Chapter 13 of 14

Updated 6 August 2026

NCERT Solutions

Statistics — NCERT Solutions

CBSE · Class 11 · Mathematics

NCERT Solutions for Statistics, CBSE Class 11 Mathematics: 28 textbook questions solved step by step. Part of the CBSE Class 11 Mathematics syllabus.

161 questions60 flashcards8 formulas & key relations5 concepts

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28 Questions Solved · 3 Sections

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Exercise 13.1

1Find the mean deviation about the mean for the data: 4, 7, 8, 9, 10, 12, 13, 17Show solution

Given: Data: 4, 7, 8, 9, 10, 12, 13, 17, $n = 8$

Step 1: Find the Mean
$\bar{x} = \frac{4+7+8+9+10+12+13+17}{8} = \frac{80}{8} = 10$

Step 2: Find $|x_i - \bar{x}|$ for each observation

$x_i$	$\|x_i - \bar{x}\|$
4	6
7	3
8	2
9	1
10	0
12	2
13	3
17	7
Total	24

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum|x_i - \bar{x}|}{n} = \frac{24}{8} = 3$

Answer: Mean Deviation about the mean $= 3$

2Find the mean deviation about the mean for the data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44Show solution

Given: Data: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44, $n = 10$

Step 1: Find the Mean
$\bar{x} = \frac{38+70+48+40+42+55+63+46+54+44}{10} = \frac{500}{10} = 50$

Step 2: Find $|x_i - \bar{x}|$ for each observation

$x_i$	$\|x_i - 50\|$
38	12
70	20
48	2
40	10
42	8
55	5
63	13
46	4
54	4
44	6
Total	84

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum|x_i - \bar{x}|}{n} = \frac{84}{10} = 8.4$

Answer: Mean Deviation about the mean $= 8.4$

3Find the mean deviation about the median for the data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17Show solution

Given: Data: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17, $n = 12$

Step 1: Arrange in ascending order
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Step 2: Find the Median
Since $n = 12$ (even),
$\text{Median} = \frac{\text{6th term} + \text{7th term}}{2} = \frac{13 + 14}{2} = 13.5$

Step 3: Find $|x_i - M|$ for each observation

$x_i$	$\|x_i - 13.5\|$
10	3.5
11	2.5
11	2.5
12	1.5
13	0.5
13	0.5
14	0.5
16	2.5
16	2.5
17	3.5
17	3.5
18	4.5
Total	28

Step 4: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum|x_i - M|}{n} = \frac{28}{12} = 2.33$

Answer: Mean Deviation about the median $\approx 2.33$

4Find the mean deviation about the median for the data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49Show solution

Given: Data: 36, 72, 46, 42, 60, 45, 53, 46, 51, 49, $n = 10$

Step 1: Arrange in ascending order
36, 42, 45, 46, 46, 49, 51, 53, 60, 72

Step 2: Find the Median
Since $n = 10$ (even),
$\text{Median} = \frac{\text{5th term} + \text{6th term}}{2} = \frac{46 + 49}{2} = 47.5$

Step 3: Find $|x_i - M|$ for each observation

$x_i$	$\|x_i - 47.5\|$
36	11.5
42	5.5
45	2.5
46	1.5
46	1.5
49	1.5
51	3.5
53	5.5
60	12.5
72	24.5
Total	70

Step 4: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum|x_i - M|}{n} = \frac{70}{10} = 7$

Answer: Mean Deviation about the median $= 7$

5Find the mean deviation about the mean for the data:

x_i

: 5, 10, 15, 20, 25

f_i

: 7, 4, 6, 3, 5Show solution

Given: Discrete frequency distribution with $N = \sum f_i = 7+4+6+3+5 = 25$

Step 1: Find the Mean

$x_i$	$f_i$	$f_i x_i$
5	7	35
10	4	40
15	6	90
20	3	60
25	5	125
Total	25	350

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{350}{25} = 14$

Step 2: Find $f_i|x_i - \bar{x}|$

$x_i$	$f_i$	$\|x_i - 14\|$	$f_i\|x_i - 14\|$
5	7	9	63
10	4	4	16
15	6	1	6
20	3	6	18
25	5	11	55
Total	25		158

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{158}{25} = 6.32$

Answer: Mean Deviation about the mean $= 6.32$

6Find the mean deviation about the mean for the data:

x_i

: 10, 30, 50, 70, 90

f_i

: 4, 24, 28, 16, 8Show solution

Given: Discrete frequency distribution with $N = \sum f_i = 4+24+28+16+8 = 80$

Step 1: Find the Mean

$x_i$	$f_i$	$f_i x_i$
10	4	40
30	24	720
50	28	1400
70	16	1120
90	8	720
Total	80	4000

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{4000}{80} = 50$

Step 2: Find $f_i|x_i - \bar{x}|$

$x_i$	$f_i$	$\|x_i - 50\|$	$f_i\|x_i - 50\|$
10	4	40	160
30	24	20	480
50	28	0	0
70	16	20	320
90	8	40	320
Total	80		1280

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{1280}{80} = 16$

Answer: Mean Deviation about the mean $= 16$

7Find the mean deviation about the median for the data:

x_i

: 5, 7, 9, 10, 12, 15

f_i

: 8, 6, 2, 2, 2, 6Show solution

Given: $N = \sum f_i = 8+6+2+2+2+6 = 26$

Step 1: Find the Median

Compute cumulative frequencies:

$x_i$	$f_i$	Cumulative Frequency
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26

$N = 26$ , so $\frac{N}{2} = 13$ .

The 13th and 14th observations both lie in the $x_i = 7$ group (cumulative frequency reaches 14 at $x_i = 7$ ).

$\text{Median} = \frac{7+7}{2} = 7$

Step 2: Find $f_i|x_i - M|$

$x_i$	$f_i$	$\|x_i - 7\|$	$f_i\|x_i - 7\|$
5	8	2	16
7	6	0	0
9	2	2	4
10	2	3	6
12	2	5	10
15	6	8	48
Total	26		84

Step 3: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{84}{26} = \frac{42}{13} \approx 3.23$

Answer: Mean Deviation about the median $\approx 3.23$

8Find the mean deviation about the median for the data:

x_i

: 15, 21, 27, 30, 35

f_i

: 3, 5, 6, 7, 8Show solution

Given: $N = \sum f_i = 3+5+6+7+8 = 29$

Step 1: Find the Median

Compute cumulative frequencies:

$x_i$	$f_i$	Cumulative Frequency
15	3	3
21	5	8
27	6	14
30	7	21
35	8	29

$\frac{N+1}{2} = \frac{30}{2} = 15$ th observation.

The 15th observation lies in the group where $x_i = 30$ (cumulative frequency 21 covers positions 15 to 21).

$\text{Median} = 30$

Step 2: Find $f_i|x_i - M|$

$x_i$	$f_i$	$\|x_i - 30\|$	$f_i\|x_i - 30\|$
15	3	15	45
21	5	9	45
27	6	3	18
30	7	0	0
35	8	5	40
Total	29		148

Step 3: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{148}{29} \approx 5.1$

Answer: Mean Deviation about the median $\approx 5.1$

9Find the mean deviation about the mean for the data:
Income per day (₹): 0-100, 100-200, 200-300, 300-400, 400-500, 500-600, 600-700, 700-800
Number of persons: 4, 8, 9, 10, 7, 5, 4, 3Show solution

Given: Continuous frequency distribution. $N = 4+8+9+10+7+5+4+3 = 50$

Step 1: Find midpoints and compute mean

Class	$x_i$ (mid)	$f_i$	$f_i x_i$
0–100	50	4	200
100–200	150	8	1200
200–300	250	9	2250
300–400	350	10	3500
400–500	450	7	3150
500–600	550	5	2750
600–700	650	4	2600
700–800	750	3	2250
Total		50	17900

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{17900}{50} = 358$

Step 2: Find $f_i|x_i - \bar{x}|$

$x_i$	$f_i$	$\|x_i - 358\|$	$f_i\|x_i - 358\|$
50	4	308	1232
150	8	208	1664
250	9	108	972
350	10	8	80
450	7	92	644
550	5	192	960
650	4	292	1168
750	3	392	1176
Total	50		7896

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{7896}{50} = 157.92$

Answer: Mean Deviation about the mean $= 157.92$

10Find the mean deviation about the mean for the data:
Height (cms): 95-105, 105-115, 115-125, 125-135, 135-145, 145-155
Number of boys: 9, 13, 26, 30, 12, 10Show solution

Given: Continuous frequency distribution. $N = 9+13+26+30+12+10 = 100$

Step 1: Find midpoints and compute mean

Class	$x_i$ (mid)	$f_i$	$f_i x_i$
95–105	100	9	900
105–115	110	13	1430
115–125	120	26	3120
125–135	130	30	3900
135–145	140	12	1680
145–155	150	10	1500
Total		100	12530

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{12530}{100} = 125.3$

Step 2: Find $f_i|x_i - \bar{x}|$

$x_i$	$f_i$	$\|x_i - 125.3\|$	$f_i\|x_i - 125.3\|$
100	9	25.3	227.7
110	13	15.3	198.9
120	26	5.3	137.8
130	30	4.7	141.0
140	12	14.7	176.4
150	10	24.7	247.0
Total	100		1128.8

Step 3: Calculate Mean Deviation
$\text{M.D.}(\bar{x}) = \frac{\sum f_i|x_i - \bar{x}|}{N} = \frac{1128.8}{100} = 11.288$

Answer: Mean Deviation about the mean $\approx 11.28$

11Find the mean deviation about median for the following data:
Marks: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60
Number of Girls: 6, 8, 14, 16, 4, 2Show solution

Given: Continuous frequency distribution. $N = 6+8+14+16+4+2 = 50$

Step 1: Find the Median

Compute cumulative frequencies:

Class	$f_i$	Cumulative Frequency
0–10	6	6
10–20	8	14
20–30	14	28
30–40	16	44
40–50	4	48
50–60	2	50

$\frac{N}{2} = 25$ . The cumulative frequency just exceeding 25 is 28, so the median class is 20–30.

Using the formula:
$\text{Median} = l + \frac{\frac{N}{2} - C}{f} \times h$

Here $l = 20$ , $C = 14$ , $f = 14$ , $h = 10$ :
$M = 20 + \frac{25 - 14}{14} \times 10 = 20 + \frac{110}{14} = 20 + 7.857 \approx 27.86$

Step 2: Find midpoints and $f_i|x_i - M|$

Class	$x_i$	$f_i$	$\|x_i - 27.86\|$	$f_i\|x_i - 27.86\|$
0–10	5	6	22.86	137.16
10–20	15	8	12.86	102.88
20–30	25	14	2.86	40.04
30–40	35	16	7.14	114.24
40–50	45	4	17.14	68.56
50–60	55	2	27.14	54.28
Total		50		517.16

Step 3: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{517.16}{50} \approx 10.34$

Answer: Mean Deviation about the median $\approx 10.34$

12Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Age (in years): 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55
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]Show solution

Given: $N = 5+6+12+14+26+12+16+9 = 100$

Step 1: Convert to continuous classes (subtract 0.5 from lower, add 0.5 to upper)

Class (continuous)	$x_i$ (mid)	$f_i$	Cumulative Frequency
15.5–20.5	18	5	5
20.5–25.5	23	6	11
25.5–30.5	28	12	23
30.5–35.5	33	14	37
35.5–40.5	38	26	63
40.5–45.5	43	12	75
45.5–50.5	48	16	91
50.5–55.5	53	9	100

Step 2: Find the Median

$\frac{N}{2} = 50$ . The cumulative frequency just exceeding 50 is 63, so the median class is 35.5–40.5.

$M = l + \frac{\frac{N}{2} - C}{f} \times h = 35.5 + \frac{50 - 37}{26} \times 5 = 35.5 + \frac{65}{26} = 35.5 + 2.5 = 38$

Step 3: Find $f_i|x_i - M|$

$x_i$	$f_i$	$\|x_i - 38\|$	$f_i\|x_i - 38\|$
18	5	20	100
23	6	15	90
28	12	10	120
33	14	5	70
38	26	0	0
43	12	5	60
48	16	10	160
53	9	15	135
Total	100		735

Step 4: Calculate Mean Deviation
$\text{M.D.}(M) = \frac{\sum f_i|x_i - M|}{N} = \frac{735}{100} = 7.35$

Answer: Mean Deviation about the median $= 7.35$ years

Exercise 13.2

1Find the mean and variance for the data: 6, 7, 10, 12, 13, 4, 8, 12Show solution

Given: Data: 6, 7, 10, 12, 13, 4, 8, 12, $n = 8$

Step 1: Find the Mean
$\bar{x} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$

Step 2: Find $(x_i - \bar{x})^2$

$x_i$	$x_i - 9$	$(x_i - 9)^2$
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		74

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{74}{8} = 9.25$

Answer: Mean $= 9$ , Variance $= 9.25$

2Find the mean and variance for the first

n

natural numbers.Show solution

Given: First $n$ natural numbers: $1, 2, 3, \ldots, n$

Step 1: Find the Mean
$\bar{x} = \frac{1+2+3+\cdots+n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$

Step 2: Find the Variance

We use the formula:
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$

$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$

$\frac{\sum x_i^2}{n} = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6}$

$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2$

$= \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$

$= (n+1)\left[\frac{2n+1}{6} - \frac{n+1}{4}\right]$

$= (n+1)\left[\frac{2(2n+1) - 3(n+1)}{12}\right]$

$= (n+1)\left[\frac{4n+2-3n-3}{12}\right]$

$= (n+1) \cdot \frac{n-1}{12} = \frac{n^2-1}{12}$

Answer: Mean $= \dfrac{n+1}{2}$ , Variance $= \dfrac{n^2-1}{12}$

3Find the mean and variance for the first 10 multiples of 3.

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

Step 1: Find the Mean
$\bar{x} = \frac{3+6+9+\cdots+30}{10} = \frac{3(1+2+\cdots+10)}{10} = \frac{3 \times 55}{10} = \frac{165}{10} = 16.5$

Step 2: Find $(x_i - \bar{x})^2$

$x_i$	$x_i - 16.5$	$(x_i - 16.5)^2$
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		742.5

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n} = \frac{742.5}{10} = 74.25$

Answer: Mean $= 16.5$ , Variance $= 74.25$

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4Find 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

Given: $N = \sum f_i = 2+4+7+12+8+4+3 = 40$

Step 1: Find the Mean

$x_i$	$f_i$	$f_i x_i$
6	2	12
10	4	40
14	7	98
18	12	216
24	8	192
28	4	112
30	3	90
Total	40	760

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{760}{40} = 19$

Step 2: Find $f_i(x_i - \bar{x})^2$

$x_i$	$f_i$	$(x_i - 19)^2$	$f_i(x_i-19)^2$
6	2	169	338
10	4	81	324
14	7	25	175
18	12	1	12
24	8	25	200
28	4	81	324
30	3	121	363
Total	40		1736

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{1736}{40} = 43.4$

Answer: Mean $= 19$ , Variance $= 43.4$

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5Find 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

Given: $N = \sum f_i = 3+2+3+2+6+3+3 = 22$

Step 1: Find the Mean

$x_i$	$f_i$	$f_i x_i$
92	3	276
93	2	186
97	3	291
98	2	196
102	6	612
104	3	312
109	3	327
Total	22	2200

$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{2200}{22} = 100$

Step 2: Find $f_i(x_i - \bar{x})^2$

$x_i$	$f_i$	$(x_i - 100)^2$	$f_i(x_i-100)^2$
92	3	64	192
93	2	49	98
97	3	9	27
98	2	4	8
102	6	4	24
104	3	16	48
109	3	81	243
Total	22		640

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{640}{22} = 29.09$

Answer: Mean $= 100$ , Variance $\approx 29.09$

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6Find 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

Given: $N = \sum f_i = 2+1+12+29+25+12+10+4+5 = 100$

Let assumed mean $A = 64$ , $h = 1$ , so $y_i = \frac{x_i - 64}{1} = x_i - 64$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

$x_i$	$f_i$	$y_i = x_i - 64$	$f_i y_i$	$f_i y_i^2$
60	2	$-4$	$-8$	32
61	1	$-3$	$-3$	9
62	12	$-2$	$-24$	48
63	29	$-1$	$-29$	29
64	25	0	0	0
65	12	1	12	12
66	10	2	20	40
67	4	3	12	36
68	5	4	20	80
Total	100		0	286

Step 2: Find Mean
$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 64 + \frac{0}{100} \times 1 = 64$

Step 3: Find Variance and Standard Deviation
$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$
$= \frac{1}{(100)^2}\left[100 \times 286 - 0^2\right] = \frac{28600}{10000} = 2.86$

$\sigma = \sqrt{2.86} \approx 1.69$

Answer: Mean $= 64$ , Standard Deviation $\approx 1.69$

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7Find the mean and variance for the 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

Given: $N = 2+3+5+10+3+5+2 = 30$

Step 1: Find midpoints and compute mean

Class	$x_i$	$f_i$	$f_i x_i$
0–30	15	2	30
30–60	45	3	135
60–90	75	5	375
90–120	105	10	1050
120–150	135	3	405
150–180	165	5	825
180–210	195	2	390
Total		30	3210

$\bar{x} = \frac{3210}{30} = 107$

Step 2: Find $f_i(x_i - \bar{x})^2$

$x_i$	$f_i$	$(x_i - 107)^2$	$f_i(x_i - 107)^2$
15	2	8464	16928
45	3	3844	11532
75	5	1024	5120
105	10	4	40
135	3	784	2352
165	5	3364	16820
195	2	7744	15488
Total	30		68280

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{68280}{30} = 2276$

Answer: Mean $= 107$ , Variance $= 2276$

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8Find the mean and variance for the frequency distribution:
Classes: 0-10, 10-20, 20-30, 30-40, 40-50
Frequencies: 5, 8, 15, 16, 6

Given: $N = 5+8+15+16+6 = 50$

Step 1: Find midpoints and compute mean

Class	$x_i$	$f_i$	$f_i x_i$
0–10	5	5	25
10–20	15	8	120
20–30	25	15	375
30–40	35	16	560
40–50	45	6	270
Total		50	1350

$\bar{x} = \frac{1350}{50} = 27$

Step 2: Find $f_i(x_i - \bar{x})^2$

$x_i$	$f_i$	$(x_i - 27)^2$	$f_i(x_i - 27)^2$
5	5	484	2420
15	8	144	1152
25	15	4	60
35	16	64	1024
45	6	324	1944
Total	50		6600

Step 3: Calculate Variance
$\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{6600}{50} = 132$

Answer: Mean $= 27$ , Variance $= 132$

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9Find the mean, variance and standard deviation using short-cut method:
Height (cms): 70-75, 75-80, 80-85, 85-90, 90-95, 95-100, 100-105, 105-110, 110-115
No. of children: 3, 4, 7, 7, 15, 9, 6, 6, 3

Given: $N = 3+4+7+7+15+9+6+6+3 = 60$

Let assumed mean $A = 92.5$ , $h = 5$ , so $y_i = \frac{x_i - 92.5}{5}$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

Class	$x_i$	$f_i$	$y_i$	$f_i y_i$	$f_i y_i^2$
70–75	72.5	3	$-4$	$-12$	48
75–80	77.5	4	$-3$	$-12$	36
80–85	82.5	7	$-2$	$-14$	28
85–90	87.5	7	$-1$	$-7$	7
90–95	92.5	15	0	0	0
95–100	97.5	9	1	9	9
100–105	102.5	6	2	12	24
105–110	107.5	6	3	18	54
110–115	112.5	3	4	12	48
Total		60		6	254

Step 2: Find Mean
$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 92.5 + \frac{6}{60} \times 5 = 92.5 + 0.5 = 93$

Step 3: Find Variance
$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$
$= \frac{25}{3600}\left[60 \times 254 - (6)^2\right] = \frac{25}{3600}\left[15240 - 36\right] = \frac{25 \times 15204}{3600} = \frac{380100}{3600} = 105.58$

Step 4: Standard Deviation
$\sigma = \sqrt{105.58} \approx 10.28$

Answer: Mean $= 93$ cm, Variance $\approx 105.58$ , Standard Deviation $\approx 10.28$ cm

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10The diameters of circles (in mm) drawn in a design are given below:
Diameters: 33-36, 37-40, 41-44, 45-48, 49-52
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, 44.5-48.5, 48.5-52.5 and then proceed.]

Given: $N = 15+17+21+22+25 = 100$

Converting to continuous classes: 32.5–36.5, 36.5–40.5, 40.5–44.5, 44.5–48.5, 48.5–52.5

Let assumed mean $A = 44.5$ , $h = 4$ , so $y_i = \frac{x_i - 44.5}{4}$

Step 1: Compute $f_i y_i$ and $f_i y_i^2$

Class	$x_i$	$f_i$	$y_i$	$f_i y_i$	$f_i y_i^2$
32.5–36.5	34.5	15	$-2.5$	$-37.5$	93.75
36.5–40.5	38.5	17	$-1.5$	$-25.5$	38.25
40.5–44.5	42.5	21	$-0.5$	$-10.5$	5.25
44.5–48.5	46.5	22	$0.5$	$11$	5.5
48.5–52.5	50.5	25	$1.5$	$37.5$	56.25
Total		100		$-25$	199

Step 2: Find Mean
$\bar{x} = A + \frac{\sum f_i y_i}{N} \times h = 44.5 + \frac{-25}{100} \times 4 = 44.5 - 1 = 43.5 \text{ mm}$

Step 3: Find Variance and Standard Deviation
$\sigma^2 = \frac{h^2}{N^2}\left[N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2\right]$
$= \frac{16}{10000}\left[100 \times 199 - (-25)^2\right] = \frac{16}{10000}\left[19900 - 625\right] = \frac{16 \times 19275}{10000} = \frac{308400}{10000} = 30.84$

$\sigma = \sqrt{30.84} \approx 5.55 \text{ mm}$

Answer: Mean diameter $= 43.5$ mm, Standard Deviation $\approx 5.55$ mm

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Miscellaneous Exercise on Chapter 13

1The 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.

Given: $n = 8$ , $\bar{x} = 9$ , $\sigma^2 = 9.25$
Six observations: 6, 7, 10, 12, 12, 13

Let the remaining two observations be $a$ and $b$ .

Step 1: Use the mean condition
$\frac{6+7+10+12+12+13+a+b}{8} = 9$
$60 + a + b = 72$
$a + b = 12 \quad \cdots (1)$

Step 2: Use the variance condition
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$
$9.25 = \frac{\sum x_i^2}{8} - 81$
$\sum x_i^2 = (9.25 + 81) \times 8 = 90.25 \times 8 = 722$

Sum of squares of known observations:
$6^2+7^2+10^2+12^2+12^2+13^2 = 36+49+100+144+144+169 = 642$

So:
$642 + a^2 + b^2 = 722$
$a^2 + b^2 = 80 \quad \cdots (2)$

Step 3: Solve equations (1) and (2)
From (1): $(a+b)^2 = 144 \Rightarrow a^2 + 2ab + b^2 = 144$

Using (2): $80 + 2ab = 144 \Rightarrow ab = 32$

So $a$ and $b$ are roots of:
$t^2 - 12t + 32 = 0$
$(t-4)(t-8) = 0$
$t = 4 \text{ or } t = 8$

Answer: The remaining two observations are $4$ and $8$ .

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2The 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.

Given: $n = 7$ , $\bar{x} = 8$ , $\sigma^2 = 16$
Five observations: 2, 4, 10, 12, 14

Let the remaining two observations be $a$ and $b$ .

Step 1: Use the mean condition
$\frac{2+4+10+12+14+a+b}{7} = 8$
$42 + a + b = 56$
$a + b = 14 \quad \cdots (1)$

Step 2: Use the variance condition
$\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$
$16 = \frac{\sum x_i^2}{7} - 64$
$\sum x_i^2 = 80 \times 7 = 560$

Sum of squares of known observations:
$4+16+100+144+196 = 460$

So:
$460 + a^2 + b^2 = 560$
$a^2 + b^2 = 100 \quad \cdots (2)$

Step 3: Solve equations (1) and (2)
$(a+b)^2 = 196 \Rightarrow a^2 + 2ab + b^2 = 196$

Using (2): $100 + 2ab = 196 \Rightarrow ab = 48$

So $a$ and $b$ are roots of:
$t^2 - 14t + 48 = 0$
$(t-6)(t-8) = 0$
$t = 6 \text{ or } t = 8$

Answer: The remaining two observations are $6$ and $8$ .

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3The 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.

Given: $n = 6$ , $\bar{x} = 8$ , $\sigma = 4$

Each observation is multiplied by 3. Let new observations be $y_i = 3x_i$ .

New Mean:
$\bar{y} = \frac{\sum y_i}{n} = \frac{\sum 3x_i}{n} = 3\bar{x} = 3 \times 8 = 24$

New Variance:
$\sigma_y^2 = \frac{1}{n}\sum(y_i - \bar{y})^2 = \frac{1}{n}\sum(3x_i - 3\bar{x})^2 = \frac{9}{n}\sum(x_i - \bar{x})^2 = 9\sigma^2 = 9 \times 16 = 144$

New Standard Deviation:
$\sigma_y = \sqrt{144} = 12$

Answer: New Mean $= 24$ , New Standard Deviation $= 12$

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4Given that

\bar{x}

is the mean and

\sigma^2

is the variance of

n

observations

x_1, x_2, \ldots, x_n

. Prove that the mean and variance of the observations

ax_1, ax_2, ax_3, \ldots, ax_n

are

a\bar{x}

and

a^2\sigma^2

, respectively,

(a \neq 0)

Given: $n$ observations $x_1, x_2, \ldots, x_n$ with mean $\bar{x}$ and variance $\sigma^2$ .

New observations: $y_i = ax_i$ for $i = 1, 2, \ldots, n$ .

Proof of New Mean:
$\bar{y} = \frac{1}{n}\sum_{i=1}^{n} y_i = \frac{1}{n}\sum_{i=1}^{n} ax_i = a \cdot \frac{1}{n}\sum_{i=1}^{n} x_i = a\bar{x}$

Hence the mean of the new observations is $a\bar{x}$ . $\quad$ Proved.

Proof of New Variance:
$\sigma_y^2 = \frac{1}{n}\sum_{i=1}^{n}(y_i - \bar{y})^2 = \frac{1}{n}\sum_{i=1}^{n}(ax_i - a\bar{x})^2$
$= \frac{1}{n}\sum_{i=1}^{n} a^2(x_i - \bar{x})^2 = a^2 \cdot \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 = a^2\sigma^2$

Hence the variance of the new observations is $a^2\sigma^2$ . $\quad$ Proved.

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5The 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.

Given: $n = 20$ , $\bar{x} = 10$ , $\sigma = 2$

So $\sum x_i = 20 \times 10 = 200$ and $\sigma^2 = 4$ .

$\frac{\sum x_i^2}{n} - \bar{x}^2 = 4 \Rightarrow \sum x_i^2 = (4+100) \times 20 = 2080$

(i) Wrong item (8) is omitted:

Correct $\sum x_i = 200 - 8 = 192$ , $n = 19$

$\text{Correct Mean} = \frac{192}{19} \approx 10.1$

Correct $\sum x_i^2 = 2080 - 64 = 2016$

$\text{Correct } \sigma^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 = \frac{2016}{19} - \frac{36864}{361}$
$= \frac{2016 \times 19 - 36864}{361} = \frac{38304 - 36864}{361} = \frac{1440}{361} \approx 3.99$

$\text{Correct } \sigma = \sqrt{\frac{1440}{361}} = \frac{\sqrt{1440}}{19} = \frac{12\sqrt{10}}{19} \approx 1.997 \approx 2$

(ii) Wrong item (8) is replaced by 12:

Correct $\sum x_i = 200 - 8 + 12 = 204$ , $n = 20$

$\text{Correct Mean} = \frac{204}{20} = 10.2$

Correct $\sum x_i^2 = 2080 - 64 + 144 = 2160$

$\text{Correct } \sigma^2 = \frac{2160}{20} - (10.2)^2 = 108 - 104.04 = 3.96$

$\text{Correct } \sigma = \sqrt{3.96} \approx 1.99 \approx 2$

Answer:

(i) Correct Mean $\approx 10.1$ , Correct S.D. $\approx 2$
(ii) Correct Mean $= 10.2$ , Correct S.D. $\approx 1.99$

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6The 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.

Given: $n = 100$ , $\bar{x} = 20$ , $\sigma = 3$

$\sum x_i = 100 \times 20 = 2000$

$\sigma^2 = 9 \Rightarrow \frac{\sum x_i^2}{100} - 400 = 9 \Rightarrow \sum x_i^2 = 40900$

Step 1: Remove incorrect observations (21, 21, 18)

Correct $n = 100 - 3 = 97$

Correct $\sum x_i = 2000 - 21 - 21 - 18 = 1940$

$\text{Correct Mean} = \frac{1940}{97} = 20$

Step 2: Correct $\sum x_i^2$

Correct $\sum x_i^2 = 40900 - 441 - 441 - 324 = 39694$

Step 3: Correct Variance
$\sigma^2 = \frac{39694}{97} - (20)^2 = 409.216... - 400 = 9.216...$

More precisely:
$\sigma^2 = \frac{39694}{97} - 400 = \frac{39694 - 38800}{97} = \frac{894}{97} \approx 9.21$

$\sigma = \sqrt{\frac{894}{97}} \approx \sqrt{9.21} \approx 3.036$

Answer: Correct Mean $= 20$ , Correct Standard Deviation $\approx 3.04$

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