The NCERT Solutions For Class 11 Maths Chapter 15 Statistics are available as a pdf on this page. The NCERT Solutions are authored by the most experienced educators in the teaching industry making the solution of every problem straightforward and justifiable. Every solution written in the pdf given underneath empowers the students to get ready for the test and accomplish it. These solutions assist a Class 11 student with mastering the idea of Limits and Derivatives.
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Get detailed solutions for all the questions listed under the below exercises:
Exercise 15.1 Solutions : 12 Questions
Exercise 15.2 Solutions : 10 Questions
Exercise 15.3 Solutions : 5 Questions
Miscellaneous Exercise Solutions: 7 Questions
Access Solutions for Class 11 Maths Chapter 15 Statistics
Exercise 15.1 Page: 360
Find the mean deviation about the mean for the data in Exercises 1 and 2.
1. 4, 7, 8, 9, 10, 12, 13, 17
Solution:-
First we have to find (xÌ…) of the given data
So, the respective values of the deviations from mean,
i.e., x_{i}Â â€“ xÌ… are, 10 â€“ 4 = 6, 10 â€“ 7 = 3, 10 â€“ 8 = 2, 10 â€“ 10 = 0,
10 â€“ 12 = â€“ 2, 10 â€“ 13 = â€“ 3, 10 â€“ 17 = â€“ 7
6, 3, 2, 1, 0, -2, -3, -7
Now absolute values of the deviations,
6, 3, 2, 1, 0, 2, 3, 7
MD = sum of deviations/ number of observations
= 24/8
= 3
So, the mean deviation for the given data is 3.
2. 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Solution:-
First we have to find (xÌ…) of the given data
So, the respective values of the deviations from mean,
i.e., x_{i}Â â€“ xÌ… are, 50 â€“ 38 = -12, 50 -70 = -20, 50 â€“ 48 = 2, 50 â€“ 40 = 10, 50 â€“ 42 = 8,
50 â€“ 55 = â€“ 5, 50 â€“ 63 = â€“ 13, 50 â€“ 46 = 4, 50 â€“ 54 = -4, 50 â€“ 44 = 6
-12, 20, -2, -10, -8, 5, 13, -4, 4, -6
Now absolute values of the deviations,
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
MD = sum of deviations/ number of observations
= 84/10
= 8.4
So, the mean deviation for the given data is 8.4.
Find the mean deviation about the median for the data in Exercises 3 and 4.
3. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Solution:-
First we have to arrange the given observations into ascending order,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.
The number of observations is 12
Then,
Median = ((12/2)^{th} observation + ((12/2)+ 1)^{th} observation)/2
(12/2)^{th} observation = 6^{th} = 13
(12/2)+ 1)^{th} observation = 6 + 1
= 7^{th} = 14
Median = (13 + 14)/2
= 27/2
= 13.5
So, the absolute values of the respective deviations from the median, i.e., |x_{i}Â â€“ 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
Mean Deviation,
= (1/12) Ã— 28
= 2.33
So, the mean deviation about the median for the given data is 2.33.
4. 36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Solution:-
First we have to arrange the given observations into ascending order,
36, 42, 45, 46, 46, 49, 51, 53, 60, 72.
The number of observations is 10
Then,
Median = ((10/2)^{th} observation + ((10/2)+ 1)^{th} observation)/2
(10/2)^{th} observation = 5^{th} = 46
(10/2)+ 1)^{th} observation = 5 + 1
= 6^{th} = 49
Median = (46 + 49)/2
= 95
= 47.5
So, the absolute values of the respective deviations from the median, i.e., |x_{i}Â â€“ M| are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Mean Deviation,
= (1/10) Ã— 70
= 7
So, the mean deviation about the median for the given data is 7.
Find the mean deviation about the mean for the data in Exercises 5 and 6.
5.
x_{i} | 5 | 10 | 15 | 20 | 25 |
f_{i} | 7 | 4 | 6 | 3 | 5 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | f_{i}x_{i} | |x_{i}Â â€“ xÌ…| | f_{i} |x_{i}Â â€“ xÌ…| |
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 |
The sum of calculated data,
The absolute values of the deviations from the mean, i.e., |x_{i}Â â€“ xÌ…|, as shown in the table.
6.
x_{i} | 10 | 30 | 50 | 70 | 90 |
f_{i} | 4 | 24 | 28 | 16 | 8 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | f_{i}x_{i} | |x_{i}Â â€“ xÌ…| | f_{i} |x_{i}Â â€“ xÌ…| |
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 |
Find the mean deviation about the median for the data in Exercises 7 and 8.
7.
x_{i} | 5 | 7 | 9 | 10 | 12 | 15 |
f_{i} | 8 | 6 | 2 | 2 | 2 | 6 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | c.f. | |x_{i}Â â€“ M| | f_{i} |x_{i}Â â€“ 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 |
Now, N = 26, which is even.
Median is the mean of the 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
Then,
Median = (13^{th} observation + 14^{th }observation)/2
= (7 + 7)/2
= 14/2
= 7
So, the absolute values of the respective deviations from the median, i.e., |x_{i}Â â€“ M| are shown in the table.
8.
x_{i} | 15 | 21 | 27 | 30 | 35 |
f_{i} | 3 | 5 | 6 | 7 | 8 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | c.f. | |x_{i}Â â€“ M| | f_{i} |x_{i}Â â€“ M| |
15 | 3 | 3 | 13.5 | 40.5 |
21 | 5 | 8 | 7.5 | 37.5 |
27 | 6 | 14 | 1.5 | 9 |
30 | 7 | 21 | 1.5 | 10.5 |
35 | 8 | 29 | 6.5 | 52 |
Now, N = 30, which is even.
Median is the mean of the 15^{th} and 16^{th} observations. Both of these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.
Then,
Median = (15^{th} observation + 16^{th }observation)/2
= (30 + 30)/2
= 60/2
= 30
So, the absolute values of the respective deviations from the median, i.e., |x_{i}Â â€“ M| are shown in the table.
Find the mean deviation about the mean for the data in Exercises 9 and 10.
9.
Income per day in â‚¹ | 0 â€“ 100 | 100 â€“ 200 | 200 â€“ 300 | 300 â€“ 400 | 400 â€“ 500 | 500 â€“ 600 | 600 â€“ 700 | 700 â€“ 800 |
Number of persons | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
Income per day in â‚¹ | Number of persons f_{i} |
Mid â€“ points
x_{i} |
f_{i}x_{i} | |x_{i}Â â€“ xÌ…| | f_{i}|x_{i}Â â€“ xÌ…| |
0 â€“ 100 | 4 | 50 | 200 | 308 | 1232 |
100 â€“ 200 | 8 | 150 | 1200 | 208 | 1664 |
200 â€“ 300 | 9 | 250 | 2250 | 108 | 972 |
300 â€“ 400 | 10 | 350 | 3500 | 8 | 80 |
400 â€“ 500 | 7 | 450 | 3150 | 92 | 644 |
500 â€“ 600 | 5 | 550 | 2750 | 192 | 960 |
600 â€“ 700 | 4 | 650 | 2600 | 292 | 1160 |
700 â€“ 800 | 3 | 750 | 2250 | 392 | 1176 |
50 | 17900 | 7896 |
10.
Height in cms | 95 â€“ 105 | 105 â€“ 115 | 115 â€“ 125 | 125 â€“ 135 | 135 â€“ 145 | 145 â€“ 155 |
Number of boys | 9 | 13 | 26 | 30 | 12 | 10 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
Height in cms | Number of boys f_{i} |
Mid â€“ points
x_{i} |
f_{i}x_{i} | |x_{i}Â â€“ xÌ…| | f_{i}|x_{i}Â â€“ xÌ…| |
95 â€“ 105 | 9 | 100 | 900 | 25.3 | 227.7 |
105 â€“ 115 | 13 | 110 | 1430 | 15.3 | 198.9 |
115 â€“ 125 | 26 | 120 | 3120 | 5.3 | 137.8 |
125 â€“ 135 | 30 | 130 | 3900 | 4.7 | 141 |
135 â€“ 145 | 12 | 140 | 1680 | 14.7 | 176.4 |
145 â€“ 155 | 10 | 150 | 1500 | 24.7 | 247 |
100 | 12530 | 1128.8 |
11. Find the mean deviation about median for the following data:
Marks | 0 -10 | 10 -20 | 20 â€“ 30 | 30 â€“ 40 | 40 â€“ 50 | 50 â€“ 60 |
Number of girls | 6 | 8 | 14 | 16 | 4 | 2 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
Marks | Number of Girls f_{i} | Cumulative frequency (c.f.) |
Mid â€“ points
x_{i} |
|x_{i}Â â€“ Med| | f_{i}|x_{i}Â â€“ Med| |
0 â€“ 10 | 6 | 6 | 5 | 22.85 | 137.1 |
10 â€“ 20 | 8 | 14 | 15 | 12.85 | 102.8 |
20 â€“ 30 | 14 | 28 | 25 | 2.85 | 39.9 |
30 â€“ 40 | 16 | 44 | 35 | 7.15 | 114.4 |
40 â€“ 50 | 4 | 48 | 45 | 17.15 | 68.6 |
50 â€“ 60 | 2 | 50 | 55 | 27.15 | 54.3 |
50 | 517.1 |
The class interval containingÂ N^{th}/2Â or 25^{th}Â item is 20-30
So, 20-30 is the median class.
Then,
Median = l + (((N/2) â€“ c)/f) Ã— h
Where, l = 20, c = 14, f = 14, h = 10 and n = 50
Median = 20 + (((25 â€“ 14))/14) Ã— 10
= 20 + 7.85
= 27.85
12. Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age
(in years) |
16 â€“ 20 | 21 â€“ 25 | 26 â€“ 30 | 31 â€“ 35 | 36 â€“ 40 | 41 â€“ 45 | 46 â€“ 50 | 51 â€“ 55 |
Number | 5 | 6 | 12 | 14 | 26 | 12 | 16 | 9 |
[Hint Convert 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]
Solution:-
The given data is converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding the 0.5 to the upper limit of each class intervals and append other columns after calculations.
Age | Number f_{i} | Cumulative frequency (c.f.) |
Mid â€“ points
x_{i} |
|x_{i}Â â€“ Med| | f_{i}|x_{i}Â â€“ Med| |
15.5 â€“ 20.5 | 5 | 5 | 18 | 20 | 100 |
20.5 â€“ 25.5 | 6 | 11 | 23 | 15 | 90 |
25.5 â€“ 30.5 | 12 | 23 | 28 | 10 | 120 |
30.5 â€“ 35.5 | 14 | 37 | 33 | 5 | 70 |
35.5 â€“ 40.5 | 26 | 63 | 38 | 0 | 0 |
40.5 â€“ 45.5 | 12 | 75 | 43 | 5 | 60 |
45.5 â€“ 50.5 | 16 | 91 | 48 | 10 | 160 |
50.5 â€“ 55.5 | 9 | 100 | 53 | 15 | 135 |
100 | 735 |
The class interval containingÂ N^{th}/2Â or 50^{th}Â item is 35.5 â€“ 40.5
So, 35.5 â€“ 40.5 is the median class.
Then,
Median = l + (((N/2) â€“ c)/f) Ã— h
Where, l = 35.5, c = 37, f = 26, h = 5 and N = 100
Median = 35.5 + (((50 â€“ 37))/26) Ã— 5
= 35.5 + 2.5
= 38
Exercise 15.2 Page: 371
Find the mean and variance for each of the data in Exercise 1 to 5.
1. 6, 7, 10, 12, 13, 4, 8, 12
Solution:-
So, xÌ… = (6 + 7 + 10 + 12 + 13 + 4 + 8 + 12)/8
= 72/8
= 9
Let us make the table of the given data and append other columns after calculations.
X_{i} |
Deviations from mean
(x_{i}Â â€“ xÌ…) |
(x_{i}Â â€“ xÌ…)^{2} |
6 | 6 â€“ 9 = -3 | 9 |
7 | 7 â€“ 9 = -2 | 4 |
10 | 10 â€“ 9 = 1 | 1 |
12 | 12 â€“ 9 = 3 | 9 |
13 | 13 â€“ 9 = 4 | 16 |
4 | 4 â€“ 9 = â€“ 5 | 25 |
8 | 8 â€“ 9 = â€“ 1 | 1 |
12 | 12 â€“ 9 = 3 | 9 |
74 |
We know that Variance,
Ïƒ^{2}Â = (1/8) Ã— 74
= 9.2
âˆ´Mean = 9 and Variance = 9.25
2. First n natural numbers
Solution:-
We know that Mean = Sum of all observations/Number of observations
âˆ´Mean, xÌ… = ((n(n + 1))2)/n
= (n + 1)/2
and also WKT Variance,
By substitute that value of xÌ… we get,
WKT, (a + b)(a â€“ b) = a^{2} â€“ b^{2}
Ïƒ^{2}Â = (n^{2} â€“ 1)/12
âˆ´Mean = (n + 1)/2 and Variance = (n^{2} â€“ 1)/12
3. First 10 multiples of 3
Solution:-
First we have to write the first 10 multiples of 3,
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
So, xÌ… = (3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30)/10
= 165/10
= 16.5
Let us make the table of the data and append other columns after calculations.
X_{i} |
Deviations from mean
(x_{i}Â â€“ xÌ…) |
(x_{i}Â â€“ xÌ…)^{2} |
3 | 3 â€“ 16.5 = -13.5 | 182.25 |
6 | 6 â€“ 16.5 = -10.5 | 110.25 |
9 | 9 â€“ 16.5 = -7.5 | 56.25 |
12 | 12 â€“ 16.5 = -4.5 | 20.25 |
15 | 15 â€“ 16.5 = -1.5 | 2.25 |
18 | 18 â€“ 16.5 = 1.5 | 2.25 |
21 | 21 â€“ 16.5 = â€“ 4.5 | 20.25 |
24 | 24 â€“ 16.5 = 7.5 | 56.25 |
27 | 27 â€“ 16.5 = 10.5 | 110.25 |
30 | 30 â€“ 16.5 = 13.5 | 182.25 |
742.5 |
Then, Variance
= (1/10) Ã— 742.5
= 74.25
âˆ´Mean = 16.5 and Variance = 74.25
4.
x_{i} | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
f_{i} | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | f_{i}x_{i} |
Deviations from mean
(x_{i}Â â€“ xÌ…) |
(x_{i}Â â€“ xÌ…)^{2} | f_{i}(x_{i}Â â€“ xÌ…)^{2} |
6 | 2 | 12 | 6 â€“ 19 = 13 | 169 | 338 |
10 | 4 | 40 | 10 â€“ 19 = -9 | 81 | 324 |
14 | 7 | 98 | 14 â€“ 19 = -5 | 25 | 175 |
18 | 12 | 216 | 18 â€“ 19 = -1 | 1 | 12 |
24 | 8 | 192 | 24 â€“ 19 = 5 | 25 | 200 |
28 | 4 | 112 | 28 â€“ 19 = 9 | 81 | 324 |
30 | 3 | 90 | 30 â€“ 19 = 11 | 121 | 363 |
N = 40 | 760 | 1736 |
5.
x_{i} | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
f_{i} | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
X_{i} | f_{i} | f_{i}x_{i} |
Deviations from mean
(x_{i}Â â€“ xÌ…) |
(x_{i}Â â€“ xÌ…)^{2} | f_{i}(x_{i}Â â€“ xÌ…)^{2} |
92 | 3 | 276 | 92 â€“ 100 = -8 | 64 | 192 |
93 | 2 | 186 | 93 â€“ 100 = -7 | 49 | 98 |
97 | 3 | 291 | 97 â€“ 100 = -3 | 9 | 27 |
98 | 2 | 196 | 98 â€“ 100 = -2 | 4 | 8 |
102 | 6 | 612 | 102 â€“ 100 = 2 | 4 | 24 |
104 | 3 | 312 | 104 â€“ 100 = 4 | 16 | 48 |
109 | 3 | 327 | 109 â€“ 100 = 9 | 81 | 243 |
N = 22 | 2200 | 640 |
6. 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 |
Solution:-
Let the assumed mean A = 64. Here h = 1
We obtain the following table from the given data.
X_{i} | Frequency f_{i} | Y_{i} = (x_{i} â€“ A)/h | Y_{i}^{2} | 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 |
0 | 286 |
Mean,
Where A = 64, h = 1
So, xÌ… = 64 + ((0/100) Ã— 1)
= 64 + 0
= 64
Then, variance,
Ïƒ^{2} = (1^{2}/100^{2}) [100(286) â€“ 0^{2}]
= (1/10000) [28600 â€“ 0]
= 28600/10000
= 2.86
Hence, standard deviation = Ïƒ = âˆš2.886
= 1.691
âˆ´Â Mean = 64 and Standard Deviation = 1.691
Find the mean and variance for the following frequency distributions in Exercises 7 and 8.
7.
Classes | 0 â€“ 30 | 30 â€“ 60 | 60 â€“ 90 | 90 â€“ 120 | 120 â€“ 150 | 150 â€“ 180 | 180 â€“ 210 |
Frequencies | 2 | 3 | 5 | 10 | 3 | 5 | 2 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
Classes | Frequency f_{i} |
Mid â€“ points
x_{i} |
f_{i}x_{i} | (x_{i}Â â€“ xÌ…) | (x_{i}Â â€“ xÌ…)^{2} | f_{i}(x_{i}Â â€“ xÌ…)^{2} |
0 â€“ 30 | 2 | 15 | 30 | -92 | 8464 | 16928 |
30 â€“ 60 | 3 | 45 | 135 | -62 | 3844 | 11532 |
60 â€“ 90 | 5 | 75 | 375 | -32 | 1024 | 5120 |
90 â€“ 120 | 10 | 105 | 1050 | -2 | 4 | 40 |
120 â€“ 150 | 3 | 135 | 405 | 28 | 784 | 2352 |
150 â€“ 180 | 5 | 165 | 825 | 58 | 3364 | 16820 |
180 â€“ 210 | 2 | 195 | 390 | 88 | 7744 | 15488 |
N = 30 | 3210 | 68280 |
8.
Classes | 0 â€“ 10 | 10 â€“ 20 | 20 â€“ 30 | 30 â€“ 40 | 40 â€“50 |
Frequencies | 5 | 8 | 15 | 16 | 6 |
Solution:-
Let us make the table of the given data and append other columns after calculations.
Classes | Frequency f_{i} |
Mid â€“ points
x_{i} |
f_{i}x_{i} | (x_{i}Â â€“ xÌ…) | (x_{i}Â â€“ xÌ…)^{2} | f_{i}(x_{i}Â â€“ xÌ…)^{2} |
0 â€“ 10 | 5 | 5 | 25 | -22 | 484 | 2420 |
10 â€“ 20 | 8 | 15 | 120 | -12 | 144 | 1152 |
20 â€“ 30 | 15 | 25 | 375 | -2 | 4 | 60 |
30 â€“ 40 | 16 | 35 | 560 | 8 | 64 | 1024 |
40 â€“50 | 6 | 45 | 270 | 18 | 324 | 1944 |
N = 50 | 1350 | 6600 |
9. Find the mean, variance and standard deviation using short-cut method
Height in cms | 70 â€“ 75 | 75 â€“ 80 | 80 â€“ 85 | 85 â€“ 90 | 90 â€“ 95 | 95 â€“ 100 | 100 â€“ 105 | 105 â€“ 110 | 110 â€“ 115 |
Frequencies | 3 | 4 | 7 | 7 | 15 | 9 | 6 | 6 | 3 |
Solution:-
Let the assumed mean, A = 92.5 and h = 5
Let us make the table of the given data and append other columns after calculations.
Height (class) |
Number of children
Frequency f_{i} |
Midpoint
X_{i} |
Y_{i} = (x_{i} â€“ A)/h | Y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} |
70 â€“ 75 | 2 | 72.5 | -4 | 16 | -12 | 48 |
75 â€“ 80 | 1 | 77.5 | -3 | 9 | -12 | 36 |
80 â€“ 85 | 12 | 82.5 | -2 | 4 | -14 | 28 |
85 â€“ 90 | 29 | 87.5 | -1 | 1 | -7 | 7 |
90 â€“ 95 | 25 | 92.5 | 0 | 0 | 0 | 0 |
95 â€“ 100 | 12 | 97.5 | 1 | 1 | 9 | 9 |
100 â€“ 105 | 10 | 102.5 | 2 | 4 | 12 | 24 |
105 â€“ 110 | 4 | 107.5 | 3 | 9 | 18 | 54 |
110 â€“ 115 | 5 | 112.5 | 4 | 16 | 12 | 48 |
N = 60 | 6 | 254 |
Mean,
Where, A = 92.5, h = 5
So, xÌ… = 92.5 + ((6/60) Ã— 5)
= 92.5 + Â½
= 92.5 + 0.5
= 93
Then, Variance,
Ïƒ^{2} = (5^{2}/60^{2}) [60(254) â€“ 6^{2}]
= (1/144) [15240 â€“ 36]
= 15204/144
= 1267/12
= 105.583
Hence, standard deviation = Ïƒ = âˆš105.583
= 10.275
âˆ´Â Mean = 93, variance = 105.583 and Standard Deviation = 10.275
10. The diameters of circles (in mm) drawn in a design are given below:
Diameters | 33 â€“ 36 | 37 â€“ 40 | 41 â€“ 44 | 45 â€“ 48 | 49 â€“ 52 |
No. of circles | 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 32.5-36.5, 36.5-40.5, 40.5-44.5, 44.5 â€“ 48.5, 48.5 â€“ 52.5 and then proceed.]
Solution:-
Let the assumed mean, A = 42.5 and h = 4
Let us make the table of the given data and append other columns after calculations.
Height (class) |
Number of children
(Frequency f_{i}) |
Midpoint
X_{i} |
Y_{i} = (x_{i} â€“ A)/h | Y_{i}^{2} | f_{i}y_{i} | f_{i}y_{i}^{2} |
32.5 â€“ 36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
36.5 â€“ 40.5 | 17 | 38.5 | -1 | 1 | -17 | 17 |
40.5 â€“ 44.5 | 21 | 42.5 | 0 | 0 | 0 | 0 |
44.5 â€“ 48.5 | 22 | 46.5 | 1 | 1 | 22 | 22 |
48.5 â€“ 52.5 | 25 | 50.5 | 2 | 4 | 50 | 100 |
N = 100 | 25 | 199 |
Mean,
Where, A = 42.5, h = 4
So, xÌ… = 42.5 + (25/100) Ã— 4
= 42.5 + 1
= 43.5
Then, Variance,
Ïƒ^{2} = (4^{2}/100^{2})[100(199) â€“ 25^{2}]
= (1/625) [19900 â€“ 625]
= 19275/625
= 771/25
= 30.84
Hence, standard deviation = Ïƒ = âˆš30.84
= 5.553
âˆ´Â Mean = 43.5, variance = 30.84 and Standard Deviation = 5.553.
Exercise 15.3 Page: 375
1. From the data given below state which group is more variable, A or B?
Marks | 10 â€“ 20 | 20 â€“ 30 | 30 â€“ 40 | 40 â€“ 50 | 50 â€“ 60 | 60 â€“ 70 | 70 â€“ 80 |
Group A | 9 | 17 | 32 | 33 | 40 | 10 | 9 |
Group B | 10 | 20 | 30 | 25 | 43 | 15 | 7 |
Solution:-
For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.
Co-efficient of variation (C.V.) = (Ïƒ/ xÌ…) Ã— 100
Where, Ïƒ = standard deviation, xÌ… = mean
For Group A
Marks |
Group A
f_{i} |
Mid-point
X_{i} |
Y_{i} = (x_{i} â€“ A)/h | (Y_{i})^{2} | f_{i}y_{i} | f_{i}(y_{i})^{2} |
10 â€“ 20 | 9 | 15 | ((15 â€“ 45)/10) = -3 | (-3)^{2} = 9 |
â€“ 27 | 81 |
20 â€“ 30 | 17 | 25 | ((25 â€“ 45)/10) = -2 | (-2)^{2} = 4 |
â€“ 34 | 68 |
30 â€“ 40 | 32 | 35 | ((35 â€“ 45)/10) = â€“ 1 | (-1)^{2} = 1 |
â€“ 32 | 32 |
40 â€“ 50 | 33 | 45 | ((45 â€“ 45)/10) = 0 | 0^{2} | 0 | 0 |
50 â€“ 60 | 40 | 55 | ((55 â€“ 45)/10) = 1 | 1^{2} = 1 |
40 | 40 |
60 â€“ 70 | 10 | 65 | ((65 â€“ 45)/10) = 2 | 2^{2} = 4 |
20 | 40 |
70 â€“ 80 | 9 | 75 | ((75 â€“ 45)/10) = 3 | 3^{2} = 9 |
27 | 81 |
Total | 150 | -6 | 342 |
Where A = 45,
and y_{i} = (x_{i} â€“ A)/h
Here h = class size = 20 â€“ 10
h = 10
So, xÌ… = 45 + ((-6/150) Ã— 10)
= 45 â€“ 0.4
= 44.6
Ïƒ^{2} = (10^{2}/150^{2}) [150(342) â€“ (-6)^{2}]
= (100/22500) [51,300 â€“ 36]
= (100/22500) Ã— 51264
= 227.84
Hence, standard deviation = Ïƒ = âˆš227.84
= 15.09
âˆ´Â C.V for group A = (Ïƒ/ xÌ…) Ã— 100
= (15.09/44.6) Ã— 100
= 33.83
Now, for group B.
Marks |
Group B
f_{i} |
Mid-point
X_{i} |
Y_{i} = (x_{i} â€“ A)/h | (Y_{i})^{2} | f_{i}y_{i} | f_{i}(y_{i})^{2} |
10 â€“ 20 | 10 | 15 | ((15 â€“ 45)/10) = -3 | (-3)^{2} = 9 |
â€“ 30 | 90 |
20 â€“ 30 | 20 | 25 | ((25 â€“ 45)/10) = -2 | (-2)^{2} = 4 |
â€“ 40 | 80 |
30 â€“ 40 | 30 | 35 | ((35 â€“ 45)/10) = â€“ 1 | (-1)^{2} = 1 |
â€“ 30 | 30 |
40 â€“ 50 | 25 | 45 | ((45 â€“ 45)/10) = 0 | 0^{2} | 0 | 0 |
50 â€“ 60 | 43 | 55 | ((55 â€“ 45)/10) = 1 | 1^{2} = 1 |
43 | 43 |
60 â€“ 70 | 15 | 65 | ((65 â€“ 45)/10) = 2 | 2^{2} = 4 |
30 | 160 |
70 â€“ 80 | 7 | 75 | ((75 â€“ 45)/10) = 3 | 3^{2} = 9 |
21 | 189 |
Total | 150 | -6 | 592 |
Where A = 45,
h = 10
So, xÌ… = 45 + ((-6/150) Ã— 10)
= 45 â€“ 0.4
= 44.6
Ïƒ^{2} = (10^{2}/150^{2}) [150(592) â€“ (-6)^{2}]
= (100/22500) [88,800 â€“ 36]
= (100/22500) Ã— 88,764
= 394.50
Hence, standard deviation = Ïƒ = âˆš394.50
= 19.86
âˆ´Â C.V for group B = (Ïƒ/ xÌ…) Ã— 100
= (19.86/44.6) Ã— 100
= 44.53
By comparing C.V. of group A and group B.
C.V of Group B > C.V. of Group A
So, Group B is more variable.
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 |
Solution:-
From the given data,
Let us make the table of the given data and append other columns after calculations.
X (x_{i}) | Y (y_{i}) | X_{i}^{2} | Y_{i}^{2} |
35 | 108 | 1225 | 11664 |
54 | 107 | 2916 | 11449 |
52 | 105 | 2704 | 11025 |
53 | 105 | 2809 | 11025 |
56 | 106 | 8136 | 11236 |
58 | 107 | 3364 | 11449 |
52 | 104 | 2704 | 10816 |
50 | 103 | 2500 | 10609 |
51 | 104 | 2601 | 10816 |
49 | 101 | 2401 | 10201 |
Total = 510 | 1050 | 26360 | 110290 |
We have to calculate Mean for x,
Mean xÌ… = âˆ‘x_{i}/n
Where, n = number of terms
= 510/10
= 51
Then, Variance for x =
= (1/10^{2})[(10 Ã— 26360) â€“ 510^{2}]
= (1/100) (263600 â€“ 260100)
= 3500/100
= 35
WKT Standard deviation = âˆšvariance
= âˆš35
= 5.91
So, co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100
= (5.91/51) Ã— 100
= 11.58
Now, we have to calculate Mean for y,
Mean È³ = âˆ‘y_{i}/n
Where, n = number of terms
= 1050/10
= 105
Then, Variance for y =
= (1/10^{2})[(10 Ã— 110290) â€“ 1050^{2}]
= (1/100) (1102900 â€“ 1102500)
= 400/100
= 4
WKT Standard deviation = âˆšvariance
= âˆš4
= 2
So, co-efficient of variation = (Ïƒ/ xÌ…) Ã— 100
= (2/105) Ã— 100
= 1.904
By comparing C.V. of X and Y.
C.V of X > C.V. of Y
So, Y is more stable than X.
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 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?
Solution:-
From the given table,
Mean monthly wages of firm A = Rs 5253
and Number of wage earners = 586
Then,
Total amount paid = 586 Ã— 5253
= Rs 3078258
Mean monthly wages of firm B = Rs 5253
Number of wage earners = 648
Then,
Total amount paid = 648 Ã— 5253
= Rs 34,03,944
(i) So, firm B pays larger amount as monthly wages.
(ii) Variance of firm A = 100
We know that, standard deviation (Ïƒ)= âˆš100
=10
Variance of firm B = 121
Then,
Standard deviation (Ïƒ)=âˆš(121 )
=11
Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.
4. 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?
Solution:-
From the given data,
Let us make the table of the given data and append other columns after calculations.
Number of goals scored x_{i} | Number of matches f_{i} | f_{i}x_{i} | X_{i}^{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 | 45 |
4 | 3 | 12 | 16 | 48 |
Total | 25 | 50 | 130 |
Since C.V. of firm B is greater
âˆ´Â Team A is more consistent.
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:
Which is more varying, the length or weight?
Solution:-
First we have to calculate Mean for Length x,
Â
Miscellaneous Exercise Page: 380
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.
Solution:-
2. The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.
Solution:-
Hence, from equation (i) and (v) we have:
When x â€“ y = 2 then x = 8 and y = 6
And, when x â€“ y = â€“ 2 then x = 6 and y = 8
âˆ´Â the remaining observations are 6 and 8
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.
Solution:-
We know that,
4. Given that xÌ… is the mean and Ïƒ^{2} is the variance of n observations x_{1}, x_{2}, â€¦,x_{n} . Prove that the mean and variance of the observations ax_{1}, ax_{2}, ax_{3}, â€¦., ax_{n} are axÌ… and a^{2}Ïƒ^{2}, respectively, (a â‰ 0).
Solution:-
From the question it is given that, n observations are x_{1}, x_{2},â€¦..x_{n}
Mean of the n observation = xÌ…
Variance of the n observation = Ïƒ^{2}
As we know that,
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
Solution:-
(i) If wrong item is omitted,
From the question it is given that,
The number of observations i.e. n = 20
The incorrect mean = 20
The incorrect standard deviation = 2
(ii) If it is replaced by 12,
From the question it is given that,
The number of incorrect sum observations i.e. n = 200
The correct sum of observations n = 200 â€“ 8 + 12
n = 204
Then, correct mean = correct sum/20
= 204/20
= 10.2
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?
Solution:-
From the question it is given that,
Mean of Mathematics = 42
Standard deviation of Mathematics = 12
Mean of Physics = 32
Standard deviation of physics = 15
Mean of Chemistry = 40.9
Standard deviation of chemistry = 20
As we know that,
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.
Solution:-
From the question it is given that,
The total number of observations (n) = 100
Incorrect mean, (xÌ…) = 20
And, Incorrect standard deviation (Ïƒ) = 3
NCERT Solutions for Class 11 Maths Chapter 15 â€“ Statistics
The topics of Class 11 Chapter 15 â€“ Statistics of NCERT Solution are as follows:
15.1 Introduction
This section talks about the concept of central tendency, mean and median [during even and the odd number of observations] with examples. It introduces the concept of measure of dispersion.
The values which cluster around the middle or centre of the distribution are measures of central tendency. They are mean, median and mode.
In a class, mean can be used to find the average marks scored by the students.
When calculating the height of students, the median can be used to find the middlemost value.
15.2 Measures of Dispersion
This section defines measures of dispersion, different measures of dispersion [range, quartile deviation, mean deviation, standard deviation]
Measures of dispersion explain the relationship with measures of central tendency. For example, the spread of data tells how well the mean represents the data. If the spread is large, then mean is not representative of data.
15.3 Range
This section defines the range, its formula and an example.
The range gives the variability of scores using the maximum and minimum values in the set.
In a cricket match, Batsman A range = 121 â€“ 0 = 121 and Batsman B range = 52 â€“ 22 = 30
Range A > Range B. So, the scores are spread in case of A whereas for B they are close to each other.
15.4 Mean Deviation
This section defines mean deviation, the formula for mean deviation.
The concept of mean deviation can be used by biologists in the comparison of different animal weights and decide what would be a healthy weight.
15.4.1Mean deviation for ungrouped data
The process of obtaining the mean deviation for ungrouped data is elaborated in this section.
Find the mean, deviations from the mean, absolute deviations and substitute the values in the mean deviation formula and arrive at the answer.
15.4.2 Mean deviation for grouped data
The process of obtaining mean deviation for discrete and continuous frequency distributions are elaborated in this section with solved examples.
15.4.3 Limitations of mean deviation
- If in a series the degree of variability is very high, then median will not be a representative of data. Hence the mean deviation calculated about such median cannot be relied on.
- If the sum of deviations from the mean is greater than the sum of deviations from the median, then the mean deviation about mean is not very specific.
- The absolute mean deviation calculated canâ€™t be subjected to further algebraic treatment. It canâ€™t be used as an appropriate measure of dispersion.
15.5 Variance and standard deviation
15.5.1 Standard Deviation
This section involves the variance and standard deviation definition, formula and solved examples on the discrete and continuous frequency distribution.
A science test was taken by a class of students. The mean score of the test was 85% on calculation. The teacher found the standard deviation of other scores and noticed that a very small standard deviation exists which suggests that most of the students scored very close to 85%.
15.5.2 Shortcut method to find Variance and standard deviation
This section deals with the simpler way of calculating the standard deviation with a few illustrations.
15.6 Analysis of Frequency Distributions
This section deals with the process of comparing the variability of two series having the same mean, coefficient of variation with few solved problems.
A few points on Chapter 15 Statistics
- Range is defined as the difference between the maximum and minimum value of the given data.
- If there exists series with equal means, then the series with lesser standard deviation is more consistent or less scattered.
- The addition or subtraction of a positive number to each data point of the data set will not affect the variance.
The solutions give substitute strategies and clarifications to take care of problems which makes the student feel sure while taking the test. Additionally, taking care of many muddled issues upgrades the information on Mathematical abilities in the students. The solutions cater to all the vital inquiries which a student should and ought to have aced to show up for the test. The BYJUâ€™S subject specialists who have written these solutions have complete knowledge about the question paper setting and the marks distributed across the chapters.