Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 1 Partition Values Ex 1.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 1 Partition Values Ex 1.1

Question 1.

Compute all the quartiles for the following series of observations:

16, 14.9, 11.5, 11.8, 11.1, 14.5, 14, 12, 10.9, 10.7, 10.6, 10.5, 13.5, 13, 12.6

Solution:

The given data can be arranged in ascending order as follows:

10.5, 10.6, 10.7, 10.9, 11.1, 11.5, 11.8, 12, 12.6, 13, 13.5, 14, 14.5, 14.9, 16

Here, n = 15

Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation

= value of 4th observation

∴ Q_{1} = 10.9

Q_{2} = 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

∴ Q_{2} = 12

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{15+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 4)th observation

= value of 12th observation

∴ Q_{3} = 14

Question 2.

The heights (in cm.) of 10 students are given below:

148, 171, 158, 151, 154, 159, 152, 163, 171, 145

Calculate Q_{1} and Q_{3} for the above data.

Solution:

The given data can be arranged in ascending order as follows:

145, 148, 151, 152, 154, 158, 159, 163, 171, 171

Here, n = 10

Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation

= value of (2.75)th observation

= value of 2nd observation + 0.75 (value of 3rd observation – value of 2nd observation)

= 148 + 0.75 (151 – 148)

= 148 + 0.75(3)

= 148 + 2.25

∴ Q_{1} = 150.25

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{10+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 2.75)th observation

= value of (8.25)th observation

= value of 8th observation + 0.25 (value of 9th observation – value of 8th observation)

= 163 + 0.25(171 – 163)

= 163 + 0.25(8)

= 163 + 2

∴ Q_{3} = 165

Question 3.

The monthly consumption of electricity (in units) of families in a certain locality is given below:

205, 201, 190, 188, 195, 172, 210, 225, 215, 232, 260, 230

Calculate electricity consumption (in units) below which 25% of the families lie.

Solution:

To find the consumption of electricity below which 25% of the families lie, we have to find Q_{1}.

Monthly consumption of electricity (in units) can be arranged in ascending order as follows:

172, 188, 190, 195, 201, 205, 210, 215, 225, 230, 232, 260.

Here, n = 12

Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{12+1}{4}\right)^{\text {th }}\) observation

= value of (3.25)th observation

= value of 3rd observation + 0.25 (value of 4th observation – value of 3rd observation)

= 190 + 0.25(195 – 190)

= 190 + 0.25(5)

= 190 + 1.25

= 191.25

∴ the consumption of electricity below which 25% of the families lie is 191.25.

Question 4.

For the following data of daily expenditure of families (in ₹), compute the expenditure below which 75% of families include their expenditure.

Solution:

To find the expenditure below which 75% of families have their expenditure, we have to find Q_{3}.

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

Here, n = 100

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{100+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 25.25)th observation

= value of (75.75)th observation

Cumulative frequency which is just greater than (or equal to) 75.75 is 87.

∴ Q_{3} = 650

∴ the expenditure below which 75% of families include their expenditure is ₹ 650.

Question 5.

Calculate all the quartiles for the following frequency distribution:

Solution:

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

Here, n = 300

Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation

= value of (75.25)th observation

Cumulative frequency which is just greater than (or equal to) 75.25 is 90.

∴ Q_{1} = 2

Q_{2} = value of 2\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 2\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation

= value of (2 × 75.25)th observation

= value of (150.50)th observation

∴ Cumulative frequency which is just greater than (or equal to) 150.50 is 185.

∴ Q_{2} = 3

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{300+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 75.25)th observation

= value of (225.75)th observation

Cumulative frequency which is just greater than (or equal to) 225.75 is 249.

∴ Q_{3} = 4

Question 6.

The following is the frequency distribution of heights of 200 male adults in a factory:

Find the central height.

Solution:

To find the central height, we have to find Q_{2}.

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

Here, N = 200

Q_{2} class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation

∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 200}{4}\) = 100

Cumulative frequency which is just greater than (or equal to) 100 is 156.

∴ Q_{2} lies in the class 165 – 170.

∴ L = 165, h = 5, f = 64, c.f. = 92

Q_{2} = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)

= 165 + \(\frac{5}{64}\) (100 – 92)

= 165 + \(\frac{5}{64}\) × 8

= 165 + \(\frac{5}{8}\)

= 165 + 0.625

= 165.625

∴ Central height is 165.625 cm.

Question 7.

The following is the data of pocket expenditure per week of 50 students in a class. It is known that the median of the distribution is ₹ 120. Find the missing frequencies.

Solution:

Let a and b be the missing frequencies of class 50 – 100 and class 150 – 200 respectively.

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

Here, N = 25 + a + b

Since, N = 50

∴ 25 + a + b = 50

∴ a + b = 25 …..(i)

Given, Median = Q_{2} = 120

∴ Q2 lies in the class 100 – 150.

∴ L = 100, h = 50, f = 15, \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 50}{4}\) = 25

∴ Q_{2} = \(\mathrm{L}+\frac{\mathrm{h}}{\mathrm{f}}\left(\frac{2 \mathrm{~N}}{4}-\text { c.f. }\right)\)

∴ 120 = 100 + \(\frac{50}{15}\) [25 – (7 + a)]

∴ 120 – 100 = \(\frac{10}{3}\) (25 – 7 – a)

∴ 20 = \(\frac{10}{3}\) (18 – a)

∴ \(\frac{60}{10}\) = 18 – a

∴ 6 = 18 – a

∴ a = 18 – 6 = 12

Substituting the value of a in equation (i), we get

12 + b = 25

∴ b = 25 – 12 = 13

∴ 12 and 13 are the missing frequencies of the class 50 – 100 and class 150 – 200 respectively.

Question 8.

The following is the distribution of 160 workers according to the wages in a certain factory:

Determine the values of all quartiles and interpret the results.

Solution:

The given table is a more than cumulative frequency.

We transform the given table into less than cumulative frequency.

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

Here, N = 160

∴ Q_{1} class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation

∴ \(\frac{\mathrm{N}}{4}=\frac{160}{4}\) = 40

Cumulative frequency which is just greater than (or equal to) 40 is 69.

∴ Q_{1} lies in the class 10000 – 11000

∴ L = 10000, h = 1000, f = 46, c.f. = 23

Q_{1} = \(L+\frac{h}{f}\left(\frac{N}{4}-\text { c.f. }\right)\)

= 10000 + \(\frac{1000}{46}\) (40 – 23)

= 10000 + \(\frac{1000}{46}\) (17)

= 10000 + \(\frac{17000}{46}\)

= 10000 + 369.57

= 10369.57

Q_{2} class = class containing \(\left(\frac{2 \mathrm{~N}}{4}\right)^{\mathrm{th}}\) observation

∴ \(\frac{2 \mathrm{~N}}{4}=\frac{2 \times 160}{4}\) = 80

Cumulative frequency which is just greater than (or equal to) 80 is 103.

∴ Q_{2} lies in the class 11000 – 12000.

∴ L = 11000, h = 1000, f = 34, c.f. = 69

∴ Q_{2} = \(L+\frac{h}{f}\left(\frac{2 N}{4}-\text { c.f. }\right)\)

= 11000 + \(\frac{1000}{34}\)(80 – 69)

= 11000 + \(\frac{1000}{34}\)(11)

= 11000 + \(\frac{11000}{34}\)

= 11000 + 323.529

= 11323.529

Q_{3} class = class containing \(\left(\frac{3 \mathrm{~N}}{4}\right)^{\text {th }}\) observation

∴ \(\frac{3 \mathrm{~N}}{4}=\frac{3 \times 160}{4}\) = 120

Cumulative frequency which is just greater than (or equal to) 120 is 137.

∴ Q_{3} lies in the class 12000 – 13000.

∴ L = 12000, h = 1000, f = 34, c.f. = 103

∴ Q_{3} = \(\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)

= 12000 + \(\frac{1000}{34}\) (120 – 103)

= 12000 + \(\frac{1000}{34}\) (17)

= 12000 + \(\frac{1000}{2}\)

= 12000 + 500

= 12500

Interpretation:

Q_{1} < Q_{2} < Q_{3}

Question 9.

Following is grouped data for the duration of fixed deposits of 100 senior citizens from a certain bank:

Calculate the limits of fixed deposits of central 50% senior citizens.

Solution:

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

To find the limits of fixed deposits of central 50% senior citizens, we have to find Q_{1} and Q_{3}.

Here, N = 100

Q_{1} class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) observation

∴ \(\frac{N}{4}=\frac{100}{4}\) = 25

Cumulative frequency which is just greater than (or equal to) 25 is 35.

∴ Q1 lies in the class 180 – 360.

∴ L = 180, h = 180, f = 20, c.f. = 15

∴ Q_{1} = \(L+\frac{h}{f}\left(\frac{N}{4}-c . f .\right)\)

= 180 + \(\frac{180}{20}\) (25 – 15)

= 180 + 9(10)

= 180 + 90

∴ Q_{1} = 270

Q_{3} class = class containing \(\left(\frac{3 \mathrm{N}}{4}\right)^{\text {th }}\) observation

∴ \(\frac{3 \mathrm{N}}{4}=\frac{3 \times 100}{4}\) = 75

Cumulative frequency which is just greater than (or equal to) 75 is 90.

∴ Q_{3} lies in the class 540 – 720.

∴ L = 540, h = 180, f = 30, c.f. = 60

∴ Q_{3} = \(L+\frac{h}{f}\left(\frac{3 N}{4}-c . f .\right)\)

= 540 + \(\frac{180}{30}\) (75 – 60)

= 540 + 6(15)

= 540 + 90

∴ Q_{3} = 630

∴ Limits of duration of fixed deposits of central 50% senior citizens is from 270 to 630.

Question 10.

Find the missing frequency given that the median of the distribution is 1504.

Solution:

Let x be the missing frequency of the class 1550 – 1750.

We construct the less than frequency table as given below:

Here, N = 199 + x

Given, Median (Q_{2}) = 1504

∴ Q_{2} lies in the class 1350 – 1550.

∴ L = 1350, h = 200, f = 100, c.f. = 63,

\(\frac{2 \mathrm{~N}}{4}=\frac{199+x}{2}\)

∴ Q_{2} = \(L+\frac{h}{f}\left(\frac{2 N}{4}-c . f .\right)\)

∴ 1504 = 1350 + \(\frac{200}{100}\left(\frac{199+x}{2}-63\right)\)

∴ 1504 – 1350 = 2\(\left(\frac{199+x-126}{2}\right)\)

∴ 154 = 199 + x – 126

∴ 154 = x + 73

∴ x = 81