NCERT Solutions For Class 11 Maths Chapter 15 Statistics are given here in a detailed and an easy to understand way. These solutions for statistics chapter in NCERT class 11 book will help the students to clear all their doubts instantly and let them know the best methods of solving these questions. These NCERT class 11 maths solutions for statistics (chapter 14) given here are available for free and any student can check these. These solutions not only help the students to clear all their doubts but also help them to know the best methods of solving the respective NCERT questions from this chapter.
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NCERT Solutions Class 11 Maths Chapter 15 Exercises
- NCERT Solutions Class 11 Maths Chapter 15 Statistics Mean Median Exercise 15.1
- NCERT Solutions Class 11 Maths Chapter 15 Statistics Mean Median Exercise 15.2
- NCERT Solutions Class 11 Maths Chapter 15 Statistics Mean Median Exercise 15.3
Exercise 15.1
Q1. Calculate the mean deviation about the mean for the given data
5, 8, 9, 10, 11, 13, 14, 18
Sol:
The given data is,
5, 8, 9, 10, 11, 13, 14, 18
Mean,
\(\bar{x} = \frac{5 + 8 + 9 + 10 + 11 + 13 + 14 + 18}{8}\)
\(\bar{x} = \frac{88}{8}\)
\(\bar{x} = 11\)
The deviations of the respective observations from the mean \(\bar{x}\),
\(x_{i} – \bar{x}\) are -6, -3, -2, -1, 0, 2, 3, 7
The absolute values of the deviations,
\(\left |x_{i} – \bar{x} \right |\) are 6, 3, 2, 1, 0 , 2, 3, 7
The required mean deviation about the mean is,
\(M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{8} \left | x_{i} – \bar{x} \right |}{8}\)
\(M.D. \;\left (\bar{x} \right ) = \frac{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}{8}\)
\(M.D. \;\left (\bar{x} \right ) = \frac{24}{8}\) = 3
Q2. Calculate the mean deviation about the mean for the given data
39, 71, 49, 41, 43, 56, 64, 47, 55, 45
Sol:
The given data is,
39, 71, 49, 41, 43, 56, 64, 47, 55, 45
Mean,
\(\bar{x} = \frac{39 + 71 + 49 + 41 + 43 + 56 + 64 + 47 + 55 + 45}{10}\)
\(\bar{x} = \frac{510}{10}\)
\(\bar{x} = 51\)
The deviations of the respective observations from the mean \(\bar{x}\),
\(x_{i} – \bar{x}\) are -12, 20, -2, -10, -8, 5, 13, -4, 4, -6.
The absolute values of the deviations,
\(\left |x_{i} – \bar{x} \right |\) are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6.
The required mean deviation about the mean is,
\(M.D. \;\left (\bar{x} \right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – \bar{x} \right |}{10}\)
\(M.D. \;\left (\bar{x} \right ) = \frac{12 + 20 + 2 + 10 + 8 + 5 + 13 + 4 + 4 + 6}{10}\)
\(M.D. \;\left (\bar{x} \right ) = \frac{84}{10}\) = 8.4
Q3. Calculate the mean deviation about the median for the given data:
13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17
Sol:
The given data is,
13, 16, 11, 10, 11, 12, 17, 14, 13, 16, 18, 17
There are 12 observations which is even number.
Arrange the data in ascending order:
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Median,
\(M = \frac{\left (\frac{12}{2} \right )^{th} observation + \left ( \frac{12}{2} + 1 \right )^{th} observation}{2}\)
= \(\frac{6 ^{th}\; observation + 7^{th}\; observation}{2}\)
= \(\frac{13 + 14}{2}\)
= \(\frac{27}{2}\)
= 13.5
The deviations of the respective observations from the median \(\bar{x}\),
\(x_{i} – M\) 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,
\(\left |x_{i} – M \right |\) 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,
\(M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{12} \left | x_{i} – M \right |}{12}\)
= \( \frac{3.5 + 2.5 + 2.5 + 1.5 + 0.5 + 0.5 + 0.5 + 2.5 + 2.5 + 3.5 + 3.5 + 4.5}{12}\)
= \( \frac{28}{12}\)
= 2.33
Q4. Calculate the mean deviation about the median for the given data:
36, 46, 60, 53, 51, 72, 42, 45, 46, 49
Sol:
The given data is,
36, 46, 60, 53, 51, 72, 42, 45, 46, 49
There are 10 observations which is even number.
Arrange the data in ascending order:
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Median,
\(M = \frac{\left (\frac{10}{2} \right )^{th} observation + \left ( \frac{10}{2} + 1 \right )^{th} observation}{2}\)
= \(\frac{5 ^{th}\; observation + 6^{th}\; observation}{2}\)
= \(\frac{46 + 49}{2}\)
= \(\frac{95}{2}\)
= 47.5
The deviations of the respective observations from the median \(\bar{x}\),
\(x_{i} – M\) 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,
\(\left |x_{i} – M \right |\) 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,
\(M.D. \;\left (M\right ) = \frac{\sum_{i = 1}^{10} \left | x_{i} – M \right |}{10}\)
= \(\frac{11.5 + 5.5 + 2.5 + 1.5 + 1.5 + 1.5 + 3.5 + 5.5 + 12.5 + 24.5}{10}\)
= \(\frac{70}{10}\)
= 7
Q5. Calculate the mean deviation about the mean for the given data
\(x_{i}\) | 5 | 10 | 15 | 20 | 25 |
\(f_{i}\) | 7 | 4 | 6 | 3 | 5 |
Sol:
\(x_{i}\) | \(x_{i}\) | \(x_{i}\) | \(|x_{i} – \bar{x}|\) | \(f_{i} |x_{i} – \bar{x}|\) |
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 |
\(N = \sum_{i = 1}^{5}f_{i} = 25\)
\(\sum_{i = 1}^{5}f_{i}x_{i} = 350\)
Therefore,
\(\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}\)
= \(\frac{1}{25} \times 350\)
= 14
\(MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |\)
= \(\frac{1}{25} \times 158\)
= 6.32
Q6. Calculate the mean deviation about the mean for the given data
\(x_{i}\) | 10 | 10 | 15 | 20 | 25 |
\(f_{i}\) | 7 | 4 | 6 | 3 | 5 |
Sol:
\(x_{i}\) | \(x_{i}\) | \(x_{i}\) | \(|x_{i} – \bar{x}|\) | \(f_{i} |x_{i} – \bar{x}|\) |
10 | 4 | 35 | 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 |
\(N = \sum_{i = 1}^{5}f_{i} = 80\)
\(\sum_{i = 1}^{5}f_{i}x_{i} = 4000\)
Therefore,
\(\bar{x} = \frac{1}{N}\sum_{i = 1}^{5}f_{i}x_{i}\)
= \(\frac{1}{80} \times 4000\)
= 50
\(MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – \bar{x} \right |\)
= \(\frac{1}{80} \times 1280\)
= 16
Q7. Calculate the mean deviation about the median for the given data
\(x_{i}\) | 10 | 10 | 15 | 20 | 25 |
\(f_{i}\) | 7 | 4 | 6 | 3 | 5 |
\(x_{i}\) | 5 | 7 | 9 | 10 | 12 | 15 |
\(f_{i}\) | 8 | 6 | 2 | 2 | 2 | 6 |
Sol:
The observations given are in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:
\(x_{i}\) | \(f_{i}\) | c.f. |
5 | 8 | 8 |
7 | 6 | 14 |
9 | 2 | 16 |
10 | 2 | 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.
\(Median = \frac{13^{th}\; observation + 14^{th} observation}{2}\)
= \(\frac{7 + 7}{2}\)
= 7
The absolute values of the deviations from median,
\(|x_{i} – M|\) | 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 |
\(\sum_{i = 1}^{6}f_{i} = 26\)
\(\sum_{i = 1}^{6}f_{i}|x_{i} – M| = 84\)
Therefore,
\(MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |\)
= \(\frac{1}{26} \times 84\)
= 3.23
Q8. Calculate the mean deviation about the median for the given data
\(x_{i}\) | 15 | 21 | 27 | 30 | 35 |
\(f_{i}\) | 3 | 5 | 6 | 7 | 8 |
Sol:
The observations given are in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the table given below:
\(x_{i}\) | \(f_{i}\) | c.f. |
15 | 3 | 3 |
21 | 5 | 8 |
27 | 6 | 14 |
30 | 7 | 21 |
35 | 8 | 29 |
Here, N = 29, which is odd.
\(Median = \left (\frac{29 + 1}{2} \right )^{th} observation\)
= 15^{th} observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
Therefore,
Median = 30
The absolute values of the deviations from median,
\(|x_{i} – M|\) | 15 | 9 | 3 | 0 | 5 |
\(f_{i}\) | 3 | 5 | 6 | 7 | 8 |
\(f_{i}|x_{i} – M|\) | 45 | 45 | 18 | 0 | 40 |
\(\sum_{i = 1}^{5}f_{i} = 29\)
\(\sum_{i = 1}^{5}f_{i}|x_{i} – M| = 148\)
Therefore,
\(MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{5}f_{i}\left |x_{i} – M\right |\)
= \(\frac{1}{29} \times 148\)
= 5.1
Q9. Calculate the mean deviation about the mean for the given 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 |
Sol:
The given table is formed.
Income per day | Number of persons \(f_{i}\) | Mid – point \(x_{i}\) | \(f_{i}\) \(x_{i}\) | \(|x_{i} – \bar{x}|\) | \(f_{i}|x_{i} – \bar{x}|\) |
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 | 1168 |
700 – 800 | 3 | 750 | 2250 | 392 | 1176 |
50 | 17900 | 7896 |
\(N = \sum_{i = 1}^{8}f_{i} = 50\)
\(\sum_{i = 1}^{8}f_{i}x_{i} = 17900\)
Therefore,
\(\bar{x} = \frac{1}{N}\sum_{i = 1}^{8}f_{i}x_{i}\)
= \(\frac{1}{50} \times 17900\)
= 358
\(MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – \bar{x} \right |\)
= \(\frac{1}{50} \times 7896\)
= 175.92
Q10. Calculate the mean deviation about the mean for the given data:
Height in cm | Number of boys |
95 – 105 | 9 |
105 – 115 | 13 |
115 – 125 | 26 |
125 – 135 | 30 |
135 – 145 | 12 |
145 – 155 | 10 |
Sol:
The given table is obtained:
Height in cm | Number of boys \(f_{i}\) | Mid – point \(x_{i}\) | \(f_{i}\) \(x_{i}\) | \(|x_{i} – \bar{x}|\) | \(f_{i}|x_{i} – \bar{x}|\) |
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 |
\(N = \sum_{i = 1}^{6}f_{i} = 100\)
\(\sum_{i = 1}^{6}f_{i}x_{i} = 12530\)
Therefore,
\(\bar{x} = \frac{1}{N}\sum_{i = 1}^{6}f_{i}x_{i}\)
= \(\frac{1}{100} \times 12530\)
= 125.3
\(MD\left (\bar{x} \right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – \bar{x} \right |\)
= \(\frac{1}{100} \times 1128.8\)
= 11.28
Q11. Calculate the mean deviation about the median for the given data:
Marks | Number of girls |
0 – 10 | 6 |
10 – 20 | 8 |
20 – 30 | 14 |
30 – 40 | 16 |
40 – 50 | 4 |
50 – 60 | 2 |
Sol:
The given table is obtained:
Marks | Number of girls \(f_{i}\) | Cumulative frequency (c.f.) | Mid – point \(x_{i}\) | \(|x_{i} – M|\) | \(f_{i}|x_{i} – \bar{x}|\) |
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 |
50 | 517.1 |
The class interval containing \(\left ( \frac{N}{2} \right )^{th}\) or 25^{th} item is 20 – 30.
Therefore, 20 -30 is the median class.
Now,
\(Median = l + \frac{\frac{N}{2} – C}{f} \times h\)
Where, l = 20
C = 14
f = 14
h = 10
N = 50
\(Median = 20 + \frac{25 – 14}{14} \times 10\)
= \(20 + \frac{110}{14}\)
= 20 + 7.85
= 27.85
Mean deviation about the median is,
\(MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{6}f_{i}\left |x_{i} – M\right |\)
= \(\frac{1}{50} \times 517.1\)
= 10.34
Q12. Calculate the mean deviation about the median for the given data:
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 |
Sol:
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 as given below:
Age | Number \(f_{i}\) | Cumulative frequency (c.f.) | Mid – point \(x_{i}\) | \(|x_{i} – M|\) | \(f_{i}|x_{i} – \bar{x}|\) |
16 – 20 | 5 | 5 | 18 | 20 | 100 |
21 – 25 | 6 | 11 | 23 | 15 | 90 |
26 – 30 | 12 | 23 | 28 | 10 | 120 |
31 – 35 | 14 | 37 | 33 | 5 | 70 |
36 – 40 | 26 | 63 | 38 | 0 | 0 |
41 – 45 | 12 | 75 | 43 | 5 | 60 |
46 – 50 | 16 | 91 | 48 | 10 | 160 |
51 – 55 | 9 | 100 | 53 | 15 | 135 |
100 |
The class interval containing \(\left ( \frac{N}{2} \right )^{th}\) or 50^{th} item is 35.5 – 40.5. Therefore, 35.5 – 40.5 is the median class.
Now,
\(Median = l + \frac{\frac{N}{2} – C}{f} \times h\)
Where, l = 35.5
C = 37
f = 26
h = 5
N = 100
\(Median = 35.5 + \frac{50 – 37}{26} \times 5\)
= \(35.5 + \frac{13 × 5}{26}\)
= 35.5 + 2.5
= 38
Mean deviation about the median is,
\(MD\left (M\right ) = \frac{1}{N}\sum_{i = 1}^{8}f_{i}\left |x_{i} – M\right |\)
= \(\frac{1}{100} \times 735\)
= 7.35
Exercise 15.2
Q1. Calculate mean and variance for the given data 6, 7, 10, 12, 13, 4, 8, 12
Sol:
6, 7, 10, 12, 13, 4, 8, 12
Mean,
\(\bar{x} = \frac{\sum_{i = 1}^{8}x_{i}}{n}\)
= \(\frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8}\)
= \(\frac{72}{8}\)
= 9
The given table is obtained,
\(x_{i}\) | \(\left ( x_{i} – \bar{x} \right )\) | \(\left ( x_{i} – \bar{x} \right )^{2}\) |
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 |
Variance,
\(\left ( \sigma ^{2} \right ) = \frac{1}{n}\sum_{i = 1}^{8}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{8} \times 74\)
= 9.25
Q2. Calculate the mean and variance for the first n natural numbers.
Sol:
The mean of first n natural number is,
\(Mean = \frac{Sum \; of \; all \; observations}{Number \; of \; observations}\)
\(Mean = \frac{\frac{n\left ( n + 1 \right )}{2}}{n}\)
= \(\frac{ n + 1 }{2}\)
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{1}{n}\sum_{i = 1}^{n}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{n}\sum_{i = 1}^{n}\left [ x_{i} – \left ( \frac{n + 1}{2} \right ) \right ]^{2}\)
= \(\frac{1}{n}\sum_{i = 1}^{n} x_{i}^{2} – \frac{1}{n}\sum_{i = 1}^{n}2\left ( \frac{n + 1}{2} \right )x_{i} + \frac{1}{n}\sum_{i = 1}^{n}\left ( \frac{n + 1}{2} \right )^{2}\)
= \(\frac{1}{n}\frac{n\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \left ( \frac{n + 1}{n} \right )\left [ \frac{n\left ( n + 1 \right )}{2} \right ] + \frac{\left ( n + 1 \right )^{2}}{4n} \times n\)
= \( \frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left (n + 1 \right )^{2}}{2} + \frac{\left ( n + 1 \right )^{2}}{4}\)
= \( \frac{\left ( n + 1 \right )\left ( 2n + 1 \right )}{6} – \frac{\left ( n + 1 \right )^{2}}{4}\)
= \(\left ( n + 1 \right )\left [\frac{ 4n + 2 – 3n – 3 }{12}\right ]\)
= \(\frac{\left ( n + 1 \right )\left ( n – 1 \right )}{12}\)
= \(\frac{n^{2} – 1}{12}\)
Q3. Calculate the mean and variance for the first 10 multiples of 3.
Sol:
The first 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Number of observations,
n = 10
Mean,
\(\bar{x} = \frac{\sum_{i = 1}^{10}}{10}\)
= \(\frac{165}{10}\)
= 16.5
The given table is obtained,
\(x_{i}\) | \((x_{i} – \bar{x})\) | \((x_{i} – \bar{x})^{2}\) |
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 |
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{1}{n}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{10} \times 742.5\)
= 74.25
Q4. Calculate 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 |
Sol:
The data obtained is given in the tabular form:
\(x_{i}\) | \(x_{i}\) | \(f_{i}x_{i}\) | \(x_{i} – \bar{x}\) | \((x_{i}-\bar{x})^{2}\) | \(f_{i}(x_{i}-\bar{x})^{2}\) |
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
\(\sum_{i = 1}^{7}f_{i}x_{i}\) = 760
\(\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}\)
= \(\frac{760}{40}\)
= 19
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{40} \times 1736\)
= 43.4
Q5. Calculate 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 |
Sol:
The data obtained is given in the tabular form:
\(x_{i}\) | \(f_{i}\) | \(f_{i}x_{i}\) | \(x_{i} – \bar{x}\) | \((x_{i}-\bar{x})^{2}\) | \(f_{i}(x_{i}-\bar{x})^{2}\) |
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
\(\sum_{i = 1}^{7}f_{i}x_{i}\) = 2200
\(\bar{x} = \frac{\sum_{i = 1}^{7}f_{i}x_{i}}{N}\)
= \(\frac{2200}{22}\)
= 100
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{1}{n}\sum_{i = 1}^{7}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{22} \times 640\)
= 29.09
Q6. Calculate 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 |
Sol:
The data is obtained in tabular form as follows.
\(x_{i}\) | \(f_{i}\) | \(f_{i}=\frac{x_{i}-64}{1}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
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 |
100 | 220 | 0 | 286 |
Mean,
\(\bar{x} = A\frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h\)
= \(64 + \frac{0}{100} \times 1\)
= 64 + 0
= 64
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]\)
= \(\frac{1}{100^{2}}[ 100 \times 286 – 0]\)
= 2.86
Standard deviation,
\(\sigma = \sqrt{2.86}\)
= 1.69
Q7. Calculate the mean and variance for the given 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 |
Sol:
Class | \(f_{i}\) | \({x_{i}}\) | \(y_{i}=\frac{x_{i}-105}{30}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
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,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{7}f_{i}y_{i}}{N} \times h\)
= \(105 + \frac{2}{30} \times 30\)
= 105 + 2
= 107
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right ]\)
= \(\frac{30^{2}}{30^{2}}\left [ 30 \times 76 – \left ( 2 \right )^{2}\right ]\)
= 2280 – 4
= 2276
Q8. Calculate 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 |
Sol:
Class | Frequency \(f_{i}\) | Mid – point \(x_{i}\) | \(y_{i}=\frac{x_{i}-25}{10}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
0 – 10 | 5 | 5 | -2 | 4 | -10 | 20 |
10 – 20 | 8 | 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,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h\)
= \(25 + \frac{10}{50} \times 10\)
= 25 + 2
= 27
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]\)
= \(\frac{10^{2}}{50^{2}}\left [ 50 \times 68 – \left ( 10 \right )^{2}\right ]\)
= \(\frac{1}{25}\left [ 3400 – 100 \right ]\)
= \(\frac{3300}{25}\)
= 132
Q9. Calculate the mean, variance and standard deviation using short – cut method:
Height in cm | Number 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 |
Sol:
Class | \(f_{i}\) | \(x_{i}\) | \(y_{i}=\frac{x_{i}-92.5}{5}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
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 | 4 | 16 | 12 | 48 |
60 | 6 | 254 |
Mean,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{9}f_{i}y_{i}}{N} \times h\)
= \(92.5 + \frac{6}{60} \times 5\)
= 92.5 + 0.5
= 93
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{9}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{9}f_{i}y_{i} \right )^{2}\right ]\)
= \(\frac{5^{2}}{60^{2}}\left [ 60 \times 254 – \left ( 6 \right )^{2}\right ]\)
= \(\frac{25}{3600}\left [ 155204 \right ]\)
= 105.58
Standard deviation,
\(\left (\sigma \right )\)
= \(\sqrt{105.58}\)
= 10.27
Q10. The diameters of circles (in mm) drawn in a design are given below:
Diameters | Number of Children |
33 – 36 | 15 |
37 – 40 | 17 |
41 – 44 | 21 |
45 – 48 | 22 |
49 – 52 | 25 |
Sol:
Class | \(f_{i}\) | \(x_{i}\) | \(y_{i}=\frac{x_{i}-42.5}{4}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
32.5 – 36.5 | 15 | 34.5 | -2 | 4 | -30 | 60 |
36.5 – 40.5 | 17 | 38.5 | -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 |
N = 100
h = 4
Let us assume that mean A be 42.5
Mean,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{5}f_{i}y_{i}}{N} \times h\)
= \(42.5 + \frac{25}{100} \times 4\)
= 43.5
Variance,
\(\left ( \sigma ^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{5}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{5}f_{i}y_{i} \right )^{2}\right ]\)
= \(\frac{4^{2}}{100^{2}}\left [ 100 \times 199 – \left ( 25 \right )^{2}\right ]\)
= \(\frac{16}{10000}\left [ 19900 – 625 \right]\)
= \(\frac{16}{10000} \times 19275\)
= 30.84
Standard deviation,
\(\left (\sigma \right )\)
= \(\sqrt{30.84}\)
= 5.55
Exercise 15.3
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 |
Sol:
The standard deviation of group A is calculated as given in the table below.
Marks | \(f_{i}\) | \(x_{i}\) | \(y_{i}=\frac{x_{i}-45}{10}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
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 | 9 | 27 | 81 |
150 | -6 | 342 |
N = 150
h = 10
A = 45
Mean,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h\)
= \(45 + \frac{(-6)}{150} \times 10\)
= 45 – 0.4
= 44.6
Variance,
\(\left ( \sigma _{1}^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right]\)
= \(\frac{10^{2}}{110^{2}}\left [ 150 \times 342 – \left ( -6 \right )^{2}\right ]\)
= \(\frac{100}{22500}\left [ 51264 \right]\)
= \(\frac{1}{225} \times 51264\)
= 227.84
Standard deviation,
\(\left (\sigma _{1}\right )\)
= \(\sqrt{227.84}\)
= 15.09
The standard deviation of group A is calculated as given in the table below.
Marks | \(f_{i}\) | \(x_{i}\) | \(y_{i}=\frac{x_{i}-45}{10}\) | \(y_{i}^{2}\) | \(f_{i}y_{i}\) | \(f_{i}y_{i}^{2}\) |
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 |
N = 150
h = 10
A = 45
Mean,
\(\bar{x}\)
= \(A + \frac{\sum_{i = 1}^{7}x_{i}}{N} \times h\)
= \(45 + \frac{(-6)}{150} \times 10\)
= 45 – 0.4
= 44.6
Variance,
\(\left ( \sigma _{2}^{2} \right )\)
= \(\frac{h^{2}}{N^{2}}\left [ N\sum_{i = 1}^{7}f_{i}y_{i}^{2} – \left ( \sum_{i = 1}^{7}f_{i}y_{i} \right )^{2}\right]\)
= \(\frac{10^{2}}{110^{2}}\left [ 150 \times 366 – \left ( -6 \right )^{2}\right ]\)
= \(\frac{100}{22500}\left [ 54864 \right]\)
= \(\frac{1}{225} \times 54864\)
= 243.84
Standard deviation,
\(\left (\sigma _{1}\right )\)
= \(\sqrt{243.84}\)
= 15.61
As 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 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 |
Sol:
The prices of the shares X are 35, 54, 52, 53, 56, 58, 52, 50, 51, and 49.
Here, the number of observations, N = 10
Mean,
\(\bar{x}\)
= \(\frac{1}{N}\sum_{i = 1}^{10}x_{i}\)
= \(\frac{1}{10} \times 510\)
= 51
The following table is obtained corresponding to shares X.
\(x_{i}\) | \((x_{i}-\bar{x})\) | \((x_{i}-\bar{x})^{2}\) |
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 |
Variance,
\(\left ( \sigma _{1}^{2} \right )\)
= \(\frac{1}{N}\sum_{i = 1}^{10}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{10} \times 350\)
= 35
Standard deviation,
\(\left (\sigma _{1}\right )\)
= \(\sqrt{35}\)
= 5.91
C.V.
\(\frac{\sigma _{1}}{x} \times 100\)
= \(\frac{5.91}{51} \times 100\)
= 11.58
The prices of share Y are 108, 107, 105, 105, 106, 107, 104, 103, 104, and 101.
Mean,
\(\bar{y}\)
= \(\frac{1}{N}\sum_{i = 1}^{10}y_{i}\)
= \(\frac{1}{10} \times 1050\)
= 105
The following table is obtained corresponding to shares Y.
\(y_{i}\) | \((y_{i}-\bar{y})\) | \((y_{i}-\bar{y})^{2}\) |
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 |
Variance,
\(\left ( \sigma _{2}^{2} \right )\)
= \(\frac{1}{N}\sum_{i = 1}^{10}\left ( y_{i} – \bar{y} \right )^{2}\)
= \(\frac{1}{10} \times 40\)
= 4
Standard deviation,
\(\left (\sigma _{2}\right )\)
= \(\sqrt{4}\)
= 2
C.V.
\(\frac{\sigma _{2}}{y} \times 100\)
= \(\frac{2}{105} \times 100\)
= 1.9
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Therefore, the prices of shares Y are more stable than the prices of shares X.
Q3. 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?
Sol:
(i)
Monthly wages of firm A = Rs. 5253
Number of wage earners in firm A = 586
Therefore,
Total amount paid = Rs. 5253 × 586
Monthly wages of firm B = Rs. 5253
Number of wage earners in firm B = 648
Therefore,
Total amount paid = Rs. 5253 × 648
Hence, 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,
\(A\left ( \sigma _{1}^{2} \right )\) = 100
Therefore,
Standard deviation of the distribution of wages in firm A,
\(A\left ( \sigma _{1} \right ) = \sqrt{100}\) = 10
Variance of the distribution of wages in firm B,
\(B\left ( \sigma _{2}^{2} \right )\) = 121
Therefore,
Standard deviation of the distribution of wages in firm B,
\(B\left ( \sigma _{2} \right ) = \sqrt{121}\) = 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.
Therefore, firm B has greater variability in the individual wages.
Q4. The following is the record of goals scored by team A in a football session:
Number of goals scored | 0 | 1 | 2 | 3 | 4 |
Number of matches | 1 | 9 | 7 | 5 | 3 |
For the team B, mean number of goals scored per match was 2 with standard deviation 1.25 goals. Find which team may be considered more consistent?
Sol:
The mean and the standard deviation of goals scored by team A are calculated as follows.
No. of goals scored | No. of matches | \(f_{i}x_{i}\) | \(x_{i}^{2}\) | \(f_{i}x_{i}^{2}\) |
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 |
23 | 50 | 130 |
Mean,
\(\frac{\sum_{i = 1}^{5}f_{i}x_{i}}{\sum_{i = 1}^{5}f_{i}}\)
= \(\frac{50}{25}\)
= 2
Therefore, the mean of both the teams is same.
Standard deviation,
\(\sigma\)
= \(\frac{1}{N}\sqrt{N\sum f_{i}x_{i}^{2} – \left ( \sum f_{i}x_{i} \right )^{2}}\)
= \(\frac{1}{25}\sqrt{25 \times 130 – \left ( 50 \right )^{2}}\)
= \(\frac{1}{25}\sqrt{750}\)
= \(\frac{1}{25} \times 27.38\)
= 1.09
The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same that is 2. Therefore, the team with lower standard deviation will be more consistent.
Therefore, team A is more consistent than team B.
Q5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are as given below:
\(\sum_{i = 1}^{50}x_{i} = 212\)
\(\sum_{i = 1}^{50}x_{i}^{2} = 902.8\)
\(\sum_{i = 1}^{50}y_{i} = 261\)
\(\sum_{i = 1}^{50}y_{i}^{2} = 1457.6\)
Which is more varying, the length or weight?
Sol:
Length
\(\sum_{i = 1}^{50}x_{i} = 212\)
\(\sum_{i = 1}^{50}x_{i}^{2} = 902.8\)
Here, N = 50
Mean,
\(\bar{x}\)
= \(\frac{\sum_{i = 1}^{50}y_{i}}{N}\)
= \(\frac{212}{50}\)
= 4.24
Variance,
\(\sigma _{1}^{2}\)
= \(\frac{1}{N}\sum_{i = 1}^{50}\left ( x_{i} – \bar{x} \right )^{2}\)
= \(\frac{1}{50}\sum_{i = 1}^{50}\left ( x_{i} – 4.24 \right )^{2}\)
= \(\frac{1}{50}\sum_{i = 1}^{50} \left [x_{i}^{2} – 8.48x_{i} + 17.97 \right ]\)
= \(\frac{1}{50} \left [\sum_{i = 1}^{50}x_{i}^{2} – 8.48\sum_{i = 1}^{50}x_{i} + 17.97 \times 50 \right ]\)
= \(\frac{1}{50} \left [902.8 – 8.48 \times \left (212 \right ) + 898.5 \right ]\)
= \(\frac{1}{50} \left [1801.3 – 1797.76 \right ]\)
= \(\frac{1}{50} \times 3.54\)
= 0.07
Standard deviation,
\(\sigma _{i}\left ( Length \right )\)
= \(\sqrt{0.07}\)
= 0.26
C.V. (Length)
= \(\frac{Standard \; deviation}{Mean} \times 100\)
= \(\frac{0.26}{4.24} \times 100\)
= 6.13
Weight
\(\sum_{i = 1}^{50}y_{i} = 261\)
\(\sum_{i = 1}^{50}y_{i}^{2} = 1457.6\)
Mean,
\(\bar{y}\)
= \(\frac{\sum_{i = 1}^{50}y_{i}}{N}\)
= \(\frac{261}{50}\)
= 5.22
Variance,
\(\sigma _{2}^{2}\)
= \(\frac{1}{N}\sum_{i = 1}^{50}\left ( y_{i} – \bar{y} \right )^{2}\)
= \(\frac{1}{50}\sum_{i = 1}^{50}\left ( y_{i} – 5.22 \right )^{2}\)
= \(\frac{1}{50}\sum_{i = 1}^{50} \left [y_{i}^{2} – 10.44 y_{i} + 27.24 \right ]\)
= \(\frac{1}{50} \left [\sum_{i = 1}^{50}y_{i}^{2} – 10.44 \sum_{i = 1}^{50}y_{i} + 27.24 \times 50 \right ]\)
= \(\frac{1}{50} \left [1457.6 – 10.44 \times \left (261 \right ) + 1362 \right ]\)
= \(\frac{1}{50} \left [2819.6 – 2724.84 \right ]\)
= \(\frac{1}{50} \times 94.76\)
= 1.89
Standard deviation,
\(\sigma _{2}\left ( Weight \right )\)
= \(\sqrt{1.89}\)
= 1.37
C.V. (Weight)
= \(\frac{Standard \; deviation}{Mean} \times 100\)
= \(\frac{1.37}{5.22} \times 100\)
= 26.24
Therefore, C.V. of weights is greater than the C.V. of lengths. Hence, weights vary more than the lengths.
In the statistics chapter of NCERT book for class 11 maths, several important concepts related to statistics is introduced. This chapter is extremely crucial as the concepts of statistics are used in several other topics and have several practical applications.
In this chapter, first the basics of statistics are discussed and the several terminologies are explained like Measures of Dispersion, Range, Mean Deviation, Variance and Standard Deviation, and Analysis of Frequency Distributions. All these topics require the students to understand the in-depth concepts thoroughly and remember the formulas properly.
Several examples are included in the statistics chapter to help the students understand the methodologies for solving the respective questions. Students are also required to solve the different exercise problems that are given in the textbook to be able to develop their problem-solving abilities and develop a deeper understanding of the concepts in statistics. Students can also check these NCERT Solutions For Class 11 Maths Chapter 15 (Statistics) in case of any doubts.
Stay tuned with BYJU’S to get all the solutions for NCERT questions of all the classes for maths and science. At BYJU’S, students are also provided with several notes, sample papers, and question papers to help them learn more effectively. Students are also suggested to download BYJU’S- The Learning App and learn in a more engaging and personalized way from the different video lessons.