Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.1
Find the mean deviation about the mean for the data in questions 1 and 2.
Question 1. 4, 7, 8, 9, 10, 12, 13, 17
Solution:
Given observations are = 4, 7, 8, 9, 10, 12, 13, 17
Therefore, Total Number of observations = n = 8
Step 1: Calculating mean(a) for given data about which we have to find the mean deviation.
Mean(a) = ∑(all data points) / Total number of data points
= (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17) / 8
= 80 / 8
a = 10
Step 2: Finding deviation of each data point from mean(x_{i }– a)
x_{i} 4 7 8 9 10 12 13 17 x_{i} – a 4 – 10 = -6 7 – 10 = -3 8 – 10 = -2 9 – 10 = -1 10 – 10 = 0 12 – 10 = 2 13 – 10 = 3 17 – 10 = 7 Step 3: Taking absolute value of deviations, we get
|x_{i} – a| = 6, 3, 2, 1, 0, 2, 3, 7
Step 4: Required mean deviation about the mean is
M.D (a) = ∑ (|x_{i} – a|) / n
= (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7) / 8
= 24 / 8
= 3
So, mean deviation for given observations is 3
Question 2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:
Given observations are = 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Therefore, Total Number of observations = n = 10
Step 1: Calculating mean(a) for given data about which we have to find the mean deviation.
Mean(a) = ∑(all data points)/ Total number of data points
= (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/ 10
= 500 / 10
a = 50
Step2 : Finding deviation of each data point from mean(x_{i }– a)
x_{i} 38 70 48 40 42 55 63 46 54 44 x_{i} – a 38 – 50 = -12 70 – 50 = 20 48 – 50 = -2 40 – 50 = -10 42 – 50 = -8 55 – 50 = 5 63 – 50 = 13 46 – 50 = -4 54 – 50 = 4 44 – 50 = -6 Step 3: Taking absolute value of deviations, we get
|x_{i} – a| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6
Step 4: Required mean deviation about the mean is
M.D(a) = ∑(|x_{i} – a|) / n
= (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6) / 10
= 84 / 10
= 8.4
So, mean deviation for given observations is 8.4
Find the mean deviation about the median for the data in questions 3 and 4.
Question 3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:
Given observations are = 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Therefore, Total Number of observations = n = 12
Step 1: Calculating median(M) for given data about which we have to find the mean deviation.
To calculate median for given data we have to arrange observations in ascending order
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
As number of observations are even median is given by following formula
Median = [(n/2)^{th }observation+ ((n/2) + 1)^{th }observation] / 2
(n/2)^{th} observation = 12 / 2 = 6^{th }observation= 13
((n/2)+1)^{th} observation = (12/2) + 1= 7^{th}observation = 14
Therefore, Median = (13 + 10) / 2 = 13.5
Step 2: Finding deviation of each data point from median(x_{i} – M)
x_{i} 10 11 11 12 13 13 14 16 16 17 17 18 x_{i} -M 10 – 13.5 = -3.5 11 – 13.5 = -2.5 11 – 13.5 = -2.5 12 – 13.5 = -1.5 13 – 13.5 = -0.5 13 – 13.5 = 0.5 14 – 13.5 = 0.5 16 – 13.5 = 2.5 16 – 13.5 = 2.5 17 – 13.5 = 3.5 17 – 13.5 = 3.5 18 – 13.5 = 4.5 Step 3: Taking absolute values of deviations we get
|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
Step 4: Required mean deviation about median is given by
M.D (M) = ∑(|x_{i} – M|) / n
= (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
= 28 / 12
= 2.33
So, mean deviation about median for given observations is 2.33
Question 4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:
Given observations are = 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Therefore, Total Number of observations = n = 10
Step 1: Calculating median(M) for given data about which we have to find the mean deviation.
To calculate median for given data we have to arrange observations in ascending order
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
As number of observations are even median is given by following formula
Median = [(n/2)^{th} observation+ ((n/2) + 1)^{th} observation] / 2
(n/2)^{th }observation = 10 / 2 = 5^{th} observation= 46
((n/2) + 1)^{th }observation = (10/2) + 1 = 6^{th} observation = 49
Therefore, Median = (46 + 49) / 2 = 47.5
Step 2: Finding deviation of each data point from median(x_{i }-M)
x_{i} 36 42 45 46 46 49 51 53 60 72 x_{i} – M -11.5 -5.5 -2.5 -1.5 -1.5 1.5 3.5 5.5 12.5 24.5 Step 3: Taking absolute values of deviations we get
|x_{i }– M| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Step 4: Required mean deviation about median is given by
M.D (M) = ∑(|x_{i} – M|) / n
= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5) / 10
= 70 / 10
= 7
So, mean deviation about median for given observations is 7
Find the mean deviation about the mean for the data in questions 5 and 6.
Question 5.
x_{i} | 5 | 10 | 15 | 20 | 25 |
f_{i} | 7 | 4 | 6 | 3 | 5 |
Solution:
Given table data is of discrete frequency distribution as
we have n = 5 distinct values(x_{i}) along with their frequencies(f_{i}).
Step 1: Let us make a table of given data and append other columns after calculation
x_{i} f_{i} x_{i }* f_{i} |x_{i }– a| f_{i }* |x_{i }– a| 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 Total 25 350 158 Step2 : Now to find the mean we have to first calculate the sum of given data
N = ∑ f_{i} = (7 + 4 + 6 + 3 + 5) = 25
∑ x_{i }* f_{i} = (35 + 40 + 90 + 60 + 125) = 350
Step 3: Find the mean using following formula
Mean (a) = ∑(x_{i} * f_{i})/ N = 350 / 25 = 14
Step 4: Using above mean find absolute values of deviations
from mean i.e |x_{i }– a| and also find values of f_{i} *|xi -a| column.
Now,
∑ f_{i} *|x_{i} – a| = (63 + 16 + 6 + 18 + 55) = 158
Using formula, M.D (a) = ∑(f_{i} * |x_{i} – a|)/ N
= 158 / 25
= 6.32
So, mean deviation for given observations is 6.32
Question 6.
x_{i} | 10 | 30 | 50 | 70 | 90 |
f_{i} | 4 | 24 | 28 | 16 | 8 |
Solution:
Given table data is of discrete frequency distribution as
we have n = 5 distinct values(x_{i}) along with their frequencies(f_{i}).
Step 1: Let us make a table of given data and append other columns after calculation
x_{i} f_{i} x_{i }* f_{i} |x_{i }– a| f_{i }* |x_{i }– a| 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 Total 80 4000 1280 Step 2: Now to find the mean we have to first calculate the sum of given data. From the above table,
N = ∑(f_{i}) = 80 & ∑(x_{i} * f_{i}) = 4000
Step 3: Find the mean using following formula
Mean (a) = ∑ (x_{i }* f_{i})/ N = 4000 / 80 = 50
Step 4: Using above mean find absolute values of deviations from mean i.e |x_{i }– a| and also find values
of f_{i} * |x_{i} -a| column.
Now, from above table,
∑ f_{i} *|x_{i} – a| = (160 + 480 + 0 + 320 + 320) = 1280
Using formula, M.D (a) = ∑ f_{i }*|x_{i} -a| / N
= 1280 / 80
= 16
So, mean deviation for given observations is 16
Find the mean deviation about the median for the data in question 7 and 8.
Question 7.
x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |
f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |
Solution:
For the given discrete frequency distribution we have to find a mean deviation about the median.
Step 1: Let us make a table of given data and append another column of cumulative frequencies.
The given observations are already in ascending order.
x_{i} f_{i} C.F |x_{i }– M| f_{i }* |x_{i }– M| 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48 Total 26 84 Step 2: Identifying the observation whose cumulative frequency is equal
to or just greater than N / 2 and then Finding the median.
N = ∑f_{i }= 26 is even. We divide N by 2. Thus, 26/2 = 13
The cumulative frequency for greater than 13 is 14, for which corresponding observation is 7
Median = [(N/2)^{th }observation + ((N/2) + 1)^{th} observation] / 2
= (13^{th} observation + 14^{th} observation) / 2
= (7 + 7) / 2 = 7
Step 3: Now, find absolute values of the deviations from median,
i.e., |x_{i }– M| and f_{i }* |x_{i }– M| as shown in above table.
∑ f_{i} * |x_{i} – M| = (16 + 4 + 6 + 10 + 48) = 84
Using formula, M.D (M) = ∑ (f_{i }* |x_{i} – M|)/ N
= 84 / 26
= 3.23
So, mean deviation about median for given observations is 3.23
Question 8.
x_{i} | 15 | 21 | 27 | 30 | 35 |
f_{i} | 3 | 5 | 6 | 7 | 8 |
Solution:
For the given discrete frequency distribution we have to find a mean deviation about the median.
Step 1: Let us make a table of given data and append another column of cumulative frequencies.
The given observations are already in ascending order.
x_{i} f_{i} C.F |x_{i }– M| f_{i }* |x_{i }– M| 15 3 3 15 45 21 5 8 9 45 27 6 14 3 18 30 7 21 0 0 35 8 29 5 40 Total 29 148 Step 2: Identifying the observation whose cumulative frequency is equal to or
just greater than N / 2 and then Finding the median.
Here, N = ∑f_{i} = 29 is odd we divide N by 2. Thus, 29 / 2 = 14.5
The cumulative frequency for greater than 14.5 is 21, for which corresponding observation is 30
Therefore, Median = [(N/2)^{th} observation+ ((N/2) + 1)^{th} observation] / 2
= (15^{th} observation + 16^{th} observation) / 2
= (30 + 30)/2 = 30
Step 3: Now, find absolute values of the deviations from median,
i.e., |x_{i }– M| and f_{i} * |x_{i} – M| as shown in above table .
∑f_{i} * |x_{i} – M| = 148
Using formula, M.D (M) = ∑(f_{i }* |x_{i} – M|) / N
= 148 / 29
= 5.1
So, mean deviation about median for given observations is 5.1
Find the mean deviation about the mean for the following data in questions 9 and 10.
Question 9.
Income per day in Rs. |
0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |
Number Of Persons |
4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
Solution:
Given data is in continuous intervals along with their frequencies so it is in the
continuous frequency distribution. In this case, we assume that the frequency
in each class is centered at its mid-point.
Step 1: So, we find a midpoint for each interval.
Then we append other columns similar to discrete frequency distribution.
Income per
day in Rs.
Number Of
Person f_{i}
Midpoints
x_{i}
f_{i}x_{i} |x_{i }– a| f_{i }* |x_{i }– a| 0-100 4 50 200 308 1232 100-200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1160 700-800 3 750 2250 392 1176 Total 50 17900 7896 Step 2: Finding the sum of frequencies f_{i}‘s and sum of f_{i}x_{i}‘s
N = ∑f_{i }= 50
∑f_{i}x_{i }= 17900
Then mean of given data is given by
a = ∑ (f_{i}x_{i})/ N
= 17900 / 50
a = 358
Step 3: Computing sum of column f_{i} * |x_{i} – a|
∑f_{i} * |x_{i} – a| = 7896
Thus mean deviation about mean is given by
M.D (a) = ∑f_{i }* |x_{i} – a| / N
= 7896 / 50
= 157.92
Therefore, mean deviation about mean for given data is 157.92
Question 10.
Height in cms |
95 – 105 | 105 – 115 | 115 – 125 | 125 – 135 | 135 – 145 | 145 – 155 |
Number of boys |
9 | 13 | 26 | 30 | 12 | 10 |
Solution:
Given data is in the continuous frequency distribution.
Step 1: We find a midpoint for each interval and then append other columns.
Height in
cms
Number of
boys
Midpoints
x_{i}
f_{i}x_{i} |x_{i }– a| f_{i }* |x_{i }– a| 95-105 9 100 900 25.3 227.7 105-115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247 Total 100 12530 1128.8 Step 2: Finding the sum of frequencies f_{i}‘s and the sum of f_{i}x_{i}‘s
N = ∑f_{i }= 100
∑ f_{i}x_{i} = 17900
Then mean of given data is given by
a = ∑(f_{i}x_{i}) / N
= 12530 / 100
a = 125.3
Step 3: Computing sum of column f_{i }* |x_{i} – a|
∑f_{i }* |x_{i} – a| = 1128.8
Thus mean deviation about mean is given by
M.D(a) = ∑(f_{i} * |x_{i} – a|)/ N
= 1128.8 / 100
= 11.288
Therefore, the mean deviation about mean for given data is 11.288
Find the mean deviation about the median for the following data in questions 11 and 12.
Question 11.
Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 |
Number of Girls |
6 | 8 | 14 | 16 | 4 | 2 |
Solution:
Given data is in the continuous frequency distribution &
here the only difference is that we have to calculate the median.
Step 1: First we have to compute cumulative frequencies, then we find a midpoint for each interval and then append other columns.
The data is already arranged in ascending order
Marks Number of
Girls f_{i}
C.F Midpoints
x_{i}
|x_{i }– M| f_{i }* |x_{i }– M| 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 Total 50 517.1 Step 2: First identifying the interval in which median lies and then applying the formula to compute median
The class interval whose cumulative frequency is greater than equal to N/2 = 25 is 20 – 30.
So, 20 – 30 is the median class.
Then applying the formula
Median (M) = l + {[(N / 2) – C] / f} * h
where, l = lower limit of the median class
h = width of median class
N = sum of frequencies
C = cumulative frequency of the class just preceding the median class
Therefore,
M = 20 + {[(25 – 14) / 14] * 10}
M = 27.85
Step 3: Finding absolute values of the deviations from median as shown in table
computing sum of column f_{i} * |x_{i} – M|
∑f_{i} * |x_{i} – M| = 517.1
Thus, mean deviation about median is given by
M.D(M) = ∑f_{i} * |x_{i} – M| / N
= 517.1 / 50
= 10.34
So, mean deviation about median for given observations is 10.34
Question12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age (in years) |
16 – 20 | 21 – 25 | 26 – 30 | 31 – 35 | 36 – 40 | 41 – 45 | 46 – 50 | 51 – 55 |
Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
Solution:
Converting the given data into continuous frequency distribution by subtracting 0.5
from the lower limit and adding 0.5 to the upper limit of each class interval.
Check if data is arranged in ascending order.
Step 1: Finding midpoints and C.Fs and then appending other columns
Age
(in years)
Number
f_{i}
C.F Midpoints
x_{i}
|x_{i }– M| f_{i }* |x_{i} – M| 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 71 43 5 60 45.5-50.5 16 95 48 10 160 50.5-55.5 9 100 53 15 135 Total 100 735 Step 2: First identifying the interval in which median lies and then applying the formula to compute median
The class interval whose cumulative frequency is greater than equal to N/2 = 50 is 35.5-40.5.
So, 35.5-40.5 is the median class.
Then applying the formula
Median (M) = l + {[(N / 2) – C] / f} * h
where l = lower limit of the median class = 35.5
h = width of median class = 5
N = sum of frequencies = 100
C = cumulative frequency of the class just preceding the median class = 37
f = frequency = 26
Therefore,
M = 35.5 + {[(50 – 37) / 26] * 5}
M = 38
Step 3: Finding absolute values of the deviations from median as shown in table
computing sum of column f_{i} * |x_{i }– M|
∑f_{i} * |x_{i} – M| = 735
Thus, mean deviation about median is given by
M.D(M) = ∑f_{i} * |x_{i }– M| / N
= 735 / 100
= 7.35
So, mean deviation about median for given observations is 7.35