GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

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.
Σ |xi – \(\bar{x}\)| = 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7
= 24.
∴ 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
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 1
= 8.4

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

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.
∴ \(\frac{12}{2}\) th item = 6th item = 13.
and (\(\frac{12}{2}\) + 1)th item = 7th item = 14.
∴ Median M = \(\frac{1}{2}\)(13 + 14) = 13.5.
∴ \(\sum_{i=1}^{12}\)|xi – 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.
∴ \(\frac{10}{2}\)th = 5th observation = 46.
(\(\frac{10}{2}\) + 1)th = 6th observation = 49.
∴ 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
∴ M.D.(M) = \(\frac{70}{10}\) = 7.

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

Find the mean deviation about the median for the data in questions 7 to 8:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 2
Solutions to questions 7 and 8:
7. The table to find cumulative frequency c.f. and median is as follows:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 3
Total of the frequencies = 26.
∴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.
∴ 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:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 4
Since N = Σfi = 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.
∴ 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:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 5
Solutions to questions 9 and 10:
9.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 6
10.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 6a

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

Question 11.
Find the mean deviation about median for the following data:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 7
Solution:
Table to find cumulative frequencies and Σfi|xi – M| is given below:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 8
Here, \(\frac{N}{2}\) = \(\frac{50}{2}\) = 25.
So, median class is 20 – 30
∴ l = 20, f = 14, h = 10.
Median = \(\frac{\frac{N}{2}-C}{f}\) × h
= 20 + \(\frac{25-14}{14}\) × 10
= 20 + \(\frac{11}{14}\) × 10
= 20 + 7.86
= 27.86.
Σfi|xi – M| = 517.16.
∴ Median deviation about median
= \(\frac{517.16}{50}\) = 10.34

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1

Question 12.
Calculate the mean deviation about median for the age distribution of 100 persons given below:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 9
Solution:
Continuous frequency distribution, cumulative frequencies and fi|xi – M| are given in the following table:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.1 img 10
Median class 35.5 – 40.5.
∴ l = 35.5 and h = 5, C = 37, f = 26.
∴ Median = l + \(\frac{\frac{N}{2}-C}{f}\) × h
= 35.5 + \(\frac{50-37}{26}\) × 5 = 35.5 + \(\frac{13}{26}\) × 5
= 35.5 + 2.5 = 38.
∴ Σfi|xi – M| = 735..
N = Σfi = 100.
Mean deviation about Median
= \(\frac{\Sigma f_{i}\left|x_{i}-M\right|}{N}\) = \(\frac{735}{100}\) = 7.35.

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