Students must practice these Maths 2A Important Questions TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2A Measures of Dispersion Important Questions Long Answer Type

Question 1.

Find the mean deviation about the mean for the following continuous distribution. [AP – Mar. 2015]

Solution:

Here, Σf_{i} = 100

Σf_{i}x_{i} = 7100

∴ Mean \(\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\mathrm{N}}=\frac{7100}{100}\) = 71

∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{1040}{100}\) = 10.4.

Question 2.

Find the mean deviation about the mean for the following continuous distribution.

Solution:

We can form the following table from the given data :

Here, N = Σf_{i} = 100

Σf_{i}x_{i} = 12530

∴ Mean, \(\bar{x}=\frac{\sum_{i=1}^n f_i x_i}{N}=\frac{12530}{100}\) = 125.3

∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}\)

= \(\frac{1128.8}{100}\)

= 11.288 = 11.28.

Question 3.

Find the mean deviation from the mean of the following data, using the step deviation method. [AP – Mar. ’18; TS – Mar. 2016]

Solution:

We can form the following table from the given data :

Let the assumed mean be A = 35

Here, C = 10

N = Σf_{i} = 50

Mean, \(\bar{x}=A+\frac{\sum_{i=1}^n f_i d_i}{N} \times C\)

= \(A+\frac{\sum_{i=1}^7 f_i d_i}{N} \times C\)

= 35 + \(\frac{(-8)}{50}\) × 10

= 35 – 1.6 = 33.4

∴ Mean deviation about mean is M.D = \(\frac{\Sigma f_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{659.2}{50}\) = 13.185.

Question 4.

Find the mean deviation about the mean for the given data using step deviation method. [AP – Mar. ’17, ’16, AP – May 2015] [TS – Mar. ’18, May ’16]

Solution:

We can form the following table from the given data:

Let the assumed mean be A = 25

Here, C = 10

N = Σf_{i} = 50

\(\bar{x}=A+\frac{\sum_{i=1}^n f_i d_i}{N} \times C\)

= \(\mathrm{A}+\frac{\sum_{\mathrm{i}=1}^5 \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{N}} \times \mathrm{C}\)

= 25 + \(\frac{10}{50}\) × 10 = 27

∴ Mean deviation about mean is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{N}}=\frac{472}{50}\) = 9.44.

Question 5.

Find the mean deviation about the median for the following continuous distribution.

Solution:

Median class = Class containing \(\frac{N}{2}^{\text {th }}\) item

= \(\frac{50}{2}\) = 25 item

= 20 – 30 class

Here, l = 20, C = 10, f = 14, m = 14, N = 50

∴ Median, M = l + \(\frac{\frac{N}{2}-m}{f}\) × C

= 20 + \(\frac{25-14}{14}\) × 10

= 20 + 7.86 = 27.86

∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}\)

= \(\frac{517.16}{50}\) = 10.34

Question 6.

Find the mean deviation about median for the following data.

Solution:

Median class = Class containing \(\frac{\mathrm{N}^{\text {th }}}{2}\) item

= \(\frac{1000}{2}\) = 500th item = 35 – 40 class

Here, l = 35, C = 5, f = 160, m = 420, N = 1000

∴ Median, M = l + \(\frac{\frac{N}{2}-m}{f}\) × C

= 35 + \(\frac{500-420}{160}\) × 5

= 35 + 2.5 = 37.5

∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{8175}{1000}\) = 8.175.

Question 7.

Calculate the variance and standard deviaon,of the following continuous frequency distribution.

[AP – May 2016; TS – Mar. 2017; Mar. ‘14, Board Paper]

Solution:

We can form the following table from the given data:

∴ Mean, \(\bar{x}=\frac{1}{N} \sum_{i=1}^7 f_i x_i=\frac{3100}{50}\) = 62

Variance, σ^{2} = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)

= \(\frac{1}{50}\) (10050) = 201

Standard deviation, σ = \(\sqrt{\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2}\)

= \(\sqrt{\text { variance }}=\sqrt{201}\)= 14.17 (approximately).

Question 8.

The following table gives the daily wages of workers in a factory. Compute the standard deviation and the coefficient of variation of the wages of the workers.

Solution:

We shall solve this problem using the step deviation method, since the midpoints of the class intervals are numerically large.

From the table, N = \(\sum_{i=1}^9\) f_{i} = 72

Let the assumed mean, A = 300

Here, C = 50

From the table, \(\sum_{i=1}^9\) f_{i}d_{i} = – 31

Mean, \(\bar{x}=A+\frac{\sum_{i=1}^9 f_i d_i}{N}\) × C

= 300 + \(\frac{-31}{72}\) × 50

= 300 – \(\frac{1550}{72}\)

= 278.5 (approximately)

Question 9.

The scores of two cricketers A and B in 10 innings are given below. Find who is a better run getter and who is a more consistent player.

Solution:

For cricketer A : Scores of A are 40, 25, 19. 80, 38, 8, 67, 121, 66, 76

∴ Mean, \(\overline{\mathrm{x}}_{\mathrm{A}}=\frac{\text { Sum of the scores }}{\text { No. of innings }}\)

= \(\frac{40+25+19+80+38+8+67+121+66+76}{10}\)

= \(\frac{540}{10}\) = 54

The deviations of the respective observations from the mean i.e., x_{i} – \(\overline{\mathrm{x}}\) are: – 14, – 29, – 35, 26, – 16, -46, 13, 67, 12, 22

Variance, σ^{2} = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)

= \(\frac{196+841+1225+676+256+2116+169+4489+144+484}{10}\)

= \(\frac{10596}{10}\) = 1059.6

Standard deviation, σ_{A} = \(\sqrt{1059.6}\) = 32.55

C.V of A = \(\frac{\sigma_{\mathrm{A}}}{\overline{\mathrm{x}}_{\mathrm{A}}}\) × 100

= \(\frac{32.55}{54}\) × 100 = 60.28.

For cricketer B:

Scores of B are 28, 70, 31, 0, 14, 111, 66, 31, 25, 4

Mean \(\overline{\mathrm{x}}_{\mathrm{B}}=\frac{\text { Sum of the scores }}{\text { No. of innings }}\)

= \(\frac{28+70+31+0+14+111+66+31+25+4}{10}=\frac{380}{10}\)

= 38

The deviations of the respective observations from mean i.e., (x_{i} – \(\overline{\mathbf{x}}\)) are: – 10, 32, – 7, – 38, – 24, 73, 28, – 7, – 13, – 34

Variance, σ_{B}^{2} = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)

= \(\frac{100+1024+49+1444+576+5329+784+49+169+1156}{10}\)

= \(\frac{10680}{10}\) = 1068

Standard deviation. σ_{B} = \(\sqrt{1068}\) = 32.68

C.V of B = \(\frac{\sigma_B}{\bar{x}_B}\) × 100

= \(\frac{32.68}{38}\) × 100 = 86

Since, \(\overline{\mathbf{x}}_{\mathrm{A}}\) is greater than \(\overline{\mathbf{x}}_{\mathrm{B}}\) then cricketer A is better run getter (scorer).

Since, C.V of A is less than C.V of B then cricketer ¡s also a more consistent player.

Question 10.

From the prices of shares X and Y given below, for 10 days of trading, find out which share is more stable?

Solution:

Share X:

Prices of share X are 35, 54, 52, 53, 56, 58, 52, 50, 51, 49

Mean, \(\overline{\mathrm{x}}_{\mathrm{X}}=\frac{1}{\mathrm{~m}} \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}\)

= \(\frac{510}{10}\) = 51

The deviations of the respective observations from mean i.e., (x_{i} – \(\overline{\mathrm{x}}\))^{2} are: – 16, 3, 1, 5, 7, 1, – 10, – 2

Variance, σ_{X}^{2} = \(\frac{1}{n} \sum_{i=1}^{10}\left(x_i-\bar{x}\right)^2\)

= \(\frac{256+9+1+25+49+1+1+0+4}{10}\)

= \(\frac{350}{10}\) = 35

Standard deviation, σ_{x} = √35 = 5.9 (approximately)

C.V of X = \(\frac{\sigma_X}{\bar{x}_X}\) × 100

= \(\frac{5.9}{51}\) × 100 = 11.5 (approximately)

Share Y:

Prices of share Y are 108, 107, 105. 105, 106, 107, 104, 103, 104, 101

Mean, \(\overline{\mathrm{x}}_{\mathrm{Y}}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}\)

= \(\frac{1050}{10}\) = 105

The deviations of the respective observations from mean i.e.. (x_{i} – \(\overline{\mathrm{x}}\)) are 3, 2, 0, 0, 1, 2, – 1, – 2, – 1, – 4

Variance, σ_{Y}^{2} = \(\frac{1}{n} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right)^2\)

= \(\frac{9+4+0+0+1+4+1+4+1+16}{10}\)

= \(\frac{40}{10}\) = 4

Standard deviation, σ_{Y} = √4 = 2

C.V. of Y = \(\frac{\sigma_Y}{\bar{x}_Y}\) x 100

= \(\frac{2 \times 100}{105}\) x 100 = 1.904

Since, C.V of X is greater than C.V of Y then the share of Y is more consistent (stable).

Question 11.

Find the mean and variance using the step deviation method of the following tabular data, giving the age distribution of 542 members.

Solution:

We form the following table From the given data:

Let the assumed mean, A = 55

Here, C = 10

### Some More Maths 2A Measures of Dispersion Important Questions

Question 12.

Students of two sections A and B of a class show the following performance in a test (conducted 100 marks).

Which section of students has greater variability In performance?

Solution:

Given that,

The variance of distribution of marks of section A is σ_{1}^{2} = 64

∴ The standard deviation of distribution of marks of section A is σ_{1} = 8

The variance of distribution of marks of section B is σ_{2}^{2} = 81

∴ The standard deviation of distribution of marks of section B is σ_{2} = 9

The average marks in the test of section A is \(\bar{x}_1\) = 45

The average marks in the test of section B is \(\bar{x}_2\) = 45

∵ The average marks in the test of both sections of students is the same i.e., 45.

The standard deviation of the distribution of marks in section B is greater than the standard deviation of (he distribution of marks in section A.

Hence, the section B has greater variability in the performance.

Question 13.

Find the variance and standard deviation of the following frequency distribution.

Solution:

From the given data, we can form the following table:

Mean of the given data is (\(\overline{\mathrm{x}}\)) = \(\frac{1}{N} \sum_{i=1}^7 f_i x_i\)

= \(\frac{760}{40}\) = 19

Variance, σ^{2} = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)

= \(\frac{1}{40}\) (1736) = 43.4

Standard deviation, σ = \(\sqrt{\text { variance }}\)

= \(\sqrt{43.4}\) = 6.58 (approximately).

Question 14.

Calculate the variance and standard deviation for a discrete frequency distribution.

Solution:

From the given data, we can form the following table:

Mean of the given data is (\(\overline{\mathrm{x}}\)) = \(\frac{1}{N} \sum_{i=1}^n f_i x_i\)

= \(\frac{1}{N} \sum_{i=1}^7 f_i x_i=\frac{420}{30}\) = 14

Variance, σ^{2} = \(\frac{1}{N} \sum_{i=1}^7 f_i\left(x_i-\bar{x}\right)^2\)

= \(\frac{1}{30}\) (1374) = 45.8

Standard deviation, σ =\(\sqrt{\text { variance }}\)

= \(\sqrt{45.8}\) = 6.76 (approximately).

Question 15.

Find the mean deviation about the median for the following data.

Solution:

We can form the following table from the given data:

Median class = Class containing \(\frac{N^{t h}}{2}\) item

= \(\frac{100}{2}\) = 50th item = 40 – 50 class

Here, l = 40, f = 28, m = 32, C = 10, N = 100

∴ Median, M = l + \(\frac{\frac{\dot{N}}{2}-\mathrm{m}}{\mathrm{f}}\) × C

= 40 + \(\frac{50-32}{28}\) × 10

= 40 + \(\frac{180}{28}\)

= 40 + 6.42 = 46.42

∴ Mean deviation about median is M.D = \(\frac{\Sigma \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\mathrm{N}}=\frac{1428.4}{100}\) = 14.284

Question 16.

Lives of two models of refrigerators A and B obtained ¡n a survey are given below.

From the above data suggest which model to purchase.

Solution:

Model – A:

To find the mean and variance of lives of model A refrigerators, we shall construct the following table:

From the table, N = \(\sum_{\mathbf{i}=1}^5\) f_{i} = 46, \(\sum_{\mathbf{i}=1}^5\) f_{i}x_{i} = 212

Mean, \(\bar{x}_A=\frac{1}{N} \sum_{i=1}^5 f_i x_i\)

= \(\frac{212}{46}\) = 4.6

From the table, \(\sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\) = 244.96

Variance, σ_{A}^{2} = \(\frac{1}{N} \sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\)

= \(\frac{244.96}{46}\) = 5.32

Standard deviation, σ_{A} = \(\sqrt{5.32}\)

= 2.3 (approximately)

Coefficient of variation of model A = \(\frac{\sigma_{\mathrm{A}}}{\mathrm{x}_{\mathrm{A}}}\) × 100

= \(\frac{2.3}{4.6}\) × 100

= 50 (approximately).

Model B:

To find the mean and variance of lives of model B refrigerators, we shall construct the following table:

From the table, N = \(\sum_{i=1}^5\) f_{i} = 49

\(\sum_{i=1}^5\) f_{i}x_{i} = 297

Mean, \(\bar{x}_B=\frac{1}{N} \sum_{i=1}^5 f_i x_i\)

= \(\frac{297}{49}\) = 6.06

From the table, \(\) = 224.8164

Variance, σ_{B}^{2} = \(\frac{1}{N} \sum_{i=1}^5 f_i\left(x_i-\bar{x}\right)^2\)

= \(\frac{224.8164}{49}\) = 4.58 (approximately)

Standard deviation, σ_{B} = \(\sqrt{4.58}\)

= 2.1 (approximately)

Coefficient of variation of model B = \(\frac{\sigma_B}{x_B}\) × 100

= \(\frac{2.1}{6.06}\) × 100 = 34.65 (approximately)

Since C.V. of model B is less than C.V. of model A then the model B is more consistent than the model A with regard to life in years.

Hence, we suggest model B for purchase.