Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

Find the mean deviation about the mean for the data in questions 1 and 2:

1. 4, 7, 8, 9, 10, 12, 13, 17

2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Solutions to questions 1 and 2:

1. Arithmetic mean \(\bar{x}\) of 4, 7, 8, 9, 10, 12, 13, 17 is

\(\bar{x}\) = \(\frac{4+7+8+9+10+12+13+17}{8}\) = \(\frac{80}{8}\)

= 10.

Î£ |x_{i} – \(\bar{x}\)| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7

= 24.

âˆ´ Mean deviation about mean

= M.D. (\(\bar{x}\)) = \(\frac{\Sigma\left|x_{i}-\bar{x}\right|}{n}\) = \(\frac{24}{8}\) = 3.

2. Mean of the given data in

= 8.4

Find the mean deviation about the mean for the data in questions 3 and 4:

3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Solutions to questions 3 and 4:

3. The given data is

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

Arranging it in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

There are 12 observations.

âˆ´ \(\frac{12}{2}\) th item = 6th item = 13.

and (\(\frac{12}{2}\) + 1)th item = 7th item = 14.

âˆ´ Median M = \(\frac{1}{2}\)(13 + 14) = 13.5.

âˆ´ \(\sum_{i=1}^{12}\)|x_{i} – M| = 3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5

= 28.

M.D. (M) = \(\frac{28}{12}\) = 2.33.

4. The given data is

36, 72, 46, 42, 60, 45, 53, 46, 51, 49.

Arranging the data in ascending order:

36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The number of observations = 10.

âˆ´ \(\frac{10}{2}\)th = 5th observation = 46.

(\(\frac{10}{2}\) + 1)th = 6th observation = 49.

âˆ´ Median = \(\frac{1}{2}\)(46 + 49) = 47.5.

= |36 – 47.51 + |42 – 47.5| + |45 – 47.5| + |46 – 47.5| + |46 – 47.5| + |49 – 47.5| + |51 – 47.5| + |53 – 47.5| + |60 – 47.5| + |72 – 47.5|

=11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 = 70.

Deviations about median

âˆ´ M.D.(M) = \(\frac{70}{10}\) = 7.

Find the mean deviation about the median for the data in questions 7 to 8:

Solutions to questions 7 and 8:

7. The table to find cumulative frequency c.f. and median is as follows:

Total of the frequencies = 26.

âˆ´Median M = \(\frac{1}{2}\)[\(\frac{26}{2}\)th value + (\(\frac{26}{2}\) + 1)th value]

= \(\frac{1}{2}\)(13th value + 14th value)

= \(\frac{1}{2}\)(7 + 7) = 7.

âˆ´ Mean deviation from the median

\(\frac{\Sigma f_{i}\left(x_{i}-M\right)}{\Sigma f_{i}}\) = \(\frac{84}{26}\) = \(\frac{42}{13}\) = 3.23.

8. Cumulative frequency distribution of the given data is:

Since N = Î£f_{i} = 29, which is odd, the median is the (\(\frac{n+1}{2}\))th observation, i.e., (\(\frac{29+1}{2}\) = 15)th observation, which is equal to 30. Thus, median is 30.

âˆ´ Mean deviation from the median = \(\frac{\Sigma f_{i}\left|x_{i}-30\right|}{\Sigma f_{i}}\) = \(\frac{148}{29}\) = 5.1.

Find the median deviation about mean for the data in questions 9 and 10:

Solutions to questions 9 and 10:

9.

10.

Question 11.

Find the mean deviation about median for the following data:

Solution:

Table to find cumulative frequencies and Î£f_{i}|x_{i} – M| is given below:

Here, \(\frac{N}{2}\) = \(\frac{50}{2}\) = 25.

So, median class is 20 – 30

âˆ´ l = 20, f = 14, h = 10.

Median = \(\frac{\frac{N}{2}-C}{f}\) Ã— h

= 20 + \(\frac{25-14}{14}\) Ã— 10

= 20 + \(\frac{11}{14}\) Ã— 10

= 20 + 7.86

= 27.86.

Î£f_{i}|x_{i} – M| = 517.16.

âˆ´ Median deviation about median

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

Question 12.

Calculate the mean deviation about median for the age distribution of 100 persons given below:

Solution:

Continuous frequency distribution, cumulative frequencies and f_{i}|x_{i} – M| are given in the following table:

Median class 35.5 – 40.5.

âˆ´ l = 35.5 and h = 5, C = 37, f = 26.

âˆ´ Median = l + \(\frac{\frac{N}{2}-C}{f}\) Ã— h

= 35.5 + \(\frac{50-37}{26}\) Ã— 5 = 35.5 + \(\frac{13}{26}\) Ã— 5

= 35.5 + 2.5 = 38.

âˆ´ Î£f_{i}|x_{i} – M| = 735..

N = Î£f_{i} = 100.

Mean deviation about Median

= \(\frac{\Sigma f_{i}\left|x_{i}-M\right|}{N}\) = \(\frac{735}{100}\) = 7.35.