Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Ex 2.1 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Ex 2.1

Question 1.

Find range of the following data:

575, 609, 335, 280, 729, 544, 852, 427, 967, 250

Solution:

Here, largest value (L) = 967, smallest value (S) = 250

∴ Range = L – S

= 967 – 250

= 717

Question 2.

The following data gives the number of typing mistakes done by Radha during a week. Find the range of the data.

Solution:

Here, largest value (L) = 21, smallest value (S) = 10

∴ Range = L – S

= 21 – 10

= 11

Question 3.

Find range for the following data:

Solution:

Here, upper limit of the highest class (L) = 72, lower limit of the lowest class (S) = 62

∴ Range = L – S

= 72 – 62

= 10

Question 4.

Find the Q. D. for the following data.

3, 16, 8, 15, 19, 11, 5, 17, 9, 5, 3.

Solution:

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

3, 3, 5, 5, 8, 9, 11, 15, 16, 17, 19

Here, n = 11

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

= value of \(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation

= value of 3rd observation

∴ Q_{1} = 5

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

= value of 3\(\left(\frac{11+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 3)th observation

= value of 9th observation

= 16

∴ Q.D.= \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)

= \(\frac{16-5}{2}\)

= \(\frac{11}{2}\)

= 5.5

Question 5.

Given below are the prices of shares of a company for the last 10 days. Find Q.D.:

172, 164, 188, 214, 190, 237, 200, 195, 208, 230.

Solution:

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

164, 172, 188, 190, 195, 200, 208, 214, 230, 237

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)

= 172 + 0.75(188 – 172)

= 172 + 0.75(16)

= 172 + 12

= 184

∴ 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)

= 214 + 0.25(230 – 214)

= 214 + 0.25(16)

= 214 + 4

= 218

∴ Q.D. = \(\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}\)

= \(\frac{218-184}{2}\)

= \(\frac{34}{2}\)

= 17

Question 6.

Calculate Q.D. for the following data.

Solution:

Since the given data is arranged in ascending order, we construct less than cumulative frequency table as follows:

Here, n = 30

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

= value of \(\left(\frac{30+1}{4}\right)^{\text {th }}\) observation

= value of (7.75)th observation

Cumulative frequency which is just greater than (or equal to) 7.75 is 11.

∴ Q_{1} = 25

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

= value of \(\left[3\left(\frac{30+1}{4}\right)\right]^{\text {th }}\) observation

= value of (3 × 7.75)th observation

= value of (23.25)th observation

Cumulative frequency which is just greater than (or equal to) 23.25 is 27.

∴ Q_{3} = 29

∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}\)

= \(\frac{29-25}{2}\)

∴ Q.D. = 2

Question 7.

Following data gives the age distribution of 240 employees of a firm. Calculate Q.D. of the distribution.

Solution:

We construct the less than cumulative frequency table as follows:

Here, N = 240

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

∴ \(\frac{N}{4}=\frac{240}{4}\) = 60

Cumulative frequency which is just greater than (or equal to) 60 is 70.

∴ Q_{1} lies in the class 25 – 30.

∴ L = 25, c.f. = 30, f = 40, h = 5

Cumulative frequency which is just greater than (or equal to) 180 is 180.

∴ Q_{3} lies in the class 35-40.

∴ L = 35, c.f. = 130, f = 50, h = 5

Question 8.

Following data gives the weight of boxes. Calculate Q.D. for the data.

Solution:

Here, N = 60

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

∴ \(\frac{\mathrm{N}}{4}=\frac{60}{4}\) = 15

Cumulative frequency which is just greater than (or equal to) 15 is 26.

∴ Q_{1} lies in the class 14 – 16.

∴ L = 14, c.f. = 10, f = 16, h = 2

Cumulative frequency which is just greater than (or equal to) 45 is 58.

∴ Q_{3} lies in the class 18 – 20.

∴ L = 18, c.f. = 40, f = 18, h = 2