NCERT Solutions for Maths Class 11th Chapter 15 – Statistics
Exercise 15.1 : Solutions of Questions on Page Number : 360
Q1 :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,
The deviations of the respective observations from the mean are
-6,-3, -2, -1, 0, 2, 3, 7
The absolute values of the deviations, i.e., are
6, 3, 2, 1, 0, 2, 3, 7
The required mean deviation about the mean is
Q2 :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 given data,
The deviations of the respective observations from the mean are
-12, 20,-2, -10, -8, 5, 13, -4, 4,-6
The absolute values of the deviations, i.e. , are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is
Q3 :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
The deviations of the respective observations from the median, i.e. 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,, 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
Q4 :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 number of observations is 10, which is even.
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. 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, 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
Q5 :Find the mean deviation about the mean for the data.
xi | 5 | 10 | 15 | 20 | 25 |
fi | 7 | 4 | 6 | 3 | 5 |
Answer :
x_{i} | f_{i} | x_{i}f_{i} | ||
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 |
25 | 350 | 158 |
Q6 :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 :
X_{i} | f_{i} | X_{i} f_{i} | ||
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 |
80 | 4000 | 1280 |
Q7 :Find the mean deviation about the median for the data.
Xi | 5 | 7 | 9 | 10 | 12 | 15 |
fi | 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 obtain the following table.
x_{i} | f_{i} | c.f |
5 | 8 | 8 |
7 | 6 | 14 |
9 | 2 | 16 |
10 | 21 | 18 |
12 | 2 | 20 |
15 | 6 | 26 |
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.
The absolute values of the deviations from median, i.e.are
|xi – M|
/ X_{i }– M / | 2 | 0 | 2 | 3 | 5 | 8 |
f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |
f_{i} / X_{i }– M / | 16 | 0 | 4 | 6 | 10 | 48 |
Q8 : Find the mean deviation about the median for the data
x_{i} | 15 | 21 | 27 | 30 | 35 |
f_{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 obtain the following table.
X_{i} | f_{i} | c.f |
15 | 3 | 3 |
21 | 5 | 8 |
27 | 6 | 14 |
30 | 7 | 21 |
35 | 8 | 29 |
Here, N = 29, which is odd.
observation = 15th observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
∴ Median = 30
The absolute values of the deviations from median, i.e.|xi – M| are
/ X_{i }– M / | 15 | 9 | 3 | 0 | 5 |
f_{i} | 3 | 5 | 16 | 7 | 8 |
f_{i} / Xi – M / | 45 | 45 | 18 | 0 | 40 |
Q9 :Find the mean deviation about the mean for the data.
Income per day | Number of Persons |
0-100 | 4 |
100-200 | 8 |
200-300 | 9 |
300-400 | 10 |
400-500 | 7 |
500-600 | 5 |
600-700 | 4 |
700–800 | 3 |
Answer :
The following table is formed.
Income per day | Number of persons f_{i} | Mid point X_{i} | f_{i}x_{i} | ||
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 | 26002 | 292 | 1168 |
700-800 | 3 | 750 | 2250 | 392 | 1176 |
50 | 17900 | 7896 |
Q10 :Find the mean deviation about the mean for the data
Heights in cm | Number of Boys |
95-105 | 9 |
105-115 | 13 |
115-125 | 26 |
125-135 | 30 |
135-145 | 12 |
145-155 | 10 |
Answer :
The following table is formed.
Heights in cm | Number of boys | Mid point X_{i} | f_{i}x_{i} | ||
95-105 | 9 | 100 | 900 | 25.3 | 227.7 |
105-115 | 13 | 110 | 1430 | 15.3 | 198.8 |
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 |
here,
Q11 :Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age | Number |
16-20 | 5 |
21-25 | 6 |
26-30 | 12 |
31-35 | 14 |
36-40 | 26 |
41-45 | 12 |
46-50 | 16 |
51-55 | 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.
Age | Number fi | c.f | Mid Point xi | ||
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 | 75 | 43 | 5 | 60 |
45.5-50.5 | 16 | 91 | 48 | 10 | 160 |
50.5-55.5 | 9 | 100 | 58 | 15 | 135 |
100 | 735 |
The class interval containing the or 50th item is 35.5 – 40.5.
Therefore, 35.5 – 40.5 is the median class.
It is known that,
Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
Thus, mean deviation about the median is given by,
Exercise 15.2 : Solutions of Questions on Page Number : 371
Q1 :Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Answer :
6, 7, 10, 12, 13, 4, 8, 12
Mean,
The following table is obtained.
xi | ||
6 | -3 | 9 |
7 | -2 | 4 |
10 | -1 | 1 |
12 | 3 | 9 |
13 | 4 | 16 |
4 | -5 | 25 |
8 | -1 | 1 |
12 | 3 | 9 |
74 |
Q2 :Find the mean and variance for the first n natural numbers
Answer :
The mean of first n natural numbers is calculated as follows.
Q3 :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
The following table is obtained.
x_{i} | ||
3 | -13.5 | 182.25 |
6 | -10.5 | 110.25 |
9 | -7.5 | 56.25 |
12 | -4.5 | 20.25 |
15 | -1.5 | 2.25 |
18 | 1.5 | 2.25 |
21 | 4.5 | 20.25 |
24 | 7.5 | 56.25 |
27 | 10.5 | 110.25 |
30 | 13.5 | 182.25 |
742.5 |
Q4 :Find the mean and variance for the data
xi | 6 | 10 | 14 | 18 | 28 | 24 | 30 |
fi | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Answer :
The data is obtained in tabular form as follows.
x_{i} | f_{i} | x_{i}fi | |||
6 | 2 | 12 | -13 | 169 | 338 |
10 | 4 | 40 | -9 | 81 | 324 |
14 | 7 | 98 | -5 | 25 | 175 |
18 | 12 | 216 | -1 | 1 | 12 |
24 | 8 | 192 | 5 | 25 | 200 |
28 | 4 | 112 | 9 | 81 | 324 |
30 | 3 | 90 | 11 | 121 | 363 |
40 | 760 | 1736 |
Here, N = 40,
Q5 :Find the mean and variance for the data
x_{i} | 92 | 93 | 97 | 983 | 102 | 104 | 109 |
fi | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Answer :
The data is obtained in tabular form as follow
x_{i} | f_{i} | x_{i}f_{i} | |||
92 | 3 | 276 | -8 | 64 | 192 |
93 | 2 | 186 | -7 | 49 | 98 |
97 | 3 | 291 | -3 | 9 | 27 |
98 | 2 | 196 | -2 | 4 | 8 |
102 | 6 | 612 | 2 | 4 | 24 |
104 | 3 | 312 | 4 | 16 | 48 |
109 | 3 | 327 | 9 | 81 | 243 |
22 | 2200 | 640 |
Here, N = 22,
Q6 :Find the mean and standard deviation using short-cut 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.
x_{i} | f_{i} | yi2 | f_{i}y_{i} | f_{i}y_{i}^{2} | |
60 | 2 | -4 | 16 | -8 | 32 |
61 | 1 | -3 | 9 | -3 | 9 |
62 | 12 | -2 | 4 | -24 | 48 |
63 | 29 | -1 | 1 | -29 | 29 |
64 | 25 | 0 | 0 | 0 | 0 |
65 | 12 | 1 | 1 | 12 | 12 |
66 | 10 | 2 | 4 | 20 | 40 |
67 | 4 | 3 | 9 | 12 | 36 |
68 | 5 | 4 | 16 | 20 | 80 |
100 | 220 | 0 | 286 |
Mean,
Q7 :Find the mean and variance for the following frequency distribution.
Classes | 0-30 | 30-60 | 60-90 | 90-120 | 120-150 | 150-180 | 180-210 |
Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Answer :
Class | Frequency f_{i} | Mid Point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i} | |
0-30 | 2 | 15 | -3 | 9 | -6 | 18 |
30-60 | 3 | 45 | -2 | 4 | -6 | 12 |
60-90 | 5 | 75 | -1 | 1 | -5 | 5 |
90-120 | 10 | 105 | 0 | 0 | 0 | 0 |
120-150 | 3 | 135 | 1 | 1 | 3 | 3 |
150-180 | 5 | 165 | 2 | 4 | 10 | 20 |
180-210 | 2 | 195 | 3 | 9 | 6 | 18 |
30 | 2 | 76 |
Mean,
Q8 :Find the mean and variance for the following frequency distribution.
Classes | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Answer :
Class | Frequency fi | Mid Point x_{i} | y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |
0-10 | 5 | 5 | -2 | 4 | -10 | 20 |
10-20 | 18 | 15 | -1 | 1 | -8 | 8 |
20-30 | 15 | 25 | 0 | 0 | 0 | 0 |
30-40 | 16 | 35 | 1 | 1 | 16 | 16 |
40-50 | 6 | 45 | 2 | 4 | 12 | 24 |
50 | 10 | 68 |
Mean,
Q9 :Find the mean, variance and standard deviation using short-cut method
Heights in cm | No. of children |
70-75 | 3 |
75-80 | 4 |
80-85 | 7 |
85-90 | 7 |
90-95 | 15 |
95-100 | 9 |
100-105 | 6 |
105-110 | 6 |
110-115 | 3 |
Answer :
Class Interval | Frequency | Mid Point | y_{i} | f_{i}y_{i} | f_{i}y_{i}^{2} | |
70-75 | 3 | 72.5 | -4 | 16 | -12 | 48 |
75-80 | 4 | 77.5 | -3 | 9 | -12 | 36 |
80-85 | 7 | 82.5 | -2 | 4 | -14 | 28 |
85-90 | 7 | 87.5 | -1 | 1 | -7 | 7 |
90-95 | 15 | 92.5 | 0 | 0 | 0 | 0 |
95-100 | 9 | 97.5 | 1 | 1 | 9 | 9 |
100-105 | 6 | 102.5 | 2 | 4 | 12 | 24 |
105-110 | 6 | 107.5 | 3 | 9 | 18 | 54 |
110-115 | 3 | 112.5 | 41 | 16 | 12 | 48 |
60 | 6 | 254 |
Mean,
Q10 :The diameters of circles (in mm) drawn in a design are given below:
Diameters | No. of children |
33-36 | 15 |
37-40 | 17 |
41-44 | 21 |
45-48 | 22 |
49-52 | 25 |
Answer :
Class Interval | Frequency f_{i} | Mid-point x_{i} | f_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} | |
32.5-36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
36.5-40.5 | 17 | 34.5 | -1 | 1 | -17 | 17 |
40.5-44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
44.5-48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
48.5-52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
100 | 25 | 199 |
Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.
Mean,
Exercise 15.3 : Solutions of Questions on Page Number : 375
Q1 :From the data given below state which group is more variable, A or B?
Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
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.
Marks | Group A fi | Mid Point xi | yi^{2} | fiy_{i} | fiyi^{2} | |
10-20 | 9 | 15 | -3 | 9 | -27 | 81 |
20-30 | 17 | 25 | -2 | 4 | -34 | 68 |
30-40 | 32 | 35 | -1 | 1 | -32 | 32 |
40-50 | 33 | 45 | 0 | 0 | 0 | 0 |
50-60 | 40 | 55 | 1 | 1 | 40 | 40 |
60-70 | 10 | 65 | 2 | 4 | 20 | 40 |
70-80 | 9 | 75 | 3 | 0 | 27 | 81 |
150 | -6 | 342 |
Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
Marks | Group B fi | Mid Point xi | yi^{2} | fiyi | fiyi^{2} | |
10-20 | 10 | 15 | -3 | 9 | -30 | 90 |
20-30 | 20 | 25 | -2 | 4 | -40 | 80 |
30-40 | 30 | 35 | -1 | 1 | -30 | 30 |
40-50 | 25 | 45 | 0 | 0 | 0 | 0 |
50-60 | 43 | 55 | 1 | 1 | 43 | 43 |
60-70 | 15 | 65 | 2 | 4 | 30 | 60 |
70-80 | 7 | 75 | 3 | 9 | 21 | 63 |
150 | -6 | 366 |
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.
Q2 :From the data given below state which group is more variable, A or B?
Answer :
Firstly, the standard deviation of group A is calculated as follows.
Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
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.
Q3 :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
Xi | ||
35 | -16 | 256 |
54 | 3 | 9 |
52 | 1 | 1 |
53 | 2 | 4 |
56 | 5 | 25 |
58 | 7 | 49 |
52 | 1 | 1 |
50 | -1 | 1 |
51 | 0 | 0 |
49 | -2 | 4 |
350 |
The following table is obtained corresponding to shares X.
The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
yi | ||
108 | 3 | 9 |
107 | 2 | 4 |
105 | 0 | 0 |
105 | 0 | 0 |
106 | 1 | 1 |
107 | 2 | 4 |
104 | -1 | 1 |
103 | -2 | 4 |
104 | -1 | 1 |
101 | -4 | 16 |
40 |
The following table is obtained corresponding to shares Y.
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.
Q4 :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 wage earners | 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
∴Total amount paid = Rs 5253 x 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 x 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 = 100
∴ Standard deviation of the distribution of wages in firm
A ((Ïƒ1) =
Variance of the distribution of wages in firm = 121
∴ Standard deviation of the distribution of wages in firm
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.
Q5 :The following is the record of goals scored by team A in a football session:
No. of goals 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.
No. of goals scored | No. of matches | fixi | xi^{2} | f_{i}x_{i}^{2} |
0 | 1 | 0 | 0 | 0 |
1 | 9 | 9 | 1 | 9 |
2 | 7 | 14 | 4 | 28 |
3 | 5 | 15 | 9 | 48 |
4 | 3 | 12 | 16 | 45 |
25 | 50 | 130 |
Thus, the mean of both the teams is same.
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.
Q6 :The sum and sum of squares corresponding to length x (in cm) and weight y
(in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Answer :
Here, N = 50
∴ Mean,
Mean,
Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.
Exercise Miscellaneous : Solutions of Questions on Page Number : 380
Q1 :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.
From (1), we obtain
x^{2} + y^{2} + 2xy = 144 — (3)
From (2) and (3), we obtain
2xy = 64 —-(4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2} – 2xy = 80 – 64 = 16
⇒ x – y = Â± 4 — (5)
Therefore, from (1) and (5), we obtain
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.
Q2 :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.
From (1), we obtain
x^{2} + y^{2} + 2xy = 196 ……… (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 ……… (4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2} – 2xy = 100 – 96
⇒ (x – y)^{2} = 4
⇒ x – y = Â± 2 ……… (5)
Therefore, from (1) and (5), we obtain
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.
Q3 :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.
If each observation is multiplied by 3 and the resulting observations are y_{i}, then
From (1) and (2), it can be observed that,
Substituting the values of xi and in (2), we obtain
Therefore, variance of new observations =
Hence, the standard deviation of new observations is
Q4 :Given that is the mean and Ïƒ2 is the variance of n observations x1, x2 â€¦ xn. Prove that the mean and variance of the observations ax1, ax2, ax3 â€¦axn are and a2 Ïƒ2, respectively (a ≠ 0).
Answer :
The given n observations are x1, x2 â€¦ xn.
Mean =
Variance = Ïƒ2
If each observation is multiplied by a and the new observations are yi, then
Therefore, mean of the observations, ax1, ax2 â€¦ axn, is .
Substituting the values of xiand in (1), we obtain
Thus, the variance of the observations, ax1, ax2 â€¦ axn, is a2 Ïƒ2.
Q5 :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
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean
(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
∴ Correct sum of observations = 200 – 8 + 12 = 204
Q6 :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.) is given by
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.
Q7 :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) = 1
Incorrect mean
Incorrect standard deviation
∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 21 -18 = 2000 – 60 = 1940