From the chapter Statistics, some of the important questions are covered in this question paper.

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Download CBSE Class 11th Standard CBSE Mathematics question papers, sample papers, important questions, and previous year solved papers in PDF format. Get free study materials, NCERT solutions, and exam preparation resources for Class 11th Standard CBSE Mathematics

Questions:

Generic 2

8 x 2 = 16

Find the variance of the following data:

Classes	0-10	10-20	20-30	30-40	40-50
Frequencies	15	20	22	27	6

The analysis of daily wages paid to workers in two firms A and B belonging to the same industry are:

	firm A	Firm B
Number of workers	812	910
Daily wages	80	92
Variance	16	25

(i) Which firm A or B pays more in total daily wages?
(ii) Which A or B shows greater variability in wages?

Find the mean deviation about the mean for the following data
36,72,46,60,45,42,53,49,51,46

Find the variance of the data 6,5,9,13,12,8 and 10.

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 2 with standard deviation 1.25 goals. Find which team may be considered more consistent?

Find the mean deviation about the median for the data 34,66,30,38,44,50,40,60,42,51.

Two plants A and B of a factory show following results about the number of workers and the wages paid to them.

	A	B
Number of workers	4000	4500
Average monthly wages	3000	3000
Variance of distribution	16	25

Which plant, A or B shows greater variability in individual wages?

Find the variance and standard deviation for the following data, 6,7,10,12,13,4,8,12.

Generic 3

3 x 3 = 9

Calculate the mean deviation from the median of the following data

Wages per week in Rs	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of workers	4	6	10	20	10	6	4

10.

Calculate the mean, variance and standard deviation for the following distribution.

Class	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequency(f)	2	3	5	10	3	5	2

11.

Find the mean deviation from the mean for the following data
6, 5, 5.25, 5.5, 4.75, 4.5, 6.25, 7.75, 9.

Generic 5

2 x 5 = 10

12.

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?

13.

The marks obtained (out of 100) by two students A and B in 10 qualifying tests were:

A:	48	53	58	41	54	52	54	49	51	50
B:	11	98	60	94	48	52	17	90	20	20

who is more consistent in performance and who is more variable?

Multiple Choice Question

4 x 1 = 4

14.

If the S. D. of a set of observation is 8 and if each observation is divided by- 2 then S.D. of new set of observation is ______.

(a)

-4

(b)

(c)

(d)

-8

15.

The mean deviation of the series a, a + d, a + 2d, .... a + nd from its mean is ______.

(a)

\({(n+1)d\over (n+2)}\)

(b)

\(nd\over 2n+1\)

(c)

\(n(n+1)d\over 2n+1\)

(d)

\((2n+1)d\over n\)

16.

If ⋋is the variance and o is the S.D. then ______.

(a)

\(⋋={1\over σ^2}\)

(b)

⋋ = σ²

(c)

\(σ={1\over ⋋}\)

(d)

\(σ={1\over ⋋^2}\)

17.

Standard deviation of a data is given by ______.

(a)

\(σ=\sqrt{{1\over N}\sum fd^2-\left({1\over N}\sum fd\right)^2}\)

(b)

\(σ=\sqrt{\left({1\over N}\sum fd\right)^2-{1\over N}\sum fd^2}\)

(c)

\(σ=\sqrt{{1\over N}\sum fd^2-{1\over N}\sum fd^2}\)

(d)

None of these

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Answers Key:

Generic 2

8 x 2 = 16

144.06

(i) Firm B
(ii) Firm B.

7.2

\(\frac { 52 }{ 7 } \)

Let us make the following table from the given data.

Number of goals(x_i)	Number of matches (f_i)	x_i2	f_ix_i	f_ix_i2
0	1	0	0	0
1	9	1	9	9
2	7	4	14	28
3	5	9	15	45
4	3	16	12	48
Total	25		50	130

\(Here,\ \Sigma { f }_{ i }=25,\ \Sigma { f }_{ i }{ x }_{ i }=50,\ and\ \Sigma { f }_{ 1 }{ x }_{ i }^{ 2 }=130\)
For team A
Mean, \(\overset { \_ }{ x } =\frac { \Sigma { f }_{ i }{ x }_{ i } }{ \Sigma { f }_{ i } } =\frac { 50 }{ 25 } =2\)
\(Standard\ division,\ \sigma =\frac { 1 }{ N } \sqrt { N{ \Sigma { f }_{ i }{ x }_{ i } }^{ 2 }-{ (\Sigma { f }_{ i }{ x }_{ i }) }^{ 2 } } \)
\(=\frac { 1 }{ 25 } \sqrt { 25\times 130-{ (50) }^{ 2 } } \)
\(\frac { 5 }{ 25 } \sqrt { 130-100 } =\frac { \sqrt { 30 } }{ 5 } =\frac { 5.477 }{ 5 } =1.095\)
\(\therefore Coefficient\ of \ variation\ =\frac { \sigma }{ \overset { \_ }{ x } } \times 100=\frac { 1.095 }{ 2 } \times 100=54.75\)
For team B
Given Mean, \(\overset { \_ }{ x } \) = 2 and SD = \(\sigma \) =1.25
\(\therefore Coefficient\ of\ variation\ =\frac { \sigma }{ \overset { \_ }{ x } } \times 100=\frac { 1.25 }{ 2 } \times 100=6.25\)
Since, the coefficient of variation of goals of team A is less than that of B. Therefeore, team A is more consistent than team b

The given data can be arranged in ascending order as 30,34,38,40,42,44,50,51,60,66.
Here, total number of observations are 10 i.e. n = 10, which is even
\(\therefore \) Median
\(M=\frac { \left( \frac { n }{ 2 } \right) th\quad observation\quad +\left( \frac { n }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { \left( \frac { 10 }{ 2 } \right) th\quad observation\quad +\left( \frac { 10 }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { 5th\quad observation\quad +6th\quad observation\quad }{ 2 } \)
\(=\frac { 42+44 }{ 2 } =\frac { 86 }{ 2 } =43\)
Let us make the table for absolute deviation

\({ x }_{ i }\)	\(\left\| { x }_{ i }-M \right\| \)
30	\(\left\| 30-43 \right\| =13\)
34	\(\left\| 34-43 \right\| =9\)
38	\(\left\| 38-43 \right\| =5\)
40	\(\left\| 40-43 \right\| =3\)
42	\(\left\| 42-43 \right\| =1\)
44	\(\left\| 44-43 \right\| =1\)
50	\(\left\| 50-43 \right\| =7\)
51	\(\left\| 51-43 \right\| =8\)
60	\(\left\| 60-43 \right\| =17\)
66	\(\left\| 66-43 \right\| =23\)
Total	\(\sum _{ i-1 }^{ 10 }{ \left\| { x }_{ i }-M \right\| } =87\)

Now, mean deviation about the median.
\(MD=\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-M \right| } }{ 10 } =\frac { 87 }{ 10 } =8.7\)

Here, we observe that average monthly wages in both the plants is same i.e 3000. Therefore, the plant with greater variance will have more variability. Hence, the plant B has greater variability in individual wages.

Given observations are 6,7,10,12,13,4,8,12
Number of observations =8
\(\therefore \ Mean(\overline { x } )=\frac { 6+7+10+12+13+4+8+12 }{ 8 } \)
\(=\frac { 72 }{ 8 } =9\)
Now, let us make the following table for deviation.

x_i	\({ x }_{ i }-\overline { x } \)	\({ { (x }_{ i }-\overline { x } ) }^{ 2 }\)	x_i	\({ { x }_{ i }-\overline { x } }\)	\({ { (x }_{ i }-\overline { x } ) }^{ 2 }\)
6	-3	9	13	4	16
7	-2	4	4	-5	25
10	1	1	8	-1	1
12	3	9	12	3	9
Total		74	Total		74

\(\therefore \) Sum of squares of deviations =\(\sum _{ i=1 }^{ 8 }{ { ({ x }_{ i }-\overline { x } ) }^{ 2 } } =74\)
Hence, variance, \({ \sigma }^{ 2 }=\frac { \sum _{ i=1 }^{ 8 }{ { ({ x }_{ i }-\overline { x } ) }^{ 2 } } }{ n } =\frac { 74 }{ 8 } \)=9.25
and standard deviation =\(\sqrt { \sigma } =\sqrt { 9.25 } \)
=3.04

Generic 3

3 x 3 = 9

Wages per weeks in Rs	Mid value x	Frequency f	.f.	\|x-45\|	f\|x-45\|
10-20	15	4	4	30	120
20-30	25	6	10	20	120
30-40	35	10	20	10	100
40-50	45	20	40	0	0
50-60	55	10	50	10	100
60-70	65	6	56	20	120
70-80	75	4	60	30	120
		60			680

Here N = 60 \(∴\ {N\over 2}={60\over 2}=30\)
∴ Median class is 40 - 50
∴ Median = \(40+{(30-20)\over 20}\times10=40+5=45\)
∴ Mean deviation from me\({f|x-45|\over N}={680\over60}=11.33\)dian =

10.

Here, h=30 and a=105
We make the table from the given data

Class interval	Frequency(f_i)	Midvalue(x_i)	\({ u }_{ i }=\frac { { x }_{ i }-105 }{ 30 } \)	f_iu_i	u_i²	f_iu_i²
0-30	2	15	-3	-6	9	18
30-60	3	45	-2	-6	4	12
60-90	5	75	-1	-5	1	5
90-120	10	105	0	0	0	0
120-150	3	135	1	3	1	3
150-180	5	165	2	10	4	20
180-210	2	195	3	6	9	18
Total	30			2		76

Here, N=30,\(\sum { { f }_{ i }{ u }_{ i }=2,\sum { { f }_{ i } } { u }_{ i }^{ 2 } } =76\quad and\quad h=30\)
\(\therefore \ Mean(\overline { x } )=a+\left( \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i } } \right) \times h\)
\(=105+2=107\)
\(Variance,{ \sigma }^{ 2 }={ h }^{ 2 }\left[ \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i }^{ 2 } } -\left( \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i } } \right) ^{ 2 } \right] \)
\(={ (30) }^{ 2 }\left[ \frac { 1 }{ 30 } \times 76-{ \left( \frac { 1 }{ 30 } \times 2 \right) }^{ 2 } \right] \)
\(={ \left( 30 \right) }^{ 2 }\left[ \frac { 76 }{ 30 } -\frac { 1 }{ 225 } \right] =30\left[ 76-\frac { 30 }{ 225 } \right] \)
\(=30\left[ 76-0.13 \right] =30\times 75.87=2276.1\)
Hence, mean is 107 and variance is 2276.1

11.

Let \(\bar { x } \) be the mean of given data.
Then, \(\bar { x } \) =\(\frac { 6+5+5.25+5.5+4.75+4.5+6.25+7.75+9 }{ 9 } =\frac { 54 }{ 9 } =6\)
We make the table from the given data.

\(x_{ i }\)	\(x_{ i }-\bar { x } =\quad x_{ i }-6\)	\(\|x_{ i }-\bar { x } \|\)
6	0	0
5.0	-1	1.00
5.25	-0.75	0.75
5.5	-0.5	0.50
4.75	-1.25	1.25
4.5	-1.50	1.50
6.25	0.25	0.25
7.75	1.75	1.75
9	3	3
Total		10.00

\(\therefore \) Mean deviation from mean,
\(MD(\bar { x } )=\frac { \sum { |x_{ i }-\bar { x } | } }{ n } =\frac { 10 }{ 9 } =1.1\)
Hence, mean deviation form mean is 1.1.

Generic 5

2 x 5 = 10

12.

For Mathematics
\(\bar { x } =42\ and\ \sigma =12\)
\(\therefore \ C.V\ of\ Mathematics=\frac { 12 }{ 42 } \times 100=28.57%\)%
For Physics
\(\bar { x } =32\ and\ \sigma =15\)
\(\therefore \ C.V.\ of\ physics=\frac { 15 }{ 32 } \times 100=46.88\)%
For chemistry
\(\bar { x } =40.9\ and\ \sigma =20\)
\(\therefore C.V.\ of\ Chemistry=\frac { 20 }{ 40.9 } \times 100=48.9\)%
Thus chemistry with hightest C.V. shows highest variability and mathematics with lowest C.V. shows lowest variability.

13.

A	B	A-\(\bar { A } \)	B-\(\bar { B } \)	(A-\(\bar { A } \))²	(B-\(\bar { B } \))²
48	11	-3	-40	9	1600
53	98	2	47	4	2209
58	60	7	9	49	81
41	94	-10	43	100	1849
54	48	3	-3	9	9
52	52	1	1	1	1
54	17	3	-34	9	1156
49	90	-2	39	4	1521
51	20	0	-31	0	961
50	20	-1	-31	1	961
510	510			186	10348

Mean marks of A (\(\bar { A } \))=\(\frac { 510 }{ 10 } =51\)
Mean marks of B(\(\bar { B } \))=\(\frac { 510 }{ 10 } =51\)
Here \(\bar { A } \)=\(\bar { B } \) \({ \sigma }_{ A }=\sqrt { \frac { \sum { (A-\bar { A } { ) }^{ 2 } } }{ n } } =\sqrt { \frac { 186 }{ 10 } } =4.3\)
\({ \sigma }_{ B }=\sqrt { \frac { \sum { (B-\bar { B } { ) }^{ 2 } } }{ n } } =\sqrt { \frac { 10348 }{ 10 } } =32.16\)
The score of A is more consistent and score of B is more variable.