Rbse Solutions for Class 11 maths Chapter 13 Exercise 13.2 | Mean and Variance

This exercise focuses on calculating the mean ($\bar{x}$), variance ($\sigma^2$), and standard deviation ($\sigma$) for different types of data, including ungrouped data, discrete frequency distributions, and continuous frequency distributions.

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Mean and Variance for Ungrouped Data (Q. 1–3)

The formulas for ungrouped data are:

$$\bar{x} = \frac{\sum x_i}{n} \quad \text{and} \quad \sigma^2 = \frac{1}{n} \sum (x_i – \bar{x})^2$$

1. Data: $6, 7, 10, 12, 13, 4, 8, 12$

Here, $n=8$.

1. Mean ($\bar{x}$):$$\bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = \mathbf{9}$$
2. Variance ($\sigma^2$):| $x_i$ | $x_i – \bar{x} = x_i – 9$ | $(x_i – \bar{x})^2$ || :—: | :—: | :—: || 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 | 0 | $\sum (x_i – \bar{x})^2 = 74$ |$$\sigma^2 = \frac{74}{8} = \mathbf{9.25}$$

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2. First $n$ natural numbers: $1, 2, 3, \dots, n$

1. Mean ($\bar{x}$):$$\bar{x} = \frac{\sum x_i}{n} = \frac{n(n+1)/2}{n} = \mathbf{\frac{n+1}{2}}$$
2. Variance ($\sigma^2$):The standard formula for the variance of the first $n$ natural numbers is:$$\sigma^2 = \mathbf{\frac{n^2 – 1}{12}}$$

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3. First 10 multiples of 3: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30$

Here, $n=10$.

1. Mean ($\bar{x}$):The sum is $3 \times (1 + 2 + \dots + 10) = 3 \times \frac{10(11)}{2} = 165$.$$\bar{x} = \frac{165}{10} = \mathbf{16.5}$$
2. Variance ($\sigma^2$):| $x_i$ | $x_i – \bar{x}$ | $(x_i – \bar{x})^2$ || :—: | :—: | :—: || 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 || Total | 0 | $\sum (x_i – \bar{x})^2 = 742.5$ |$$\sigma^2 = \frac{742.5}{10} = \mathbf{74.25}$$

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Mean and Variance for Discrete Frequency Distribution (Q. 4–5)

The formulas for discrete frequency distribution are:

$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} \quad \text{and} \quad \sigma^2 = \frac{1}{N} \sum f_i (x_i – \bar{x})^2$$

where $N = \sum f_i$.

4. Mean and Variance

xi​	fi​	fi​xi​	xi​−xˉ=xi​−19.5	(xi​−xˉ)2	fi​(xi​−xˉ)2
6	2	12	-13.5	182.25	364.5
10	4	40	-9.5	90.25	361.0
14	7	98	-5.5	30.25	211.75
18	12	216	-1.5	2.25	27.0
24	8	192	4.5	20.25	162.0
28	4	112	8.5	72.25	289.0
30	3	90	10.5	110.25	330.75
Total	$N=40$	$\sum f_i x_i = 760$			$\sum f_i(x_i – \bar{x})^2 = 1746.0$

1. Mean ($\bar{x}$):$$\bar{x} = \frac{760}{40} = \mathbf{19}$$(Correction: Calculations above use $\bar{x}=19.5$ based on a previous attempt, recalculating with $\bar{x}=19$)xi​fi​xi​−19(xi​−19)2fi​(xi​−19)262-13169338104-981324147-5251751812-111224852520028498132430311121363Total40$\sum f_i(x_i – \bar{x})^2 = 1736$
2. Variance ($\sigma^2$):$$\sigma^2 = \frac{1736}{40} = \mathbf{43.4}$$

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5. Mean and Variance

xi​	fi​	fi​xi​	xi​−xˉ=xi​−100	(xi​−xˉ)2	fi​(xi​−xˉ)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
Total	$N=22$	$\sum f_i x_i = 2200$			$\sum f_i(x_i – \bar{x})^2 = 640$

1. Mean ($\bar{x}$):$$\bar{x} = \frac{2200}{22} = \mathbf{100}$$
2. Variance ($\sigma^2$):$$\sigma^2 = \frac{640}{22} \approx \mathbf{29.09}$$

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Mean and Standard Deviation using Short-Cut Method (Q. 6)

The short-cut method (or Step Deviation Method) uses an assumed mean ($A$) and a common factor ($h$ or $c$) for simplified calculation.

$$\bar{x} = A + \frac{\sum f_i d_i}{N}$$

$$\sigma = \sqrt{\frac{1}{N} \sum f_i d_i^2 – \left(\frac{1}{N} \sum f_i d_i\right)^2}$$

where $d_i = x_i – A$. Let $A=64$.

xi​	fi​	di​=xi​−64	fi​di​	di2​	fi​di2​
60	2	-4	-8	16	32
61	1	-3	-3	9	9
62	12	-2	-24	4	48
63	29	-1	-29	1	29
64	25	0	0	0	0
65	12	1	12	1	12
66	10	2	20	4	40
67	4	3	12	9	36
68	5	4	20	16	80
Total	$N=100$		$\sum f_i d_i = 0$		$\sum f_i d_i^2 = 286$

1. Mean ($\bar{x}$):$$\bar{x} = 64 + \frac{0}{100} = \mathbf{64}$$
2. Standard Deviation ($\sigma$):$$\sigma^2 = \frac{286}{100} – \left(\frac{0}{100}\right)^2 = 2.86$$$$\sigma = \sqrt{2.86} \approx \mathbf{1.69}$$

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Mean and Variance for Continuous Frequency Distribution (Q. 7–8)

7. Mean and Variance

Class	Midpoint (xi​)	fi​	fi​xi​	xi​−xˉ=xi​−105	(xi​−xˉ)2	fi​(xi​−xˉ)2
0-30	15	2	30	-90	8100	16200
30-60	45	3	135	-60	3600	10800
60-90	75	5	375	-30	900	4500
90-120	105	10	1050	0	0	0
120-150	135	3	405	30	900	2700
150-180	165	5	825	60	3600	18000
180-210	195	2	390	90	8100	16200
Total		$N=30$	$\sum f_i x_i = 3210$			$\sum f_i(x_i – \bar{x})^2 = 68400$

1. Mean ($\bar{x}$):$$\bar{x} = \frac{3210}{30} = \mathbf{107}$$
2. Variance ($\sigma^2$):(Correction: Recalculating using the actual mean $\bar{x}=107$.)xi​fi​xi​−107(xi​−107)2fi​(xi​−107)2152-92846416928453-62384411532755-321024512010510-24401353287842352165558336416820195288774415488Total30$\sum f_i(x_i – \bar{x})^2 = 68280$$$\sigma^2 = \frac{68280}{30} = \mathbf{2276}$$

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8. Mean and Variance

Class	Midpoint (xi​)	fi​	fi​xi​	xi​−xˉ=xi​−27	(xi​−xˉ)2	fi​(xi​−xˉ)2
0-10	5	5	25	-22	484	2420
10-20	15	8	120	-12	144	1152
20-30	25	15	375	-2	4	60
30-40	35	16	560	8	64	1024
40-50	45	6	270	18	324	1944
Total		$N=50$	$\sum f_i x_i = 1350$			$\sum f_i(x_i – \bar{x})^2 = 6600$

1. Mean ($\bar{x}$):$$\bar{x} = \frac{1350}{50} = \mathbf{27}$$
2. Variance ($\sigma^2$):$$\sigma^2 = \frac{6600}{50} = \mathbf{132}$$

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Mean, Variance, and Standard Deviation using Short-Cut Method (Q. 9–10)

For continuous data, $d_i = \frac{x_i – A}{h}$, where $h$ is the class size.

9. Short-Cut Method (Height in cms)

Let $A=92.5$ and $h=5$.

Class	xi​	fi​	di​=5xi​−92.5​	fi​di​	di2​	fi​di2​
70-75	72.5	3	-4	-12	16	48
75-80	77.5	4	-3	-12	9	36
80-85	82.5	7	-2	-14	4	28
85-90	87.5	7	-1	-7	1	7
90-95	92.5	15	0	0	0	0
95-100	97.5	9	1	9	1	9
100-105	102.5	6	2	12	4	24
105-110	107.5	6	3	18	9	54
110-115	112.5	3	4	12	16	48
Total		$N=60$		$\sum f_i d_i = 6$		$\sum f_i d_i^2 = 254$

1. Mean ($\bar{x}$):$$\bar{x} = A + h \left(\frac{\sum f_i d_i}{N}\right) = 92.5 + 5 \left(\frac{6}{60}\right) = 92.5 + 5(0.1) = \mathbf{93}$$
2. Variance ($\sigma^2$):$$\sigma^2 = h^2 \left[ \frac{\sum f_i d_i^2}{N} – \left(\frac{\sum f_i d_i}{N}\right)^2 \right]$$$$\sigma^2 = 5^2 \left[ \frac{254}{60} – \left(\frac{6}{60}\right)^2 \right] = 25 \left[ 4.2333 – (0.1)^2 \right]$$$$\sigma^2 = 25 [4.2333 – 0.01] = 25 \times 4.2233 \approx \mathbf{105.58}$$
3. Standard Deviation ($\sigma$):$$\sigma = \sqrt{105.58} \approx \mathbf{10.27}$$

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10. Short-Cut Method (Diameters of Circles)

First, make the data continuous: 32.5-36.5, 36.5-40.5, etc. The class size $h=4$.

Let $A=42.5$.

Class (Continuous)	xi​	fi​	di​=4xi​−42.5​	fi​di​	di2​	fi​di2​
32.5-36.5	34.5	15	-2	-30	4	60
36.5-40.5	38.5	17	-1	-17	1	17
40.5-44.5	42.5	21	0	0	0	0
44.5-48.5	46.5	22	1	22	1	22
48.5-52.5	50.5	25	2	50	4	100
Total		$N=100$		$\sum f_i d_i = 25$		$\sum f_i d_i^2 = 199$

1. Mean ($\bar{x}$):$$\bar{x} = 42.5 + 4 \left(\frac{25}{100}\right) = 42.5 + 4(0.25) = 42.5 + 1 = \mathbf{43.5 \text{ mm}}$$
2. Standard Deviation ($\sigma$):$$\sigma^2 = 4^2 \left[ \frac{199}{100} – \left(\frac{25}{100}\right)^2 \right] = 16 \left[ 1.99 – (0.25)^2 \right]$$$$\sigma^2 = 16 [1.99 – 0.0625] = 16 \times 1.9275 = 30.84$$$$\sigma = \sqrt{30.84} \approx \mathbf{5.55 \text{ mm}}$$