Inter 1st Year Maths Statistics Solutions Exercise 13a

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Intermediate 1st Year Maths Statistics Solutions Exercise 13a

Statistics Exercise 13a Solutions

Statistics Class 11 Exercise 13a Solutions – Statistics 13a Exercise Solutions

I.

Question 1.
Find the mean deviation about the mean for the following data.
(a) 4, 7, 8, 9, 10, 12, 13, 17
(b) 38, 70, 46, 40, 42, 55, 63, 46, 54, 44
Solution:
(a) The given data is 4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, \(\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}\) = 10
The deviations of the respective observations from the mean \(\overline{\mathrm{x}}\), i.e., x_i – \(\overline{\mathrm{x}}\) are -6, -3, -2, -1, 0, 2, 3, 7
The required mean deviation about the mean is,

(b) The given data are 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data, \(\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}\) = 50
The deviations of the respective observations from the mean \(\overline{\mathrm{x}}\), i.e., \(\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\) are -12, 20, -2, -10, -8, 5, 1, -4, 4, -6.
The absolute values of the deviations, i.e., \(\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|\) are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.
The required mean deviation about the mean is, M.D.(\(\overline{\mathrm{x}}\)) = \(\frac{\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{10}\)
= \(\frac{12+20+2+10+8+5+13+4+4+6}{10}\)
= \(\frac {84}{10}\)
= 8.4

Question 2.
Find the mean deviation about the median for the following data.
(a) 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
(b) 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:
(a) The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the number of observations is 12, which is even.
Arranging the data in ascending order,
we obtain 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

The deviations of the respective observations from the median,
i.e., x_i – M, are -3.5, -2.5, -2.5, -1.5, -0.5, -0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The absolute values of the deviations, |x_i – M|, are 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is M.D.(M) = \(\frac{\sum_{i=1}^{12}\left|x_i-M\right|}{12}\)
= \(\frac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}\)
= \(\frac {28}{12}\)
= 2.33

(b) The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Here, the number of observations is 10.
Arranging the data in ascending order,
We obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72.

The deviations of the respective observations from the median, i.e., x_i – M, are -11.5, -5.5, -2.5, -1.5, -1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations, |x_i – M|, are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Thus, The required mean deviation about the median is M.D.(M) = \(\frac{\sum_{i=1}^{10}\left|x_i-M\right|}{10}\)
= \(\frac{11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5}{10}\)
= \(\frac {70}{10}\)
= 7

II.

Question 1.
Find the mean deviation about the median for the following data.
(a)

Solution:

(b)

Solution:

Question 2.
Find the mean deviation about the median for the following data.
(a)

Solution:
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data,
we obtain the following tables.

Here, N = 26, which is an even number.
Median is the mean of the 13th and 14th observations.
Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
∴ Median = \(\frac{13^{\text {th }} \text { observation }+14^{\text {th }} \text { observation }}{2}\)
= \(\frac{7+7}{2}\)
= 7
The absolute values of the deviations from the median, i.e., |x_i – M|, are

(b)

Solution:
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data,
we obtain the following table.

Here, N = 29, which is an odd number.
∴ Median = \(\left(\frac{29+1}{2}\right)^{\text {th }}\) observation = 15th observation
This observation corresponds to a cumulative frequency of 21, for which the corresponding observation is 30.
∴ Median = 30
The absolute values of the deviations from the median, i.e., |x_i – M|, are

III.

Question 1.
Find the mean deviation about the mean for the following data.
(a)

Solution:

(b)

Solution:

Question 2.
Find the mean deviation about the median for the following data:

Solution:

Question 3.
Calculate the mean deviation about the median age for the age distribution of 100 persons given below:

[Hint: Convert the given data into a continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]
Solution:

N = 100
Therefore class is 35.5 – 40.5
∴ \(\frac {N}{2}\) = 50
∴ Mean Deviation (M) = \(\frac {735}{100}\) = 7.35