Class 11  Maths  Statistics
Exercise 15.1
Question 1:
Find the mean deviation about the mean for the data
4, 7, 8, 9, 10, 12, 13, 17
Answer:
The given data is
4, 7, 8, 9, 10, 12, 13, 17
Mean of the data, X = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/10 = 80/10 = 8
The deviations of the respective observations from the mean X i.e. x_{i} – X are
–6, – 3, –2, –1, 0, 2, 3, 7
The absolute values of the deviations, i.e. x_{i} – X, are
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
MD(X) = (Σ_{i=1}^{8}x_{i} – X)/8 = (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7)/8 = 24/8 = 3
Question 2:
Find the mean deviation about the mean for the data
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Answer:
The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the data, X = (38 + 70 + 48 + 40 + 42 + 55 + 63 + 46 + 54 + 44)/10 = 500/10 = 50
The deviations of the respective observations from the mean X i.e. x_{i} – X are
–12, 20, –2, –10, –8, 5, 13, –4, 4, –6
The absolute values of the deviations, i.e. x_{i} – X, are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is
MD(X) = (Σ_{i=1}^{10}x_{i} – X)/10 = (12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6)/10 = 84/10 = 8.4
Question 3:
Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Answer:
The given data is
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Median, M = [(12/2)^{th} observation + (12/2 + 1)^{th} observation]/2
= [6^{th} observation + 7^{th} observation]/2
= (13 + 14)/2
= 27/2 = 13.5
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
MD(X) = (Σ_{i=1}^{12}x_{i} – M)/12
= (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
Question 4:
Find the mean deviation about the median for the data.
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Answer:
The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, the numbers of observations are 10, which is even.
Arranging the data in ascending order, we obtain
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Median, M = [(10/2)^{th} observation + (10/2 + 1)^{th} observation]/2
= [5^{th} observation + 6^{th} observation]/2
= (46 + 49)/2
= 95/2 = 47.5
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
The required mean deviation about the median is
MD(X) = (Σ_{i=1}^{10}x_{i} – M)/10
= (11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5)/10 = 70/10 = 7
Question 5:
Find the mean deviation about the mean for the data.
x_{i} 
5 
10 
15 
20 
25 
f_{i} 
7 
4 
6 
3 
5 
Answer:
N = Σ_{i=1}^{5} f_{i} = 25
Σ_{i=1}^{5} f_{i} x_{i} = 350
Now, Mean x = Σ_{i=1}^{5} f_{i} x_{i}/ Σ_{i=1}^{5} f_{i}
= 350/25
= 14
So, MD(x) = (1/N) * Σ_{i=1}^{5} f_{i}x_{i} – x= 158/25 = 6.32
Question 6:
Find the mean deviation about the mean for the data.
x_{i} 
10 
30 
50 
70 
90 
f_{i} 
4 
24 
28 
16 
8 
Answer:
N = Σ_{i=1}^{5} f_{i} = 80
Σ_{i=1}^{5} f_{i} x_{i} = 4000
Now, Mean x = Σ_{i=1}^{5} f_{i} x_{i}/ Σ_{i=1}^{5} f_{i}
= 4000/80
= 50
So, MD(x) = (1/N) * Σ_{i=1}^{5} f_{i} x_{i} – x= 1280/80 = 16
Question 7:
Find the mean deviation about the median for the data.
x_{i} 
5 
7 
9 
10 
12 
15 
y_{i} 
8 
6 
2 
2 
2 
6 
Answer
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we get the
following table:
Here, N = 26, which is even.
Median is the mean of 13^{th} and 14^{th} observations.
Both of these observations lie in the cumulative frequency 14,
for which the corresponding observation is 7.
So, Median = (13^{th} observations + 14^{th} observations)/2
= (7 + 7)/2 = 14/2 = 7
The absolute values of the deviations from median, i.e. x_{i}  Mare
Σ_{i=1}^{6} f_{i} = 26
Σ_{i=1}^{6} f_{i} x_{i} – M= 84
So, MD(X) = (1/N) * Σ_{i=1}^{6} f_{i} x_{i} – M= 84/26 = 3.23
Question 8:
Find the mean deviation about the median for the data.
x_{i} 
15 
21 
27 
30 
35 
y_{i} 
3 
5 
6 
7 
8 
Answer
The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we get the
following table:
Here, N = 29, which is odd.
Now, Median = (29 + 1)/2 ^{th} observations = 15^{th} observations
This observation lies in the cumulative frequency 21,
for which the corresponding observation is 30.
So, Median = 30
The absolute values of the deviations from median, i.e. x_{i}  Mare
Σ_{i=1}^{5} f_{i} = 29
Σ_{i=1}^{5} f_{i} x_{i} – M= 148
So, MD(X) = (1/N) * Σ_{i=1}^{5} f_{i} x_{i} – M= 148/29 = 5.1
Question 9:
Find the mean deviation about the mean for the data:
Income per day 
Number of persons 
0100 
4 
100200 
8 
200300 
9 
300400 
10 
400500 
7 
500600 
5 
600700 
4 
700800 
3 
Answer:
The following table is formed:
Here, N = Σ_{i=1}^{8} f_{i} = 50
Σ_{i=1}^{8} f_{i} x_{i} = 17900
Now, x = (1/N) * Σ_{i=1}^{8} f_{i} x_{i} = 17900/50 = 358
So, MD(x) = (1/N) * Σ_{i=1}^{8} f_{i} x_{i} – x= 7896/50 = 157.92
Question 10:
Find the mean deviation about the mean for the data:
Heights in cms 
Number of boys 
95105 
9 
105115 
13 
115125 
26 
125135 
30 
135145 
12 
145155 
10 
Answer:
The following table is formed:
Here, N = Σ_{i=1}^{6} f_{i} = 100
Σ_{i=1}^{6} f_{i} x_{i} = 12530
Now, x = (1/N) * Σ_{i=1}^{6} f_{i} x_{i} = 12530/100 = 125.3
So, MD(x) = (1/N) * Σ_{i=1}^{6} f_{i} x_{i} – x= 1128.8/100 = 11.28
Question 11:
Find the mean deviation about median for the following data:
Marks 
Number of girls 
010 
6 
1020 
8 
2030 
14 
3040 
16 
4050 
4 
5060 
2 
Answer:
The following table is formed:
The class interval containing the (N/2)^{th} or 25^{th} item is 20 – 30.
Therefore, 20 – 30 is the median class.
We know that
Median M = l + {(N/2  C)/f} * h
Here, l = 20, C = 14, f = 14, h = 10, and N = 50
Median, M = 20 + {(25  14)/14} * 10
=> M = 20 + 110/14
=> M = 20 + 7.85
=> M = 27.85
Thus, mean deviation about the median is given by,
MD(X) = (1/N) * Σ_{i=1}^{6} f_{i} x_{i} – M= 517.1/50 = 10.34
Question 12:
Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age 
1620 
2125 
2630 
3135 
3640 
4145 
4650 
5155 
Number 
5 
6 
12 
14 
26 
12 
16 
9 
Answer:
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 (N/2)^{th} or 50^{th} item is 35.5 – 40.5
Therefore, 35.5 – 40.5 is the median class.
We know that
Median M = l + {(N/2  C)/f} * h
Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
Median, M = 35.5 + {(50  37)/26} * 5
=> M = 35.5 + (13 * 5)/26
=> M = 35.5 + 2.5
=> M = 38
Thus, mean deviation about the median is given by,
MD(X) = (1/N) * Σ_{i=1}^{6} f_{i} x_{i} – M= 735/100 = 7.35
Exercise 15.2
Question 1:
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Answer:
Given data is: 6, 7, 10, 12, 13, 4, 8, 12
Mean x = Σ_{i=1}^{8} x_{i} = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
The following table is obtained:
Now, varience σ^{2} = (1/N) * Σ_{i=1}^{8} (x_{i} – x)^{2} = 74/8 = 9.25
Question 2:
Find the mean and variance for the first n natural numbers.
Answer:
The mean of first n natural numbers is calculated as follows:
Mean = Sum of all observations/Number of observations
=> Mean, X = {n(n + 1)/2}/n = (n + 1)/2
Now, varience σ^{2} = (1/n) * Σ_{i=1}^{n} (x_{i} – X)^{2}
= (1/n) * Σ_{i=1}^{n} x_{i}^{2} – (1/n) * Σ_{i=1}^{n} 2 * {(n + 1)/2}x_{i} + (1/n) * Σ_{i=1}^{n} {(n + 1)/2}^{2}
= (1/n){n(n + 1)(2n + 1)/6} – {(n + 1)/n}[n(n + 1)/2] + {(n + 1)^{2}/4n} * n
= (n + 1)(2n + 1)/6 – (n + 1)^{2}/2 + (n + 1)^{2}/4
= (n + 1)(2n + 1)/6 – (n + 1)^{2}/4
= (n + 1)[(4n + 2 – 3n  3)/12]
= {(n + 1)(n  1)}/12
= (n^{2}  1)/12
Question 3:
Find the mean and variance for the first 10 multiples of 3
Answer:
The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
Mean x = Σ_{i=1}^{10} x_{i} = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
The following table is obtained:
Now, varience σ^{2} = (1/N) * Σ_{i=1}^{10} (x_{i} – x)^{2} = 742.5/10 = 74.25
Question 4:
Find the mean and variance for the data
x_{i} 
6 
10 
14 
18 
24 
28 
30 
f_{i} 
2 
4 
7 
12 
8 
4 
3 
Answer:
The data is obtained in tabular form as follows:
Here, N = 40, Σ_{i=1}^{7} f_{i} x_{i} = 760
Mean x = (1/N) * Σ_{i=1}^{7} f_{i} x_{i} = 760/40 = 19
Now, varience σ^{2} = (1/N) * Σ_{i=1}^{7} (x_{i} – x)^{2} = 1736/19 = 43.4
Question 5:
Find the mean and variance for the data
x_{i} 
92 
93 
97 
98 
102 
104 
109 
f_{i} 
3 
2 
3 
2 
6 
3 
3 
Answer:
The data is obtained in tabular form as follows:
Here, N = 22, Σ_{i=1}^{7} f_{i} x_{i} = 2200
Mean x = (1/N) * Σ_{i=1}^{7} f_{i} x_{i} = 2200/22 = 100
Now, varience σ^{2} = (1/N) * Σ_{i=1}^{7} (x_{i} – x)^{2} = 640/22 = 29.09
Question 6:
Find the mean and standard deviation using shortcut method.
x_{i} 
60 
61 
62 
63 
64 
65 
66 
67 
68 
f_{i} 
2 
1 
12 
29 
25 
12 
10 
4 
5 
Answer:
The data is obtained in tabular form as follows:
Mean x = A + [Σ_{i=1}^{9} f_{i} y_{i} /N] * h = 64 + 0/100 * 1 = 64 + 0 = 64
Now, varience σ^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{9} f_{i} y_{i}^{2} – (Σ_{i=1}^{9} f_{i} y_{i})^{2}]
= (1/100^{2})[100 * 286  0]
= 286/100
= 2.86
So, standard deviation (σ) = √(2.86) = 1.69
Question 7:
Find the mean and variance for the following frequency distribution.
Classes 
030 
3060 
6090 
90120 
120150 
150180 
180210 
Frequencies 
2 
3 
5 
10 
3 
5 
2 
Answer:
The data is obtained in tabular form as follows:
Mean x = A + [Σ_{i=1}^{7} f_{i} y_{i} /N] * h = 105 + (2/30) * 30
= 105 + 2 = 107
Now, varience σ^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{7} f_{i} y_{i}^{2} – (Σ_{i=1}^{7} f_{i} y_{i})^{2}]
= (30^{2}/30^{2})[30 * 76  2^{2}]
= 2284  4
= 2276
Question 8:
Find the mean and variance for the following frequency distribution.
Classes 
010 
1020 
2030 
3040 
4050 
Frequencies 
5 
8 
15 
16 
6 
Answer:
The data is obtained in tabular form as follows:
Mean x = A + [Σ_{i=1}^{5} f_{i} y_{i} /N] * h = 25 + (10/50) * 10
= 25 + 2 = 27
Now, varience σ^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{5} f_{i} y_{i}^{2} – (Σ_{i=1}^{5} f_{i} y_{i})^{2}]
= (10^{2}/50^{2})[50 * 68  10^{2}]
= (3400  100)/25
= 3300/25
= 132
Question 9:
Find the mean, varience and standard deviation using shortcut method.
Height in cms 
7075 
7580 
8085 
8590 
9095 
95100 
100105 
105110 
110115 
No. of children 
3 
4 
7 
7 
15 
9 
6 
6 
3 
Answer:
The data is obtained in tabular form as follows:
Mean x = A + [Σ_{i=1}^{9} f_{i} y_{i} /N] * h = 92.5 + (6/60) * 5
= 92.5 + 0.5
= 93
Now, varience σ^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{9} f_{i} y_{i}^{2} – (Σ_{i=1}^{9} f_{i} y_{i})^{2}]
= (5^{2}/60^{2})[60 * 254  6^{2}]
= (25/3600) * 15204
= 105.58
So, standard deviation (σ) = √(105.58) = 10.27
Question 10:
The diameters of circles (in mm) drawn in a design are given below:
Diameter 
3336 
3740 
4144 
4548 
4952 
No. of children 
15 
17 
21 
22 
25 
Calculate the standard deviation and mean diameter of the circles.
[Hint: First make the data continuous by making the classes as 32.536.5, 36.540.5, 40.544.5, 44.5  48.5, 48.5  52.5 and then proceed.]
Answer:
Here, N = 100, h = 4
The data is obtained in tabular form as follows:
Let the assumed mean, A, be 42.5
Mean x = A + [Σ_{i=1}^{9} f_{i} y_{i} /N] * h = 42.5 + (25/100) * 4
= 42.5 + 1 = 43.5
Now, varience σ^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{9} f_{i} y_{i}^{2} – (Σ_{i=1}^{9} f_{i} y_{i})^{2}]
= (16/10000)[100 * 199  25^{2}]
= (16/10000)[19900  625]
= (16/10000) * 19275
= 30.84
So, standard deviation (σ) = √(30.84) = 5.55
Exercise 15.3
Question 1:
From the data given below state which group is more variable, A or B?
Marks 
1020 
2030 
3040 
4050 
5060 
6070 
7080 
Group A 
9 
17 
32 
33 
40 
10 
9 
Group B 
10 
20 
30 
25 
43 
15 
7 
Answer:
Firstly, the standard deviation of group A is calculated as follows:
Here, h = 10, N = 150, A = 45
Mean x = A + [Σ_{i=1}^{7} x_{i} /N] * h
= 45 + (6/150) * 10
= 45  0.4
= 44.6
Now, varience σ_{1}^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{7} f_{i} y_{i}^{2} – (Σ_{i=1}^{7} f_{i} y_{i})^{2}]
= (100/22500)[150 * 342 – (6)^{2}]
= (1/225) * 51264
= 227.84
So, standard deviation (σ) = √(227.84) = 15.09
The standard deviation of group B is calculated as follows:
Mean x = A + [Σ_{i=1}^{7} x_{i} /N] * h
= 45 + (6/150) * 10
= 45  0.4
= 44.6
Now, varience σ_{2}^{2} = (h^{2}/N^{2}) * [N * Σ_{i=1}^{7} f_{i} y_{i}^{2} – (Σ_{i=1}^{7} f_{i} y_{i})^{2}]
= (100/22500)[150 * 366 – (6)^{2}]
= (1/225) * 54864
= 243.84
So, standard deviation (σ) = √(243.84) = 15.61
Since the mean of both the groups is same, the group with greater standard deviation will be
more variable.
Thus, group B has more variability in the marks.
Question 2:
From the prices of shares X and Y below, find out which is more stable in value:
X 
35 
54 
52 
53 
56 
58 
52 
50 
51 
49 
Y 
108 
107 
105 
105 
106 
107 
104 
103 
104 
101 
Answer:
The prices of the shares X are:
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10
Mean x = Σ_{i=1}^{10} x_{i} /N
= 510/10
= 51
The following table is obtained corresponding to shares X.
Now, varience σ_{1}^{2} = (1/N) * Σ_{i=1}^{10} (x_{i} – x)^{2} = 350/10 = 35
So, Standard deviation (σ_{1}) = √35 = 5.91
CV (Shares X) = (σ_{1}/x) * 100 = (5.91/51) * 100 = 11.58
The prices of share Y are:
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
Mean y = Σ_{i=1}^{10} y_{i} /N
= 1050/10
= 105
The following table is obtained corresponding to shares Y.
Now, varience σ_{2}^{2} = (1/N) * Σ_{i=1}^{10} (y_{i} – y)^{2} = 40/10 = 4
So, Standard deviation (σ_{2}) = √4 = 2
CV (Shares X) = (σ_{2}/y) * 100 = (2/105) * 100 = 1.9 < 11.58
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.
Question 3:
An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

Firm A 
Firm B 
No. of wages earner 
586 
648 
Mean of monthly wages 
Rs 5253 
Rs 5253 
Variance of the distribution of wages 
100 
121 
(i) Which firm A or B pays larger amount as monthly wages?
(ii) Which firm, A or B, shows greater variability in individual wages?
Answer:
(i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
So, total amount paid = Rs 5253 * 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
So, total amount paid = Rs 5253 * 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm
B are more than the number of wage earners in firm A.
(ii) Variance of the distribution of wages in firm A (σ_{1}^{2}) = 100
Standard deviation of the distribution of wages in firm
A (σ_{1}) = √100 = 10
Variance of the distribution of wages in firm B (σ_{2}^{2}) = 121
Standard deviation of the distribution of wages in firm
B (σ_{2}) = √121 = 11
The mean of monthly wages of both the firms is same i.e., 5253.
Therefore, the firm with greater standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.
Question 4:
The following is the record of goals scored by team A in a football session:
No. of goal scored 
0 
1 
2 
3 
4 
No. of matches 
1 
9 
7 
5 
3 
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals.
Find which team may be considered more consistent?
Answer:
The mean and the standard deviation of goals scored by team A are calculated as follows:
Mean = Σ_{i=1}^{5} f_{i} x_{i} / Σ_{i=1}^{5} f_{i}
= 50/2
= 2
Thus, the mean of both the teams is same.
Now, Standard deviation of team A is
σ_{1} = (1/N)√[N Σf_{i} x_{i}^{2} – (Σ f_{i} x_{i})^{2}]
= (1/25)√[25 * 130 – (50)^{2}]
= √750/25
= 27.38/25
= 1.09
The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2.
Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.
Question 5:
The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Σ_{i=1}^{50} x_{i} = 212, Σ_{i=1}^{50} x_{i}^{2} = 902.8, Σ_{i=1}^{50} y_{i} = 261, Σ_{i=1}^{50} y_{i}^{2} = 1457.6
Which is more varying, the length or weight?
Answer:
Given, Σ_{i=1}^{50} x_{i} = 212, Σ_{i=1}^{50} x_{i}^{2} = 902.8
Here, N = 50
So, Mean, x = Σ_{i=1}^{50} x_{i}/N = 212/50 = 4.24
Variance σ_{1}^{2} = (1/N)* Σ_{i=1}^{50} (x_{i }– x)^{2}
= (1/50) * Σ_{i=1}^{50} (x_{i }– 4.24)^{2}
= (1/50) * Σ_{i=1}^{50} [x_{i }^{2} – 8.48 x_{i} + 17.97]
= (1/50) * [Σ_{i=1}^{50} x_{i }^{2} – 8.48 Σ_{i=1}^{50} x_{i} + 17.97 * 50]
= (1/50) * [902.8 – 8.48 * 212 + 898.5]
= (1/50) * [1801.3 – 1797.76]
= 3.54/50
= 0.07
So, standard deviation (σ_{1}) = √0.07 = 0.26
CV (Length) = (standard deviation/Mean) * 100
= (0.26/4.24) * 100
= 6.13
Again, given Σ_{i=1}^{50} y_{i} = 261, Σ_{i=1}^{50} y_{i}^{2} = 1457.6
Here, N = 50
So, Mean, y = Σ_{i=1}^{50} y_{i}/N = 261/50 = 5.22
Variance σ_{2}^{2} = (1/N)* Σ_{i=1}^{50} (y_{i }– y)^{2}
= (1/50) * Σ_{i=1}^{50} (y_{i }– 5.22)^{2}
= (1/50) * Σ_{i=1}^{50} [y_{i }^{2} – 10.44 y_{i} + 27.24]
= (1/50) * [Σ_{i=1}^{50} y_{i }^{2} – 10.44 Σ_{i=1}^{50} y_{i} + 27.24 * 50]
= (1/50) * [1457.6 – 10.44 * 261 + 1362]
= (1/50) * [2819.6 – 2724.84]
= 94.76/50
= 1.89
So, standard deviation (σ_{2}) = √1.89 = 1.37
CV (Length) = (standard deviation/Mean) * 100
= (1.37/5.22) * 100
= 26.24
Thus, C.V. of weights is greater than the C.V. of lengths.
Therefore, weights vary more than the lengths.
Miscellaneous Exercise of Chapter 12
Question 1:
The mean and variance of eight observations are 9 and 9.25, respectively.
If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Answer:
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
Mean, x = (6 + 7 + 10 + 12 + 12 + 13 + x + y)/8 = 9
=> 60 + x + y = 9 * 8
=> 60 + x + y = 72
=> x + y = 72 – 69
=> x + y = 12 ………………1
Variance = (1/n)* Σ_{i=1}^{8} (x_{i }– x)^{2} = 9.25
=> 9.25 = (1/8)[(3)^{2} + (2)^{2} + 1^{2} + 3^{2} + 3^{2} + 4^{2} + x^{2} + y^{2} – 2 * 9 *(x + y) + 2 * 9^{2}]
=> 9.25 = (1/8)[9 + 4 + 1 + 9 + 9 + 16 + x^{2} + y^{2} – 18 * 12 + 162]
=> 9.25 = (1/8)[48 + x^{2} + y^{2} – 216 + 162]
=> 9.25 = (1/8)[x^{2} + y^{2} – 6]
=> x^{2} + y^{2} – 6 = 74
=> x^{2} + y^{2} = 80 ………………..2
From equation 1, we get
x^{2} + y^{2} + 2xy = 144 ...........3
From equation 2 and 3, we get
80 + 2xy = 144
=> 2xy = 64 ......................4
Subtracting equation 4 from 2, we get
x ^{2} + y^{2} – 2xy = 80 – 64 = 16
=> (x – y)^{2} = 16
=> x – y = ± 4 ...............5
Therefore, from equation 1 and 5, we get
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.
Question 2:
The mean and variance of 7 observations are 8 and 16, respectively.
If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Answer:
Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.
Mean, x = (2 + 4 + 10 + 12 + 14 + x + y)/7 = 8
=> 42 + x + y = 8 * 7
=> 42 + x + y = 56
=> x + y = 56 – 42
=> x + y = 14 ………………1
Variance = (1/n)* Σ_{i=1}^{7} (x_{i }– x)^{2} = 16
=> 16 = (1/7)[(6)^{2} + (4)^{2} + 2^{2} + 4^{2} + 6^{2} + x^{2} + y^{2} – 2 * 8 *(x + y) + 2 * 8^{2}]
=> 16 = (1/7)[36 + 16 + 4 + 16 + 36 + x^{2} + y^{2} – 16 * 14 + 128]
=> 16 = (1/7)[108 + x^{2} + y^{2} – 224 + 128]
=> 16 = (1/7)[x^{2} + y^{2} + 12]
=> x^{2} + y^{2} + 12 = 112
=> x^{2} + y^{2} = 100 ………………..2
From equation 1, we get
x^{2} + y^{2} + 2xy = 196 ...........3
From equation 2 and 3, we get
100 + 2xy = 196
=> 2xy = 96 ......................4
Subtracting equation 4 from 2, we get
x ^{2} + y^{2} – 2xy = 100 – 96
=> (x – y)^{2} = 4
=> x – y = ± 2 ...............5
Therefore, from equation 1 and 5, we get
x = 8 and y = 6, when x – y = 2
x = 6 and y = 8, when x – y = –2
Thus, the remaining observations are 6 and 8.
Question 3:
The mean and standard deviation of six observations are 8 and 4, respectively.
If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Answer:
Let the observations be x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, and x_{6}.
It is given that mean is 8 and standard deviation is 4.
Mean, x = (x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6})/6 = 8 ………………1
If each observation is multiplied by 3 and the resulting observations are y_{i}, then
y_{i} = 3x_{i}
i.e. x_{i} = y_{i}/3 for i = 1 to 6
So, New Mean, y = (y_{1} + y_{2} + y_{3} + y_{4} + y_{5} + y_{6})/6 = 8
= 3(x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6})/6
= 3 * 8
= 24
Standard deviation σ = √[(1/n) * Σ_{i=1}^{6} (x_{i }– x)^{2}]
=> 4^{2} = Σ_{i=1}^{6} (x_{i }– x)^{2}/6
=> Σ_{i=1}^{6} (x_{i }– x)^{2} = 96 ………….2
From equation 1 and 2, it can be observed that,
y = 3x
=> x = y/3
Substituting the values of x_{i} and x in equation 2, we obtain
=> Σ_{i=1}^{6} (y_{i }/3 _{ }– y/3)^{2} = 96
=> Σ_{i=1}^{6} (y_{i} _{ }– y)^{2} = 864
Therefore, variance of new observations = 864/6 = 144
Hence, the standard deviation of new observations is √144 = 12
Question 4:
Given that x is the mean and σ^{2} is the variance of n observations x_{1}, x_{2} ... x_{n}.
Prove that the mean and variance of the observations ax_{1}, ax_{2}, ax_{3} ...ax_{n} are ax and a^{2}σ^{2} respectively (a ≠ 0).
Answer:
The given n observations are x_{1}, x_{2} ... x_{n}.
Mean = x
Variance = σ^{2}
So, σ^{2} = (1/n) * Σ_{i=1}^{n} (x_{i }– x)^{2} …………1
If each observation is multiplied by a and the new observations are y_{i}, then
y_{i} = ax_{i}
=> x_{i} = y_{i}/a
So, y = (1/n) * Σ_{i=1}^{n} x_{i }= (1/n) * Σ_{i=1}^{n} ax_{i }= (a/n) * Σ_{i=1}^{n} x_{i } = ax
Therefore, mean of the observations, ax_{1}, ax_{2} ... ax_{n}, is ax.
Substituting the values of x_{i} and x in equation 1, we obtain
σ^{2} = (1/n) * Σ_{i=1}^{n} (y_{i }/a – y/a)^{2 }
=> a^{2} σ^{2} = (1/n) * Σ_{i=1}^{n} (y_{i }– y)^{2}
Thus, the variance of the observations ax_{1}, ax_{2} ... ax_{n}, is a^{2} σ^{2}
Question 5:
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively.
On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted. (ii) If it is replaced by 12.
Answer:
(i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
Mean, x = (1/n) * Σ_{i=1}^{20} x_{i }
=> 10 = (1/20) * Σ_{i=1}^{20} x_{i}
=> Σ_{i=1}^{20} x_{i }= 200
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
Correct mean = correct sum/19 = 192/19 = 10.1
Standard deviation σ = √[(1/n) * Σ_{i=1}^{n} x_{i}^{2} – (Σ_{i=1}^{n} x_{i})^{2 }/n^{2}]
=> 2 = √[(1/n) * incorrect Σ_{i=1}^{n} x_{i}^{2} – (x)^{2}]
=> 2 = √[(1/20) * incorrect Σ_{i=1}^{n} x_{i}^{2} – (10)^{2}]
=> 4 = (1/20) * incorrect Σ_{i=1}^{n} x_{i}^{2} – 100
=> incorrect Σ_{i=1}^{n} x_{i}^{2} = 2080
So, correct Σ_{i=1}^{n} x_{i}^{2} = incorrect Σ_{i=1}^{n} x_{i}^{2} – 8^{2}
= 2080 – 64
= 2016
So, Correct Standard deviation = √[correct Σ x_{i}^{2}/n – (correct mean)^{2}]
= √[2016/19 – (10.1)^{2}]
= √[106.1 – 102.01]
= √4.09
= 2.02
(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
So, Correct sum of observations = 200 – 8 + 12 = 204
Correct mean = correct sum/20 = 204/20 = 10.2
Standard deviation σ = √[(1/n) * Σ_{i=1}^{n} x_{i}^{2} – (Σ_{i=1}^{n} x_{i})^{2 }/n^{2}]
=> 2 = √[(1/n) * incorrect Σ_{i=1}^{n} x_{i}^{2} – (x)^{2}]
=> 2 = √[(1/20) * incorrect Σ_{i=1}^{n} x_{i}^{2} – (10)^{2}]
=> 4 = (1/20) * incorrect Σ_{i=1}^{n} x_{i}^{2} – 100
=> (1/20) * incorrect Σ_{i=1}^{n} x_{i}^{2} = 104
=> incorrect Σ_{i=1}^{n} x_{i}^{2} = 2080
So, correct Σ_{i=1}^{n} x_{i}^{2} = incorrect Σ_{i=1}^{n} x_{i}^{2} – 8^{2 }+ 12^{2}
= 2080 – 64 + 144
= 2160
So, Correct Standard deviation = √[correct Σ x_{i}^{2}/n – (correct mean)^{2}]
= √[2160/20 – (10.2)^{2}]
= √[108 – 104.04]
= √3.96
= 1.98
Question 6:
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Subject 
Mathematics 
Physics 
Chemistry 
Mean 
42 
32 
40.9 
Standard Deviation 
12 
15 
20 
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Answer:
Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) = (Standard Deviation/Mean) * 100
CV (in Mathematics) = (12/42) * 100 = 28.57
CV (in Physics) = (15/32) * 100 = 46.87
CV (in Chemistry) = (20/40.9) * 100 = 48.89
The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is
in Mathematics.
Question 7:
The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively.
Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18.
Find the mean and standard deviation if the incorrect observations are omitted.
Answer:
Number of observations (n) = 100
Incorrect mean (x) = 20
Incorrect standard deviation (σ) = 3
=> 20 = (1/20) * Σ_{i=1}^{100} x_{i}
=> Σ_{i=1}^{100} x_{i} = 20 * 100 = 2000
So, incorrect sum of observations = 2000
Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Correct mean = correct sum/(100  3) = 1940/97 = 20
Standard deviation σ = √[(1/n) * Σ_{i=1}^{n} x_{i}^{2} – (Σ_{i=1}^{n} x_{i})^{2 }/n^{2}]
=> 3 = √[(1/n) * incorrect Σ x_{i}^{2} – (x)^{2}]
=> 3 = √[(1/100) * incorrect Σ x_{i}^{2} – (20)^{2}]
=> 9 = (1/100) * incorrect Σ x_{i}^{2} – 400
=> incorrect Σ x_{i}^{2} = 100(9 + 400)
=> incorrect Σ x_{i}^{2} = 40900
So, correct Σ_{i=1}^{n} x_{i}^{2} = incorrect Σ_{i=1}^{n} x_{i}^{2} – 21^{2 } 21^{2}  18^{2}
= 40900 – 441 – 441  324
= 39694
So, Correct Standard deviation = √[correct Σ x_{i}^{2}/n – (correct mean)^{2}]
= √[39694/97 – (20)^{2}]
= √[409.216 – 400]
= √9.216
= 3.036
.