the shops lie between the limits ₹ 1,625 to ₹ 4,400.

Question 10.
In a particular factory, workers produce various types of output units. The following distribution was obtained:

Find the percentage of workers who have produced less than 82 output units.
Solution:
Since the given data is not continuous, we have to convert it into a continuous form by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of every class interval.
∴ the class intervals will be 69.5 – 74.5, 74.5 – 79.5, etc.
We construct the less than cumulative frequency table as given below:

Here, N = 445
Let P_x = 82
The value 82 lies in the class 79.5 – 84.5
∴ L = 79.5, h = 5, f = 50, c.f. = 85

∴ 24.72% of workers produced less than 82 output units.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 8 Continuity Ex 8.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 8 Continuity Ex 8.1

Question 1.
Examine the continuity of
(i) f(x) = x³ + 2x² – x – 2 at x = -2
Solution:
f(x) = x³ + 2x² – x – 2
Here f(x) is a polynomial function and hence it is continuous for all x ∈ R.
∴ f(x) is continuous at x = -2

(ii) f(x) = \(\frac{x^{2}-9}{x-3}\) on R
Solution:
f(x) = \(\frac{x^{2}-9}{x-3}\); x ∈ R
f(x) is a rational function and is continuous for all x ∈ R, except at the points where denominator becomes zero.
Here, denominator x – 3 = 0 when x = 3.
∴ Function f is continuous for all x ∈ R, except at x = 3, where it is not defined.

Question 2.
Examine whether the function is continuous at the points indicated against them.
(i) f(x) = x³ – 2x + 1, for x ≤ 2
= 3x – 2, for x > 2, at x = 2
Solution:

∴ Function f is discontinuous at x = 2

(ii) f(x) = \(\frac{x^{2}+18 x-19}{x-1}\) for x ≠ 1
= 20, for x = 1, at x = 1
Solution:

∴ f(x) is continuous at x = 1

Question 3.
Test the continuity of the following functions at the points indicated against them.
(i) f(x) = \(\frac{\sqrt{x-1}-(x-1)^{\frac{1}{3}}}{x-2}\) for x ≠ 2
= \(\frac{1}{5}\) for x = 2, at x = 2
Solution:

(ii) f(x) = \(\frac{x^{3}-8}{\sqrt{x+2}-\sqrt{3 x-2}}\) for x ≠ 2
= -24 for x = 2, at x = 2
Solution:

(iii) f(x) = 4x + 1 for x ≤ \(\frac{8}{3}\)
= \(\frac{59-9 x}{3}\), for x > \(\frac{8}{3}\), at x = \(\frac{8}{3}\)
Solution:

(iv) f(x) = \(\frac{x^{3}-27}{x^{2}-9}\) for 0 ≤ x < 3
= \(\frac{9}{2}\), for 3 ≤ x ≤ 6, at x = 3
Solution:

Question 4.
(i) If f(x) = \(\frac{24^{x}-8^{x}-3^{x}+1}{12^{x}-4^{x}-3^{x}+1}\), for x ≠ 0
= k, for x = 0
is continuous at x = 0, find k.
Solution:
Function f is continuous at x = 0

(ii) If f(x) = \(\frac{5^{x}+5^{-x}-2}{x^{2}}\), for x ≠ 0
= k for x = 0
is continuous at x = 0, find k.
Solution:
Function f is continuous at x = 0

(iii) For what values of a and b is the function
f(x) = ax + 2b + 18 for x ≤ 0
= x² + 3a – b for 0 < x ≤ 2 = 8x – 2 for x > 2,
continuous for every x?
Solution:
Function f is continuous for every x.
∴ Function f is continuous at x = 0 and x = 2
As f is continuous at x = 0.
∴ \(\lim _{x \rightarrow 0^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 0^{+}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 0^{-}}(a x+2 b+18)=\lim _{x \rightarrow 0^{+}}\left(x^{2}+3 a-b\right)\)
∴ a(0) + 2b + 18 = (0)2 + 3a – b
∴ 3a – 3b = 18
∴ a – b = 6 …..(i)
Also, Function f is continous at x = 2
∴ \(\lim _{x \rightarrow 2^{-}} \mathrm{f}(x)=\lim _{x \rightarrow 2^{-}} \mathrm{f}(x)\)
∴ \(\lim _{x \rightarrow 2^{-}}\left(x^{2}+3 a-b\right)=\lim _{x \rightarrow 2^{-}}(8 x-2)\)
∴ (2)² + 3a – b = 8(2) – 2
∴ 4 + 3a – b = 14
∴ 3a – b = 10 …..(ii)
Subtracting (i) from (ii), we get
2a = 4
∴ a = 2
Substituting a = 2 in (i), we get
2 – b = 6
∴ b = -4
∴ a = 2 and b = -4

(iv) For what values of a and b is the function
f(x) = \(\frac{x^{2}-4}{x-2}\) for x < 2
= ax² – bx + 3 for 2 ≤ x < 3
= 2x – a + b for x ≥ 3
continuous in its domain.
Solution:
Function f is continuous for every x on R.
∴ Function f is continuous at x = 2 and x = 3.
As f is continuous at x = 2.
∴ \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)\)

∴ 2 + 2 = a(2)² – b(2) + 3
∴ 4 = 4a – 2b + 3
∴ 4a – 2b = 1 …..(i)
Also function f is continuous at x = 3
∴ \(\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)\)
∴ \(\lim _{x \rightarrow 3^{-}}\left(a x^{2}-b x+3\right)=\lim _{x \rightarrow 3^{+}}(2 x-a+b)\)
∴ a(3)² – b(3) + 3 = 2(3) – a + b
∴ 9a – 3b + 3 = 6 – a + b
∴ 10a – 4b = 3 …..(ii)
Multiplying (i) by 2, we get
8a – 4b = 2 …..(iii)
Subtracting (iii) from (ii), we get
2a = 1
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in (i), we get
4(\(\frac{1}{2}\)) – 2b = 1
∴ 2 – 2b = 1
∴ 1 = 2b
∴ b = \(\frac{1}{2}\)
∴ a = \(\frac{1}{2}\) and b = \(\frac{1}{2}\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Miscellaneous Exercise 5

Question 1.
Find the slopes of the lines passing through the following points:
(i) (1, 2), (3, -5)
(ii) (1, 3), (5, 2)
(iii) (-1, 3), (3, -1)
(iv) (2, -5), (3, -1)
Solution:
(i) Let A = (1, 2) = (x₁, y₁) and B = (3, -5) = (x₂, y₂) say.
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-5-2}{3-1}=\frac{-7}{2}\)

(ii) Let C = (1, 3) = (x₁, y₁) and D = (5, 2) = (x₂, y₂) say.
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-3}{5-1}=\frac{-1}{4}\)

(iii) Let E = (-1, 3) = (x₁, y₁) and F = (3, -1) = (x₂, y₂) say.
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{3-(-1)}=\frac{-4}{4}\) = -1

(iv) Let P = (2, -5) = (x₁, y₁) and Q = (3, -1) = (x₂, y₂) say.
Slope of line PQ = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-(-5)}{3-2}\) = \(\frac{-1+5}{1}\) = 4

Question 2.
Find the slope of the line which
(i) makes an angle of 120° with the positive X-axis.
(ii) makes intercepts 3 and -4 on the axes.
(iii) passes through the points A(-2, 1) and the origin.
Solution:
(i) θ = 120°
Slope of the line = tan 120°
= tan (180° – 60°)
= -tan 60° …..[tan(180° – θ) = -tan θ]
= -√3

(ii) Given, x-intercept of line is 3 and y-intercept of line is -4
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, -4).
∴ The line passes through (3, 0) = (x₁, y₁) and (0, -4) = (x₂, y₂) say.
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-4-0}{0-3}=\frac{-4}{-3}=\frac{4}{3}\)

(iii) Required line passes through O(0, 0) = (x₁, y₁) and A(-2, 1) = (x₂, y₂) say.
Slope of line OA = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-0}{-2-0}=\frac{1}{-2}\) = \(\frac{-1}{2}\)

Question 3.
Find the value of k:
(i) if the slope of the line passing through the points (3, 4), (5, k) is 9.
(ii) the points (1, 3), (4, 1), (3, k) are collinear.
(iii) the point P(1, k) lies on the line passing through the points A(2, 2) and B(3, 3).
Solution:
(i) Let P(3, 4), Q(5, k).
Slope of PQ = 9 …….[Given]
∴ \(\frac{\mathrm{k}-4}{5-3}\) = 9
∴ \(\frac{\mathrm{k}-4}{2}\) = 9
∴ k – 4 = 18
∴ k = 22

(ii) The points A(1, 3), B(4, 1) and C(3, k) are collinear.
∴ Slope of AB = Slope of BC
∴ \(\frac{1-3}{4-1}=\frac{k-1}{3-4}\)
∴ \(\frac{-2}{3}=\frac{\mathrm{k}-1}{-1}\)
∴ 2 = 3k – 3
∴ k = \(\frac{5}{3}\)

(iii) Given, point P(1, k) lies on the line joining A(2, 2) and B(3, 3).
∴ Slope of AB = Slope of BP
∴ \(\frac{3-2}{3-2}=\frac{3-k}{3-1}\)
∴ 1 = \(\frac{3-k}{2}\)
∴ 2 = 3 – k
∴ k = 1

Question 4.
Reduce the equation 6x + 3y + 8 = 0 into slope-intercept form. Hence, find its slope.
Solution:
Given equation is 6x + 3y + 8 = 0, which can be written as
3y = -6x – 8
∴ y = \(\frac{-6 x}{3}-\frac{8}{3}\)
∴ y = -2x – \(\frac{8}{3}\)
This is of the form y = mx + c with m = -2
∴ y = -2x – \(\frac{8}{3}\) is in slope-intercept form with slope = -2

Question 5.
Verify that A(2, 7) is not a point on the line x + 2y + 2 = 0.
Solution:
Given equation is x + 2y + 2 = 0.
Substituting x = 2 and y = 7 in L.H.S. of given equation, we get
L.H.S. = x + 2y + 2
= 2 + 2(7) + 2
= 2 + 14 + 2
= 18
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 6.
Find the X-intercept of the line x + 2y – 1 = 0.
Solution:
Given equation of the line is x + 2y – 1 = 0
To find the x-intercept, put y = 0 in given equation of the line
∴ x + 2(0) – 1 = 0
∴ x + 0 – 1 = 0
∴ x = 1
∴ X-intercept of the given line is 1.
Alternate method:
Given equation of the line is x + 2y – 1 = 0
i.e. x + 2y = 1
∴ \(\frac{x}{1}+\frac{y}{\frac{1}{2}}=1\)
Comparing with \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1\), we get a = 1
X-intercept of the line is 1.

Question 7.
Find the slope of the line y – x + 3 = 0.
Solution:
Equation of given line is y – x + 3 = 0
i.e. y = x – 3
Comparing with y = mx + c, we get
m = Slope = 1

Question 8.
Does point A(2, 3) lie on the line 3x + 2y – 6 = 0? Give reason.
Solution:
Given equation is 3x + 2y – 6 = 0.
Substituting x = 2 and y = 3 in L.H.S. of given equation, we get
L.H.S. = 3x + 2y – 6
= 3(2)+ 2(3) – 6
= 6
≠ R.H.S.
∴ Point A does not lie on the given line.

Question 9.
Which of the following lines passes through the origin?
(a) x = 2
(b) y = 3
(c) y = x + 2
(d) 2x – y = 0
Solution:
Any line passing through origin is of the form y = mx or ax + by = 0.
Here in the given option, 2x – y = 0 is in the form ax + by = 0.

Question 10.
Obtain the equation of the line which is:
(i) parallel to the X-axis and 3 units below it.
(ii) parallel to the Y-axis and 2 units to the left of it.
(iii) parallel to the X-axis and making an intercept of 5 on the Y-axis.
(iv) parallel to the Y-axis and making an intercept of 3 on the X-axis.
Solution:
(i) Equation of a line parallel to X-axis is y = k.
Since, the line is at a distance of 3 units below X-axis.
∴ k = -3
∴ the equation of the required line is y = -3
i.e., y + 3 = 0.

(ii) Equation of a line parallel to Y-axis is x = h.
Since, the line is at a distance of 2 units to the left of Y-axis.
∴ h = -2
∴ the equation of the required line is x = -2
i.e., x + 2 = 0.

(iii) Equation of a line parallel to X-axis with y-intercept ‘k’ is y = k.
Here, y-intercept = 5
∴ the equation of the required line is y = 5.

(iv) Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 3
∴ the equation of the required line is x = 3.

Question 11.
Obtain the equation of the line containing the point:
(i) (2, 3) and parallel to the X-axis.
(ii) (2, 4) and perpendicular to the Y-axis.
(iii) (2, 5) and perpendicular to the X-axis.
Solution:
(i) Equation of a line parallel to X-axis is of the form y = k.
Since, the line passes through (2, 3).
∴ k = 3
∴ the equation of the required line is y = 3.

(ii) Equation of a line perpendicular to Y-axis
i.e., parallel to X-axis, is of the form y = k.
Since, the line passes through (2, 4).
∴ k = 4
∴ the equation of the required line is y = 4.

(iii) Equation of a line perpendicular to X-axis
i.e., parallel to Y-axis, is of the form x = h.
Since, the line passes through (2, 5).
∴ h = 2
∴ the equation of the required line is x = 2.

Question 12.
Find the equation of the line:
(i) having slope 5 and containing point A(-1, 2).
(ii) containing the point (2, 1) and having slope 13.
(iii) containing the point T(7, 3) and having inclination 90°.
(iv) containing the origin and having inclination 90°.
(v) through the origin which bisects the portion of the line 3x + 2y = 2 intercepted between the co-ordinate axes.
Solution:
(i) Given, slope (m) = 5 and the line passes through A(-1, 2).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ the equation of the required line is y – 2 = 5(x + 1)
∴ y – 2 = 5x + 5
∴ 5x – y + 7 = 0

(ii) Given, slope (m) = 13 and the line passes through (2, 1).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ the equation of the required line is y – 1 = 13(x – 2)
∴ y – 1 = 13x – 26
∴ 13x – y = 25.

(iii) Given, Inclination of line = θ = 90°
∴ the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form x = h.
Since, the line passes through (7, 3).
∴ h = 7
∴ the equation of the required line is x = 7.

(iv) Given, Inclination of line = θ = 90°
∴ the required line is parallel to Y-axis (or lies on the Y-axis.)
Equation of a line parallel to Y-axis is of the form x = h.
Since, the line passes through origin (0, 0).
∴ h = 0
∴ the equation of the required line is x = 0.

(v) Given equation of the line is 3x + 2y = 2.
∴ \(\frac{3 x}{2}+\frac{2 y}{2}=1\)
∴ \(\frac{x}{\frac{2}{3}}+\frac{y}{1}=1\)

This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1\), with
a = \(\frac{2}{3}\), b = 1
∴ the line 3x + 2y = 2 intersects the X-axis at A(\(\frac{2}{3}\), 0) and Y-axis at B(0, 1).
Required line is passing through the midpoint of AB.
∴ Midpoint of AB = \(\left(\frac{\frac{2}{3}+0}{2}, \frac{0+1}{2}\right)=\left(\frac{1}{3}, \frac{1}{2}\right)\)
∴ Required line passes through (0, 0) and \(\left(\frac{1}{3}, \frac{1}{2}\right)\).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ the equation of the required line is
\(\frac{y-0}{\frac{1}{2}-0}=\frac{x-0}{\frac{1}{3}-0}\)
∴ 2y = 3x
∴ 3x – 2y = 0

Question 13.
Find the equation of the line passing through the points A(-3, 0) and B(0, 4).
Solution:
Since, the required line passes through the points A(-3, 0) and B(0, 4).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁, y₁) = (-3, 0) and (x₂, y₂) = (0, 4)
∴ the equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-(-3)}{0-(-3)}\)
∴ \(\frac{y}{4}=\frac{x+3}{3}\)
∴ 4x + 12 = 3y
∴ 4x – 3y + 12 = 0

Question 14.
Find the equation of the line:
(i) having slope 5 and making intercept 5 on the X-axis.
(ii) having an inclination 60° and making intercept 4 on the Y-axis.
Solution:
(i) Since, the x-intercept of the required line is 5.
∴ it passes through (5, 0).
Also, slope(m) of the line is 5
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ the equation of the required line is y – 0 = 5(x – 5)
∴ y = 5x – 25
∴ 5x – y – 25 = 0

(ii) Given, Inclination of line = θ = 60°
∴ Slope of the line (m) = tan θ
= tan 60°
= √3
and the y-intercept of the required line is 4.
∴ it passes through (0, 4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ the equation of the required line is y – 4 = √3(x – 0)
∴ y – 4 = √3x
∴ √3x – y + 4 = 0

Question 15.
The vertices of a triangle are A(1, 4), B(2, 3), and C(1, 6). Find equations of
(i) the sides
(ii) the medians
(iii) Perpendicular bisectors of sides
(iv) altitudes of ∆ABC
Solution:
Vertices of ∆ABC are A(1, 4), B(2, 3), and C(1, 6)
(i) Equation of the line in two-point form is

Since, both the points A and C have same x co-ordinates i.e. 1
∴ the points A and C lie on a line parallel to Y-axis.
∴ the equation of side AC is x = 1.

(ii) Let D, E, and F be the midpoints of sides BC, AC, and AB respectively of ∆ABC.

(iii) Slope of side BC = \(\left(\frac{6-3}{1-2}\right)=\left(\frac{3}{-1}\right)\) = -3
∴ Slope of perpendicular bisector of BC is \(\frac{1}{3}\) and the line passes through \(\left(\frac{3}{2}, \frac{9}{2}\right)\)
∴ Equation of the perpendicular bisector of side BC is \(\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)\)
∴ \(\frac{2 y-9}{2}=\frac{1}{3}\left(\frac{2 x-3}{2}\right)\)
∴ 3(2y – 9) = (2x – 3)
∴ 2x – 6y + 24 = 0
∴ x – 3y + 12 = 0
Since, both the points A and C have the same x co-ordinates i.e. 1
∴ the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, the perpendicular bisector of side AC is parallel to X-axis.
Since, the perpendicular bisector of side AC passes through E(1, 5).
∴ the equation of the perpendicular bisector of side AC is y = 5.
Slope of side AB = \(\left(\frac{3-4}{2-1}\right)\) = -1
∴ Slope of perpendicular bisector of AB is 1 and the line passes through \(\left(\frac{3}{2}, \frac{7}{2}\right)\).
∴ Equation of the perpendicular bisector of side AB is \(\left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right)\)
∴ \(\frac{2 y-7}{2}=\frac{2 x-3}{2}\)
∴ 2y – 7 = 2x – 3
∴ 2x – 2y + 4 = 0
∴ x – y + 2 = 0

(iv) Let AX, BY and CZ be the altitudes through the vertices A, B, and C respectively of ∆ABC.

Slope of BC = -3
∴ Slope of AX = \(\frac{1}{3}\) …..[∵ AX ⊥ BC]
Since, altitude AX passes through (1, 4) and has slope \(\frac{1}{3}\)
∴ equation of altitude AX is y – 4 = \(\frac{1}{3}\)(x – 1)
∴ 3y – 12 = x – 1
∴ x – 3y + 11 = 0
Since, both the points A and C have the same x co-ordinates i.e. 1
∴ the points A and C lie on the line x = 1.
AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis.
Since, the altitude BY passes through B(2, 3).
∴ the equation of altitude BY is y = 3.
Also, slope of AB = -1
∴ Slope of CZ = 1 …..[∵ CZ ⊥ AB]
Since, altitude CZ passes through (1, 6) and has slope 1
∴ equation of altitude CZ is y – 6 = 1(x – 1)
∴ y – 6 = x – 1
∴ x – y + 5 = 0

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Locus and Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Locus and Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines.
(a) 2x + 3y – 6 = 0
(b) x + 2y = 0
Solution:
(a) Given equation of the line is 2x + 3y – 6 = 0
Comparing this equation with ax + by + c = 0, we get
a = 2, b = 3, c = -6
∴ Slope of the line = \(\frac{-a}{b}=\frac{-2}{3}\)
x-intercept = \(\frac{-\mathrm{c}}{\mathrm{a}}=\frac{-(-6)}{2}\) = 3
y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-(-6)}{3}\) = 2

(b) Given equation of the line is x + 2y = 0
Comparing this equation with ax + by + c = 0, we get
a = 1, b = 2, c = 0
∴ Slope of the line = \(\frac{-a}{b}=\frac{-1}{2}\)
x-intercept = \(\frac{-c}{a}=\frac{-0}{1}\) = 0
y-intercept = \(\frac{-\mathrm{c}}{\mathrm{b}}=\frac{-0}{2}\) = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
(a) y = 2x – 4
(b) y = 4
(c) \(\frac{x}{2}+\frac{y}{4}=1\)
(d) \(\frac{x}{3}=\frac{y}{2}\)
Solution:
(a) y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

(b) y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

(c) \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}=1\)
∴ 2x + y = 4
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

(d) \(\frac{x}{3}=\frac{y}{2}\)
∴ 2x = 3y
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 5 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 5 = 0.
∴ m₂ = \(\frac{-2}{-4}=\frac{1}{2}\)
Since, m₁ = m₂
∴ The given lines are parallel to each other.

Question 4.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y-axes at points A and B respectively.

3x + 4y = p
∴ \(\frac{3 x}{\mathrm{p}}+\frac{4 y}{\mathrm{p}}=1\)
∴ \(\frac{x}{\frac{p}{3}}+\frac{y}{\frac{p}{4}}=1\)
The equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\), with a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A = (a, 0) = (\(\frac{p}{3}\), 0) and B = (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A(∆OAB) = 24 sq. units
∴ \(\frac{1}{2}\) × OA × OB = 24
∴ \(\frac{1}{2}\) × \(\frac{p}{3}\) × \(\frac{p}{4}\) = 24
∴ p² = 576
∴ p = ±24

Question 5.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(-2, 3), B(6, -1), C(4, 3).
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of ∆ABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC respectively.
∴ D = \(\left(\frac{6+4}{2}, \frac{-1+3}{2}\right)\) = (5, 1)
and E = \(\left(\frac{-2+4}{2}, \frac{3+3}{2}\right)\) = (1, 3)
Now, slope of BC = \(\frac{-1-3}{6-4}\) = -2
∴ slope of FD = \(\frac{1}{2}\) …..[∵ FD ⊥ BC]
Since, FD passes through (5, 1) and has slope \(\frac{1}{2}\)
∴ Equation of FD is y – 1 = \(\frac{1}{2}\)(x – 5)
∴ 2(y – 1) = x – 5
∴ x – 2y – 3 = 0 ……(i)
Since, both the points A and C have same y co-ordinates i.e. 3
∴ the points A and C lie on the liney = 3.
Since, FE passes through E(1, 3).
∴ the equation of FE is x = 1. …….(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y – 3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F = (1, -1).

Question 6.
Find the equation of the line whose x-intercept is 3 and which is perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ slope of the required line which is perpendicular to 3x – y + 23 = 0 is \(\frac{-1}{3}\).
Since, the x-intercept of the required line is 3.
∴ it passes through (3, 0).
∴ the equation of the required line is
y – 0 = \(\frac{-1}{3}\) (x – 3)
∴ 3y = -x – 3
∴ x – 3y = 3

Question 7.
Find the distance of the point A(-2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(-2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = -5, c = -13, x₁ = -2, y₁ = 3

Question 8.
Find the distance between parallel lines 9x + 6y – 1 = 0 and 9x + 6y – 32 = 0.
Solution:
Equations of the given parallel lines are 9x + 6y – 7 = 0 and 9x + 6y – 32 = 0.
Here, a = 9, b = 6, C₁ = -7 and C₂ = -32
∴ Distance between the parallel lines

Question 9.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2x – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Given equations of lines are
x + y – 2 = 0 ……(i)
and 2x – 3y – 4 = 0 ……(ii)
Multiplying equation (i) by 3, we get
3x – 3y – 6 = 0 …..(iii)
Adding equation (ii) and (iii), we get
5x – 2 = 0
∴ x = \(\frac{2}{5}\)
Substituting x = \(\frac{2}{5}\) in equation (i), we get
\(\frac{2}{5}\) + y – 2 = 0
∴ y = 2 – \(\frac{2}{5}\) = \(\frac{8}{5}\)
∴ The required line passes through point (\(\frac{2}{5}\), \(\frac{8}{5}\)).
Also, the line makes intercept of 3 on X-axis
∴ it also passes through point (3, 0).
∴ required equation of line passing through points (\(\frac{2}{5}\), \(\frac{8}{5}\)) and (3, 0) is

∴ 13(5y – 8) = -8(5x – 2)
∴ 65y – 104 = -40x + 16
∴ 40x + 65y – 120 = 0
∴ 8x + 13y – 24 = 0 which is the equation of the required line.

Question 10.
D(-1, 8), E(4, -2), F(-5, -3) are midpoints of sides BC, CA and AB of ∆ABC. Find
(i) equations of sides of ∆ABC.
(ii) co-ordinates of the circumcentre of ∆ABC.
Solution:
(i) Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ∆ABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ∆ABC.

For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = -4
∴ x₁ + x₂ + x₃ = -2 …..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = -10
Solving (v) and (vii), we get x₃ = 8
For y-coordinates:
Adding (ii), (iv) and (vi), we get
2y₁ + 2y₂ + 2y₃ = 6
∴ y₁ + y₂ + y₃ = 3 …..(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of ∆ABC are A(0, -13), B(-10, 7), C(8, 9)

(ii) Here, A(0, -13), B(-10, 7), C(8, 9) are the vertices of ∆ABC.
Let F be the circumcentre of ∆ABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC.
∴ D = \(\left(\frac{-10+8}{2} \cdot \frac{7+9}{2}\right)\) = (-1, 8)
and E = \(\left(\frac{0+8}{2}, \frac{-13+9}{2}\right)\) = (4, -2)
Now, slope of BC = \(\frac{7-9}{-10-8}=\frac{1}{9}\)
∴ slope of FD = -9 ……[∵ FD ⊥ BC]
Since, FD passes through (-1, 8) and has slope -9
∴ Equation of FD is y – 8 = -9(x + 1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1 …..(i)
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) ….[∵ FE ⊥ AC]
Since, FE passes through (4, -2) and has slope \(\frac{-4}{11}\)
∴ Equation of FE is y + 2 = \(\frac{-4}{11}\)(x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = -4x + 16
∴ 4x + 11y = -6 ….(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of y in (ii), we get
∴ 4x + 11(-9x – 1) = -6
∴ 4x – 99x – 11 = -6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F = \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 5 Correlation Miscellaneous Exercise 5 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 5 Correlation Miscellaneous Exercise 5

Question 1.
Two series of x and y with 50 items each have standard deviations of 4.8 and 3.5 respectively. If the sum of products of deviations of x and y series from respective arithmetic means is 420, then find the correlation coefficient between x and y.
Solution:

Question 2.
Find the number of pairs of observations from the following data,
r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40.
Solution:
Given, r = 0.15, σ_y = 4, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 12, \(\Sigma\left(x_{i}-\bar{x}\right)^{2}\) = 40

Question 3.
Given that r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900. Find the number of pairs of observations.
Solution:
Given, r = 0.4, σ_y = 3, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)\left(y_{\mathrm{i}}-\bar{y}\right)\) = 108, \(\sum\left(x_{\mathrm{i}}-\bar{x}\right)^{2}\) = 900

Question 4.
Given the following information, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5, where x_i and y_i are the deviations from their respective means, find the number of items.
Solution:
Here, Σ\(x_{\mathrm{i}}^{2}\) = 90, Σx_iy_i = 60, r = 0.8, σ_y = 2.5
Here, x_i and y_i are the deviations from their respective means.
∴ If X_i, Y_i are elements of x and y series respectively, then
X_i – \(\bar{x}\) = x_i and Y_i – \(\bar{y}\) = y_i
∴ Σx_iy_i = \(\Sigma\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)\left(\mathrm{Y}_{\mathrm{i}}-\bar{y}\right)\) = 60, \(\sum x_{\mathrm{i}}^{2}=\sum\left(\mathrm{X}_{\mathrm{i}}-\bar{x}\right)^{2}\) = 90

Question 5.
A sample of 5 items is taken from the production of a firm. The length and weight of 5 items are given below. [Given: √0.8823 = 0.9393]

Calculate the correlation coefficient between length and weight and interpret the result.
Solution:
Let length = x_i (in cm), Weight = y_i (in gm)


∴ the value of r indicates a high degree of positive correlation between length and weight of items.

Question 6.
Calculate the correlation coefficient from the following data and interpret it.

Solution:


∴ the value of r indicates a perfect negative correlation between x and y.

Question 7.
Calculate the correlation coefficient from the following data and interpret it.

Solution:



∴ the value of r indicates a perfect positive correlation between x and y.

Question 8.
If the correlation coefficient between X and Y is 0.8, what is the correlation coefficient between
(i) 2X and Y
(ii) \(\frac{X}{2}\) and Y
(iii) X and 3Y
(iv) X – 5 and Y – 3
(v) X + 7 and Y + 9
(vi) \(\frac{X-5}{7}\) and \(\frac{Y-3}{8}\)?
Solution:
The correlation coefficient remains unaffected by the change of origin and scale.
i.e., if u_i = \(\frac{x_{i}-\mathrm{a}}{\mathrm{h}}\) and v_i = \(\frac{y_{i}-\mathrm{b}}{\mathrm{k}}\), then Corr(U, V) = ±Corr(X, Y).
according to the same or opposite signs of h and k.
(i) u_i = \(\frac{2\left(x_{i}-0\right)}{1}\), v_i = \(\frac{y_{i}-0}{1}\)
Here, h = 1 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(ii) u_i = \(\frac{x_{i}-0}{2}\), v_i = \(\frac{y_{\mathrm{i}}-0}{1}\)
Here, h = 2 and k = 1 are of the same signs.
∴ Corr (U, V) = Corr (X, Y) = 0.8

(iii) Corr (X, 3Y) = Corr (X, Y) = 0.8

(iv) Corr (X – 5, Y – 3) = Corr(X, Y) = 0.8

(v) Corr (X + 7, Y + 9) = Corr(X, Y) = 0.8

(vi) Corr(\(\frac{X-5}{7}, \frac{Y-3}{8}\)) = Corr(X, Y) = 0.8

Question 9.
In the calculation of the correlation coefficient between the height and weight of a group of students of a college, one investigator took the measurements in inches and pounds while the other investigator took the measurements in cm. and kg. Will they get the same value of the correlation coefficient or different values? Justify your answer.
Solution:
The coefficient of correlation is a ratio of covariance and standard deviations.
Since covariance and standard deviations are independent of units of measurement.
∴ coefficient of correlation is also independent of units of measurement.
∴ values of coefficient of correlation obtained by first and second investigators are the same.

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2

Question 1.
Find the range for the following data.
116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144
Solution:
Here, largest value (L) = 203, smallest value (S) = 114
∴ Range = L – S
= 203 – 114
= 89

Question 2.
Given below the frequency distribution of weekly w ages of 400 workers. Find the range.

Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

Question 3.
Find the range of the following data.

Solution:
Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115
∴ Range = L – S
= 175 – 115
= 60

Question 4.
The city traffic police issued challans for not observing the traffic rules:

Find Q.D.
Solution:
The given data can be arranged in ascending order as follows:
24, 36, 40, 58, 62, 80
Here, n = 6
Q₁ = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of \(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (1.75)th observation
= value of 1st observation + 0.75(value of 2nd observation – value of 1st observation)
= 24 + 0.75(36 – 24)
= 24 + 0.75(12)
= 24 + 9
∴ Q₁ = 33
Q₃ = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation
= value of 3\(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation
= value of (3 × 1.75)th observation
= value of (5.25)th observation
= value of 5th observation + 0.25(value of 6th observation – value of 5th observation)
= 62 + 0.25(80 – 62)
= 62 + 0.25(18)
= 62 + 4.5
= 66.5
∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}=\frac{66.5-33}{2}=\frac{33.5}{2}\) = 16.75

Question 5.
Calculate Q.D. from the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 35
Q₁ class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{35}{4}\) = 8.75
Cumulative frequency which is just greater than (or equal to) 8.75 is 15.
∴ Q₁ lies in the class 20-30.
∴ L = 20, c.f. = 8, f = 7, h = 10

Question 6.
Calculate the appropriate measure of dispersion for the following data.

Solution:
Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.
We construct the less than cumulative frequency table as follows:

Here N = 250
Q₁ class class containing \(\left(\frac{N}{4}\right)^{t h}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{250}{4}\) = 62.5
Cumulative frequency which is just greater than (or equal to) 62.5 is 65.
∴ Q₁ lies in the class 35-40.
∴ L = 35, c.f. = 15, f = 50, h = 5

The cumulative frequency which is just greater than (or equal to) 187.5 is 190.
∴ Q₃ lies in the class 45-50.
∴ L = 45, c.f. = 150, f = 40, h = 5

Question 7.
Calculate Q.D. of the following data.

Solution:
We construct the less than cumulative frequency table as follows:

Here, N = 120
Q₁ class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation
∴ \(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30
Cumulative frequency which is just greater than (or equal to) 30 is 35.
∴ Q₁ lies in the class 4-6.
∴ L = 4, c.f. = 15, f = 20, h = 2

Cumulative frequency which is just greater than (or equal to) 90 is 90.
∴ Q₃ lies in the class 10-12.
∴ L = 10, c.f. = 72, f = 18, h = 2

Question 8.
Find variance and S.D. for the following set of numbers.
25, 21, 23, 29, 27, 22, 28, 23, 27, 25 (Given √6.6 = 2.57)
Solution:
We prepare the following table for the calculation of variance and S.D.:

Question 9.
Following data gives no. of goals scored by a team in 90 matches.

Compute the variance and standard deviation for the above data.
Solution:
We prepare the following table for the calculation of variance and S.D:

Question 10.
Compute the arithmetic mean and S.D. and C.V. (Given √296 = 17.20)

Solution:
We prepare the following table for calculation of arithmetic mean and S.D.:

Question 11.
The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.
Solution:
Here, n = 200, \(\bar{x}\) = Mean = 60, S.D. = 20
Wrongly taken items are 3 and 67.
Correct items are 13 and 17.
Now, \(\bar{x}\) = 60

Correct value of \(\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i}\) (sum of wrongly taken items) + (sum of correct items)
= 12000 – (3 + 67) + (13 + 17)
= 12000 – 70 + 30
= 11960
Correct value of mean = \(\frac{1}{n}\) × correct value of \(\sum_{i=1}^{n} x_{i}\)
= \(\frac{1}{200}\) × 11960
= 59.8
Now, S.D. = 20
Variance = (S.D.)² = 20²
∴ Variance = 400
∴ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^{2}-(\bar{x})^{2}=400\)

Correct value of \(\sum_{i=1}^{n} x_{i}^{2}\)
= \(\sum_{i=1}^{n} x_{i}^{2}\) – (Sum of squares of wrongly taken items) + (Sum of squares of correct items)
= 800000 – (3² + 67²) + (13² + 17²)
= 800000 – (9 + 4489) + (169 + 289)
= 800000 – 4498 + 458
= 795960
∴ Correct value of Variance = (\(\frac{1}{n}\) × \(\sum_{i=1}^{n} x_{i}^{2}\)) – (correct value of \(\bar{x}\))²
= \(\frac{1}{200}\) × 795960 – (59.8)²
= 3979.8 – 3576.04
= 403.76
∴ The correct mean is 59.8 and correct variance is 403.76.

Question 12.
The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.
Solution:

Question 13.
The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.
Solution:
Let n₁ and n₂ be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{\mathrm{c}}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{1}\) = 62, σ₁ = 8, σ₂ = 10
Here, n₁ + n₂ = n
∴ n₁ + n₂ = 200 …….(i)
Combined mean is given by

∴ 70n₁ + 62n₂ = 13000
∴ 35n₁ + 31n₂ = 6500 ……..(ii)
Solving (i) and (ii), we get
n₁ = 75, n₂ = 125
Combined standard deviation is given by,

Question 14.
From the following data available for 5 pairs of observations of two variables x and y, obtain the combined S.D. for all 10 observations,
where \(\sum_{i=1}^{n} x_{i}=30, \sum_{i=1}^{n} y_{i}=40, \sum_{i=1}^{n} x_{i}^{2}=225, \sum_{i=1}^{n} y_{i}^{2}=340\)
Solution:

Question 15.
The mean and standard deviations of two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.
Solution:

Since C.V. (I) > C.V. (II)
∴ the brand I is more variable.

Question 16.
Calculate the coefficient of variation for the data given below. [Given √3.3 = 1.8166]

Solution:

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.3 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Ex 2.3

Question 1.
The mean and standard deviation of two distributions of 100 and 150 items are 50, 5, and 40, 6 respectively. Find the mean and standard deviation of all the 250 items taken together.
Solution:


∴ The mean and standard deviation of all 250 items taken together are 44 and √55.6 respectively.

Question 2.
For certain bivariate data, the following information is available.

Obtain the combined standard deviation.
Solution:

Question 3.
Calculate the coefficient of variation of marks secured by a student in the exam, where the marks are: 2, 4, 6, 8, 10. (Given: √8 = 2.8284)
Solution:

Question 4.
Find the coefficient of variation of a sample that has a mean equal to 25 and a standard deviation of 5.
Solution:
Given, \(\bar{x}\) = 25, σ = 5
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
= 100 × \(\frac{5}{25}\)
= 20%

Question 5.
A group of 65 students of class XI has their average height as 150.4 cm with a coefficient of variation of 2.5%. What is the standard deviation of their height?
Solution:
Given, n = 65, \(\bar{x}\) = 150.4, C.V. = 2.5%
C.V. = 100 × \(\frac{\sigma}{\bar{x}}\)
∴ 2.5 = 100 × \(\frac{\sigma}{150.4}\)
∴ \(\frac{2.5 \times 150.4}{100}\) = σ
∴ σ = 3.76
∴ the standard deviation of students’ height is 3.76.

Question 6.
Two workers on the same job show1 the following results:

(i) Regarding the time required to complete the job, which worker is more consistent?
(ii) Which worker seems to be faster in completing the job?
Solution:

(i) Since, C.V. (P) < C.V.(Q)
∴ Worker P is more consistent regarding the time required to complete the job.

(ii) Since, \(\bar{p}\) > \(\bar{q}\)
i.e., the expected time for completing the job is less for worker Q.
∴ Worker Q seems to be faster in completing the job.

Question 7.
A company has two departments with 42 and 60 employees respectively. Their average weekly wages are ₹ 750 and ₹ 400. The standard deviations are 8 and 10 respectively.
(i) Which department has a larger bill?
(ii) Which department has larger variability in wages?
Solution:

(i) Since, \(\bar{x}_{1}>\bar{x}_{2}\)
i.e., average weekly wages are more for the first department.
∴ the first department has a larger bill.
(ii) Since, C.V. (1) < C.V. (2)
∴ the second department is less consistent.
∴ the second department has larger variability in wages.

Question 8.
The following table gives the weights of the students of class A. Calculate the coefficient of variation (Given √8 = 0.8944)

Solution:

Question 9.
Compute coefficient of variation for team A and team B. (Given: √2.5162 = 1.5863, √2.244 = 1.4980)

Which team is more consistent?
Solution:
Let f₁ denote no. of goals of team A and f₂ denote no. of goals of team B.


Since C.V. of team A > C.V. of team B.
∴ team B is more consistent.

Question 10.
Given below is the information about marks obtained in Mathematics and Statistics by 100 students in a class. Which subject shows the highest variability in marks?

Solution:

Since C.V. (S) > C.V. (M)
∴ The subject statistics show higher variability in marks.