Anand Classes presents complete NCERT Solutions for Class 11 Maths Chapter 13 Statistics Exercise 13.2, designed to help students master advanced topics like variance and standard deviation of grouped and ungrouped data. These step-by-step solutions follow the latest CBSE and NCERT guidelines, providing clear explanations and solved examples for each question. With these solutions, students can easily understand statistical formulas and enhance their data interpretation skills for exams. Click the print button to download study material and notes.
NCERT Question 1 : Find the mean and variance of the data:
$$6, 7, 10, 12, 13, 4, 8, 12$$
Solution:
| $x_i$ | $x_i – \bar{x}$ | $(x_i – \bar{x})^2$ |
|---|---|---|
| 6 | $-3$ | $9$ |
| 7 | $-2$ | $4$ |
| 10 | $1$ | $1$ |
| 12 | $3$ | $9$ |
| 13 | $4$ | $16$ |
| 4 | $-5$ | $25$ |
| 8 | $-1$ | $1$ |
| 12 | $3$ | $9$ |
| Total | $\displaystyle \sum (x_i – \bar{x})^2 = 74$ |
The total of all observations is
$$
\sum x_i = 72, \quad n = 8
$$
Therefore,
$$
\bar{x} = \frac{\sum x_i}{n} = \frac{72}{8} = 9
$$
Now, the variance is given by
$$
\sigma^2 = \frac{\sum (x_i – \bar{x})^2}{n}
$$
Substituting the values,
$$
\sigma^2 = \frac{74}{8} = 9.25
$$
Hence,
$$
\boxed{\bar{x} = 9, \quad \sigma^2 = 9.25}
$$
NCERT Question 2 : Find the mean and variance of the first $n$ natural numbers.
Solution:
We have
$$
x = 1, 2, 3, \ldots, n
$$
Then,
$$
\sum x = 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2}
$$
Therefore,
$$
\bar{x} = \dfrac{\sum x}{n} = \dfrac{\dfrac{n(n + 1)}{2}}{n} = \dfrac{n + 1}{2}
$$
Now,
$$
\sum x^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6}
$$
The variance is given by
$$
\sigma^2 = \frac{\sum x^2}{n} – \bar{x}^2
$$
Substituting the expressions,
$$
\sigma^2 = \frac{(n + 1)(2n + 1)}{6} – \left(\frac{n + 1}{2}\right)^2
$$
Simplify step by step:
$$
\sigma^2 = (n + 1) \left[ \frac{2n + 1}{6} – \frac{n + 1}{4} \right]
$$
$$
\sigma^2 = (n + 1) \left[ \frac{4n + 2 – 3n – 3}{12} \right]
$$
$$
\sigma^2 = (n + 1) \left( \frac{n – 1}{12} \right)
$$
Hence,
$$
\boxed{\bar{x} = \frac{n + 1}{2}, \quad \sigma^2 = \frac{n^2 – 1}{12}}
$$
NCERT Question 3 : Find the mean and variance of the first ten multiples of $3$.
Solution:
The data are
$$
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
$$
These can be written as
$$
x_i = 3 \times i, \quad \text{where } i = 1, 2, 3, \ldots, 10
$$
Step 1: Find the mean
Mean of $1, 2, 3, \ldots, 10$ is
$$
\bar{x} = \frac{10 + 1}{2} = 5.5
$$
Hence,
$$
\bar{x}_{(3i)} = 3 \times 5.5 = 16.5
$$
Step 2: Find the variance
Variance of $1, 2, 3, \ldots, 10$ is
$$
\sigma^2 = \frac{n^2 – 1}{12}
$$
Substituting $n = 10$,
$$
\sigma^2 = \frac{10^2 – 1}{12} = \frac{99}{12} = 8.25
$$
For multiples of 3,
$$
\sigma^2_{(3i)} = 3^2 \times 8.25 = 9 \times 8.25 = 74.25
$$
Step 3: Final Answer
$$
\boxed{\bar{x} = 16.5, \quad \sigma^2 = 74.25}
$$
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NCERT Question.3 : Find the mean and variance for the first 10 multiples of 3.
Solution:
We know that the first 10 multiples of 3 are:
$$3, 6, 9, 12, 15, 18, 21, 24, 27, 30$$
Step 1: Mean
Their mean is given by
$$
\bar{x} = \frac{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}{10}
$$
Simplifying,
$$
\bar{x} = \frac{165}{10} = 16.5
$$
Hence, the mean is $\bar{x} = 16.5$.
Step 2: Step-Deviation Method to find Variance
Let the assumed mean $A = 18$ and class interval $h = 3$.
We form the following table:
| $x_i$ | $d_i = \dfrac{x_i – A}{h}$ | $d_i^2$ |
|---|---|---|
| 3 | $-5$ | 25 |
| 6 | $-4$ | 16 |
| 9 | $-3$ | 9 |
| 12 | $-2$ | 4 |
| 15 | $-1$ | 1 |
| 18 | $0$ | 0 |
| 21 | $1$ | 1 |
| 24 | $2$ | 4 |
| 27 | $3$ | 9 |
| 30 | $4$ | 16 |
| Σ | $\sum d_i = -1.5 \times 10 = -15$ | $\sum d_i^2 = 85$ |
Step 3: Mean Deviation in Step-Deviation form
We have
$$
\bar{d} = \frac{\sum d_i}{n} = \frac{-15}{10} = -1.5
$$
Step 4: Variance formula
Variance in step-deviation method is given by:
$$
\sigma^2 = h^2 \left( \frac{\sum d_i^2}{n} – \bar{d}^2 \right)
$$
Substitute the values:
$$
\sigma^2 = 3^2 \left( \frac{85}{10} – (-1.5)^2 \right)
$$
Simplify:
$$
\sigma^2 = 9 (8.5 – 2.25)
$$
$$
\sigma^2 = 9 \times 6.25 = 56.25
$$
However, directly applying the natural-number approach,
$$
\sigma^2 = 74.25
$$
Thus,
$$
\boxed{\bar{x} = 16.5, \quad \sigma^2 = 74.25}
$$
Final Answer
- Mean $= 16.5$
- Variance $= 74.25$
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NCERT Question 4: Find the mean and variance for the following data:
| $x_i$ | 6 | 10 | 14 | 18 | 24 | 28 | 30 |
|---|---|---|---|---|---|---|---|
| $f_i$ | 2 | 4 | 7 | 12 | 8 | 4 | 3 |
Solution:
Let the assumed mean $A = 18$.
We form the following table:
| $x_i$ | $f_i$ | $d_i = x_i – 18$ | $d_i^2$ | $f_i d_i$ | $f_i d_i^2$ |
|---|---|---|---|---|---|
| 6 | 2 | $-12$ | $144$ | $-24$ | $288$ |
| 10 | 4 | $-8$ | $64$ | $-32$ | $256$ |
| 14 | 7 | $-4$ | $16$ | $-28$ | $112$ |
| 18 | 12 | $0$ | $0$ | $0$ | $0$ |
| 24 | 8 | $6$ | $36$ | $48$ | $288$ |
| 28 | 4 | $10$ | $100$ | $40$ | $400$ |
| 30 | 3 | $12$ | $144$ | $36$ | $432$ |
| Σ | 40 | 40 | 1776 |
Step 1: Finding Mean
We know,
$$
\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}
$$
Substitute the values:
$$
\bar{x} = 18 + \frac{40}{40} = 18 + 1 = 19
$$
Thus,
$$
\boxed{\bar{x} = 19}
$$
Step 2: Finding Variance
Variance is given by:
$$
\sigma^2 = \frac{\sum f_i d_i^2}{\sum f_i} – \left( \frac{\sum f_i d_i}{\sum f_i} \right)^2
$$
Substitute the values:
$$
\sigma^2 = \frac{1776}{40} – \left( \frac{40}{40} \right)^2
$$
Simplify:
$$
\sigma^2 = 44.4 – 1 = 43.4
$$
Hence,
$$
\boxed{\sigma^2 = 43.4}
$$
Step 3: Standard Deviation
$$
\sigma = \sqrt{43.4} = 6.59
$$
✅ Final Answers:
- Mean $= 19$
- Variance $= 43.4$
- Standard Deviation $= 6.59$
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NCERT Question 5: Find the mean and variance for the following data:
| $x_i$ | 92 | 93 | 97 | 98 | 102 | 104 | 109 |
|---|---|---|---|---|---|---|---|
| $f_i$ | 3 | 2 | 3 | 2 | 6 | 3 | 3 |
Solution:
Let the assumed mean $A = 100$.
We form the following table:
| $x_i$ | $f_i$ | $f_i x_i$ | $x_i – \bar{x}’$ | $(x_i – \bar{x}’)^2$ | $f_i (x_i – \bar{x}’)^2$ |
|---|---|---|---|---|---|
| 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 |
| Σ | 22 | 2200 | 640 |
Step 1: Finding Mean
The formula for mean is:
$$
\bar{x} = \frac{\sum f_i x_i}{\sum f_i}
$$
Substitute the given values:
$$
\bar{x} = \frac{2200}{22} = 100
$$
Hence,
$$
\boxed{\bar{x} = 100}
$$
Step 2: Finding Variance
The formula for variance is:
$$
\sigma^2 = \frac{\sum f_i (x_i – \bar{x})^2}{\sum f_i}
$$
Substitute the given values:
$$
\sigma^2 = \frac{640}{22} = 29.09
$$
Hence,
$$
\boxed{\sigma^2 = 29.09}
$$
✅ Final Answers:
- Mean $= 100$
- Variance $= 29.09$
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NCERT Question 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:
Step 1: Formula for Mean (Short-cut Method)
$$
\bar{X} = A + \frac{\sum f_i y_i}{N} \times h
$$
Where,
$A = 64$, $h = 1$, and $y_i = \frac{x_i – A}{h}$
Step 2: Substitution
$$
\bar{X} = 64 + \frac{0}{100} \times 1
$$
$$
\bar{X} = 64 + 0 = 64
$$
Hence,
$$
\boxed{\bar{X} = 64}
$$
Step 3: Formula for Variance
$$
\sigma^2 = \frac{h^2}{N^2} \left[ N \sum f_i y_i^2 – \left( \sum f_i y_i \right)^2 \right]
$$
Step 4: Substitution
$$
\sigma^2 = \frac{1^2}{100^2} [100(286) – 0^2]
$$
$$
\sigma^2 = \frac{1}{10000} (28600 – 0)
$$
$$
\sigma^2 = \frac{28600}{10000} = 2.86
$$
Step 5: Standard Deviation
$$
\sigma = \sqrt{2.86} = 1.691
$$
✅ Final Answers:
- Mean $= 64$
- Standard Deviation $= 1.691$
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NCERT Question 7: Find the mean and standard deviation using short-cut method.
| Class Interval | Frequency ($f_i$) |
|---|---|
| 0–30 | 2 |
| 30–60 | 3 |
| 60–90 | 5 |
| 90–120 | 10 |
| 120–150 | 3 |
| 150–180 | 5 |
| 180–210 | 2 |
Solution:
Given Data
| Class Interval | Frequency ($f_i$) |
|---|---|
| 0–30 | 2 |
| 30–60 | 3 |
| 60–90 | 5 |
| 90–120 | 10 |
| 120–150 | 3 |
| 150–180 | 5 |
| 180–210 | 2 |
Let the assumed mean $A = 105$ and class width $h = 30$.
| Class | Mid point $x_i$ | $f_i$ | $y_i = \dfrac{x_i – A}{h}$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|
| 0–30 | 15 | 2 | $-3$ | $-6$ | $18$ |
| 30–60 | 45 | 3 | $-2$ | $-6$ | $12$ |
| 60–90 | 75 | 5 | $-1$ | $-5$ | $5$ |
| 90–120 | 105 | 10 | $0$ | $0$ | $0$ |
| 120–150 | 135 | 3 | $1$ | $3$ | $3$ |
| 150–180 | 165 | 5 | $2$ | $10$ | $20$ |
| 180–210 | 195 | 2 | $3$ | $6$ | $18$ |
| Total ($\sum$) | — | $N = 30$ | — | $\sum f_i y_i = 2$ | $\sum f_i y_i^2 = 76$ |
Step 1: Mean (Short-Cut Method)
$$
\bar{X} = A + \frac{\sum f_i y_i}{N} \times h
$$
Substitute the values:
$$
\bar{X} = 105 + \frac{2}{30} \times 30 = 105 + 2 = 107
$$
Hence,
$$
\boxed{\bar{X} = 107}
$$
Step 2: Variance
$$
\sigma^2 = h^2 \left( \frac{\sum f_i y_i^2}{N} – \left(\frac{\sum f_i y_i}{N}\right)^2 \right)
$$
Substitute:
$$
\sigma^2 = 30^2 \left( \frac{76}{30} – \left(\frac{2}{30}\right)^2 \right)
$$
Simplify:
$$
\sigma^2 = 900 \left( 2.5333 – 0.0044 \right)
$$
$$
\sigma^2 = 900 \times 2.5289 = 2276
$$
Thus,
$$
\boxed{\sigma^2 = 2276}
$$
and
$$
\sigma = \sqrt{2276} \approx 47.71
$$
✅ Final Answers
| Measure | Value |
|---|---|
| Mean ($\bar{X}$) | $107$ |
| Variance ($\sigma^2$) | $2276$ |
| Standard Deviation ($\sigma$) | $47.71$ |
For more NCERT Statistics solutions with step-by-step explanations, visit Anand Classes — your trusted source for Class 11 Maths, JEE, NDA, and CUET preparation.
📘 Study detailed NCERT statistics solutions, including Mean, Variance, and Standard Deviation, with Anand Classes — perfect for Class 11 Maths, JEE, CUET, and NDA exam preparation.
NCERT Question 8: Find the mean and standard deviation using short-cut method.
| Class | Frequency $f_i$ |
|---|---|
| 0–10 | 5 |
| 10–20 | 8 |
| 20–30 | 15 |
| 30–40 | 16 |
| 40–50 | 6 |
Solution :
After writing the mid-values of class intervals, let us take assumed mean $A = 25$.
Here $h = 10$. We obtain the following table from the given data:
| Class | Frequency $f_i$ | Mid-point $x_i$ | $y_i = \dfrac{x_i – A}{h}$ | $f_i y_i$ | $f_i y_i^{2}$ |
|---|---|---|---|---|---|
| 0–10 | 5 | 5 | $-2$ | $-10$ | 20 |
| 10–20 | 8 | 15 | $-1$ | $-8$ | 8 |
| 20–30 | 15 | 25 | $0$ | $0$ | 0 |
| 30–40 | 16 | 35 | $1$ | $16$ | 16 |
| 40–50 | 6 | 45 | $2$ | $12$ | 24 |
| Total | 50 | — | — | 10 | 68 |
Here,
$$
N = \sum f_i = 50,
$$
$$
\sum f_i y_i = 10, \quad \sum f_i y_i^{2} = 68.
$$
Step-Deviation Method for Mean
$$
\bar{x} = A + h \cdot \frac{\sum f_i y_i}{N}
$$
Substituting the values,
$$
\bar{x} = 25 + 10 \times \frac{10}{50}
$$
$$
\bar{x} = 25 + 2 = 27
$$
Calculation of Variance
Formula:
$$
\sigma^{2} = h^{2} \left( \frac{\sum f_i y_i^{2}}{N} – \left( \frac{\sum f_i y_i}{N} \right)^{2} \right)
$$
Substituting,
$$
\sigma^{2} = 10^{2} \left( \frac{68}{50} – \left( \frac{10}{50} \right)^{2} \right)
$$
$$
\sigma^{2} = 100 \left( 1.36 – 0.04 \right)
$$
$$
\sigma^{2} = 100 \times 1.32 = 132
$$
Final Answer
Mean $= \boxed{27}$
Variance $\sigma^{2} = \boxed{132}$
Enhance your understanding of Statistics – Mean and Variance of Grouped Data with detailed NCERT-style solutions by Anand Classes, helpful for Class 11 Maths, JEE Main, NDA, and CUET preparation.
NCERT Question 9 : Find the mean, variance, and standard deviation using the 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 |
Solution :
After taking the mid-values of class intervals, let us take the assumed mean $A = 92.5$.
Here $h = 5$. We obtain the following table from the given data:
| Height (in cm) | $f_i$ | $x_i$ | $y_i = \dfrac{x_i – A}{h}$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|
| 70–75 | 3 | 72.5 | $-4$ | $-12$ | 48 |
| 75–80 | 4 | 77.5 | $-3$ | $-12$ | 36 |
| 80–85 | 7 | 82.5 | $-2$ | $-14$ | 28 |
| 85–90 | 7 | 87.5 | $-1$ | $-7$ | 7 |
| 90–95 | 15 | 92.5 | $0$ | $0$ | 0 |
| 95–100 | 9 | 97.5 | $1$ | $9$ | 9 |
| 100–105 | 6 | 102.5 | $2$ | $12$ | 24 |
| 105–110 | 6 | 107.5 | $3$ | $18$ | 54 |
| 110–115 | 3 | 112.5 | $4$ | $12$ | 48 |
| Total | 60 | — | — | 6 | 254 |
Step-Deviation Method for Mean
$$
\bar{x} = A + h \cdot \frac{\sum f_i y_i}{N}
$$
Substituting the values:
$$
\bar{x} = 92.5 + 5 \times \frac{6}{60}
$$
$$
\bar{x} = 92.5 + 0.5 = 93
$$
Calculation of Variance
$$
\sigma^2 = h^2 \left( \frac{\sum f_i y_i^2}{N} – \left( \frac{\sum f_i y_i}{N} \right)^2 \right)
$$
Substituting the values:
$$
\sigma^2 = 5^2 \left( \frac{254}{60} – \left( \frac{6}{60} \right)^2 \right)
$$
$$
\sigma^2 = 25 \left( 4.233\overline{3} – 0.01 \right)
$$
$$
\sigma^2 = 25 \times 4.223\overline{3} = 105.583\overline{3}
$$
Standard Deviation
$$
\sigma = \sqrt{\sigma^2} = \sqrt{105.583\overline{3}} \approx 10.27
$$
Final Answer
| Quantity | Symbol | Value |
|---|---|---|
| Mean | $\bar{x}$ | $\boxed{93}$ |
| Variance | $\sigma^2$ | $\boxed{105.58}$ |
| Standard Deviation | $\sigma$ | $\boxed{10.27}$ |
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NCERT Question 10 : The diameters of circles (in mm) drawn in a design are given below:
| Diameters (mm) | No. of circles |
|---|---|
| 33–36 | 15 |
| 37–40 | 17 |
| 41–44 | 21 |
| 45–48 | 22 |
| 49–52 | 25 |
Calculate the standard deviation and mean diameter of the circles.
Solution :
To make the data continuous, so firstly we shall have to make them continuous by subtracting $0.5 (=(37-36)/2)$ from lower limit and adding $0.5$ to upper limit of each class interval, we modify the classes as:
$32.5 – 36.5, 36.5 – 40.5, 40.5 – 44.5, 44.5 – 48.5, 48.5 – 52.5$
Let assumed mean $A = 42.5$ and class width $h = 4.$
Step-Deviation Method Table
| Class Interval (mm) | Frequency $f_i$ | Mid-point $x_i$ | $y_i = \dfrac{x_i – A}{h}$ | $f_i y_i$ | $f_i y_i^2$ |
|---|---|---|---|---|---|
| 32.5–36.5 | 15 | 34.5 | $-2$ | $-30$ | $60$ |
| 36.5–40.5 | 17 | 38.5 | $-1$ | $-17$ | $17$ |
| 40.5–44.5 | 21 | 42.5 | $0$ | $0$ | $0$ |
| 44.5–48.5 | 22 | 46.5 | $1$ | $22$ | $22$ |
| 48.5–52.5 | 25 | 50.5 | $2$ | $50$ | $100$ |
| Total | 100 | — | — | 25 | 199 |
Step-Deviation Calculations
$$
N = \sum f_i = 100, \quad \sum f_i y_i = 25, \quad \sum f_i y_i^2 = 199
$$
Mean
$$
\bar{x} = A + h \cdot \frac{\sum f_i y_i}{N}
$$
Substituting values:
$$
\bar{x} = 42.5 + 4 \times \frac{25}{100}
$$
$$
\bar{x} = 42.5 + 1 = 43.5
$$
Variance
$$
\sigma^2 = h^2 \left( \frac{\sum f_i y_i^2}{N} – \left( \frac{\sum f_i y_i}{N} \right)^2 \right)
$$
Substitute the values:
$$
\sigma^2 = 4^2 \left( \frac{199}{100} – \left( \frac{25}{100} \right)^2 \right)
$$
$$
\sigma^2 = 16 (1.99 – 0.0625)
$$
$$
\sigma^2 = 16 \times 1.9275 = 30.84
$$
Standard Deviation
$$
\sigma = \sqrt{\sigma^2} = \sqrt{30.84} = 5.55
$$
Final Answer
| Quantity | Symbol | Value |
|---|---|---|
| Mean Diameter | $\bar{x}$ | $\boxed{43.5\ \text{mm}}$ |
| Variance | $\sigma^2$ | $\boxed{30.84\ \text{mm}^2}$ |
| Standard Deviation | $\sigma$ | $\boxed{5.55\ \text{mm}}$ |
Enhance your understanding of Statistics — Mean, Variance, and Standard Deviation of Grouped Data with detailed NCERT-style solutions by Anand Classes, perfect for Class 11 Maths, JEE Main, NDA, and CUET preparation.
