Referring to the AP Inter 1st Year Maths Study Material Chapter 13 Statistics Exercise 13a Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Statistics Solutions Exercise 13a
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, 48, 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, \(\frac{1}{n} \sum_{i=1}^n x_i\) = x̄ = \(\frac{4+7+8+9+10+12+13+17}{8}=\frac{80}{8}\) = 10
The deviations of the respective observations from the mean x̄ i.e., xi – x̄ are -6, -3, -2, -1, 0, 2, 3, 7.
The required mean deviation about the mean is 8
M.D. (x̄) = \(\frac{\sum_{i=1}^8\left|x_i-\bar{x}\right|}{8}=\frac{6+3+2+1+0+2+3+7}{8}=\frac{24}{8}\) = 3
b) The given data is 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.
Mean of the given data, x̄ = \(\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}\) = 50
The deviations of the respective observations from the mean Xs i.e.,
xi – x̄ are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.
\(\sum_{i=1}^{10}\)|xi – x̄| = 12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6 = 84
The required mean deviation about the mean is
M.D. (x̄) = \(\frac{\sum_{j=1}^{10}\left|x_i-\bar{x}\right|}{10}=\frac{84}{10}\) = 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) Here, the number of observations are 12, which is even.
Arranging the data in ascending order, we have,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
\(\sum_{i=1}^{12}\)|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
∴ The required mean deviation about the median is
M.D. (M) = \(\frac{\sum_{i=1}^{12}\left|x_i-M\right|}{12}=\frac{28}{12}\) = 2.33
b) The given data is 36, 72, 46, 42, 60, 45, 53, 46, 51, 49.
Here, the number of observations is 10, which is even.
Arranging the data in an ascending order, we have
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
\(\sum_{i=1}^{10}\)|xi – M| = 11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5 = 70
Thus, the required mean deviation about the median is
M.D (M) = \(\frac{\sum_{i=1}^{10}\left|x_i-M\right|}{10}=\frac{70}{10}\) = 7
II.
Question 1.
Find the mean deviation about the mean for the following data.
Solution:
We make following table from the given data.
N = \(\sum_{i=1}^5\) fi = 25 and \(\sum_{i=1}^5\)fixi = 350;
∴ x̄ = \(\frac{1}{N_i} \sum_{i=1}^5\) fixi = \(\frac{1}{25}\) × 350 = 14
∴ M.D.(x̄) = \(\frac{1}{2}\) fi |xi – x̄| = \(\frac{1}{25}\) x 158 = 6.32
Solution:
We make following table from the given data.
Question 2.
Find the mean deviation about the median for the following data.
Solution:
The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we have the following table.
Here, N = 26, which is even. Median is the Mean of 13th and 14th observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
∴ Median = \(\frac{\left(13^{\text {th }} \text { observation }+14^{\text {th }} \text { observation }\right)}{2}=\frac{7+7}{2}=\frac{14}{2}\)
Median = 7
\(\sum_{i=1}^6\)fi = 26 and \(\sum_{i=1}^6\)fi|xi – M| = 84;
M.D (M) = \(\frac{1}{N_i} \sum_{i=1}^6\)fi|xi – M| = \(\frac{1}{26\) × 84 = 3.23
Solution:
The given observations are already in ascending order. Adding a column corresponding to cumulative frequencies of the given data, we have the following table.
Here, N = 29, which is odd.
Median = \(\left(\frac{29+1}{2}\right)^{\text {th }}\) observation = 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
M = 30
\(\sum_{i=1}^5\) fi = 29, \(\sum_{i=1}^5\) fi|xi – M| = 148
∴ M.D. (M) = \(\frac{1}{N} \sum_{i=1}^5\)fi|xi – M| = \(\frac{1}{29}\) × 148 = 5.1
III.
Question 1.
Find the mean deviation about the mean for the following data.
Solution:
The following table is formed for the given data.
Here, N = \(\frac{1}{2}\) fi = 50 and fixi = 17900
x = \(\frac{1}{2}\)fixi = \(\frac{1}{50}\) × 17900 = 358
and M.D. (x) = \(\frac{1}{N} \sum_{i=1}^8\) fi|xi – x̄| = \(\frac{7896}{50}\) = 157.92
Solution:
The following table is formed for the given data.
Here, N = \(\sum_{i=1}^6\) fi = 100 and \(\sum_{i=1}^6\) fixi = 12530
x = \(\frac{1}{N_i} \sum_{i=1}^6\)fixi = \(\frac{1}{100}\) × 12530 = 125.3
and M.D (x̄) = \(\frac{1}{N_i} \sum_{i=1}^6\) fi|xi – x̄| = \(\frac{1}{100}\) × 1128.8 = 11.28
Question 2.
Find the mean deviation about median for the folloeing data:
Solution:
The following table is formed for the given data.
The class interval containing the \(\left[\frac{\mathrm{N}}{2}\right]^{\mathrm{th}}\) or 25th item is 20 – 30.
Therefore, 20 – 30 is the median class. It is known that
Median = l + \(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\) × h
Here, 1= 20, C = 14, f = 14, h = 10 and N = 50
Median = 20 + \(\frac{25-14}{14}\) × 10 = 20 + \(\frac{110}{14}\) = 20 + 7.85 = 27.85
Thus, mean deviation about the median is given by,
M.D. (M) = \(\frac{1}{N_i} \sum_{i=1}^6\)fi|xi – x̄| = \(\frac{1}{50}\) × 517.1 = 10.34
Question 3.
Calculate the mean deviation about median age for the age distribution of 100 persons given below.
Solution:
The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.
The class interval containing the \(\left[\frac{\mathrm{N}}{2}\right]^{\text {th }}\) or 50th item is 35.5 – 50.5
Therefore, 35.5 – 40.5 is the median class. It is known that
Median (M) = l + \(\frac{\frac{\mathrm{N}}{2}-\mathrm{C}}{\mathrm{f}}\) × h
Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100
∴ Median = 35.5 + \(\frac{50-37}{26}\) × 5 = 35.5 + \(\frac{13 \times 5}{26}\) = 35.5 + 2.5 = 38
Thus, mean deviation about the median is given by,
M.D. (M) = \(\frac{1}{N_i} \sum_{i=1}^8\)fi|xi – x̄| = \(\frac{1}{100}\) × 735 = 7.35