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SSC > Statistics & Data Interpretation

Explore popular questions from Statistics & Data Interpretation for SSC. This collection covers Statistics & Data Interpretation previous year SSC questions hand picked by experienced teachers.

Q 1.

Correct2

Incorrect-0.5

If the sum of 11 consecutive natural numbers is 2761, then the middle number is:

A

249

B

250

251

D

252

Explanation

Let the first natural number be x

According to the question

x+x+1x+2+x+3+x+4+x+5+x+6+x+7+x+8+x+9+x+10=2761

=> 11x + 55 = 2761

x = {tex} \frac{2761 - 55}{11} = 246 {/tex}

Middle number = x + 5 = 246 + 5 = 251

Q 2.

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Incorrect-0.5

The mean weight of 9 items is 15. If one more item is added to the series, the mean 16. The value of 10th item is

A

35

B

30

25

D

20

Explanation

Total weight of 9 items = 15 x 9 = 135

and total weigth of 10 items = 16 X 10 = 160

Therefore, weight of 10 items = 160 - 135 = 25

Q 3.

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Incorrect-0.5

The mean of 12 items is {tex} \bar{X} {/tex} . If the first term is increased by 1, second by 2 and so on, then the new mean is

A

{tex} \overline{X}+n {/tex}

B

{tex} \overline{X}+\frac{n}{2} {/tex}

{tex} \overline{X}+\frac{n+1}{2} {/tex}

D

none of these

Q 4.

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Incorrect-0.5

If the mean of n observation {tex} 1^{2},\ 2^{2},\ 3^{2},........n^{2} {/tex}, is {tex} \frac{46}{11}n {/tex}, then n is equal to

11

B

12

C

23

D

22

Q 5.

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Incorrect-0.5

The value of mean, median and mode coincides, then the distribution is

A

Positive skewness

Symmetrical distribution

C

Negative skewness

D

All of these

Explanation

If mean, median and mode coincides, then there is a symmetrical distribution.

Q 6.

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Incorrect-0.5

In a class of 100 students, the average amount of pocket money is Rs.35 per student. If the average is Rs.25 for girls and Rs.50 for boys, then the number of girls in the class is

A

20

B

40

60

D

80

Explanation

Let the number of girls in the class = y

Therefore, Number of boys in the class = 100 - y

Now, {tex} \overline{x}_{1}=25,n_{1}=y,\overline{x}_{2}=50,n_{2}=100-y\ {/tex}

and {tex} \overline{x}=35,n_{1}+n_{2}=100 {/tex}

{tex} \therefore\ 35 = \frac{25 \times y+50 \times \left(100-y\right) }{100} {/tex}

=> 3500 = 25y + 5000 - 50y

=> 25y = 1500 => y = 60

=> Number of girls in the class = 60

Q 7.

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Incorrect-0.5

For a series the value of mean deviation is 15, the most likely value of its quartile deviation is

12.5

B

11.6

C

13

D

9.7

Explanation

Since,

{tex} MD = \frac{4}{5}\sigma,\ QD=\frac{2}{3}\sigma {/tex}

{tex} \therefore\ \frac{MD}{QD} = \frac{6}{5} {/tex}

=> {tex} QD = \frac{5}{6} \left(MD\right)=\frac{5}{6} \left(15\right)=12.5 {/tex}

Q 8.

Correct2

Incorrect-0.5

If {tex} \overline{x} {/tex} is the arithmetic mean of n independent variates {tex} x_{1},\ x_{2},\ x_{3}.....,\ x_{n} {/tex} each of the standard deviation {tex} \sigma {/tex}, the variance {tex} \overline{x} {/tex} is

{tex} \frac{\sigma^{2}}{n} {/tex}

B

{tex} \frac{n\sigma^{2}}{2} {/tex}

C

{tex} \frac{\left(n+1\right)\sigma^{2}}{3} {/tex}

D

None of these

Explanation

We have, {tex} \overline{x} = \frac{1}{n}\sum^{n}_{i=1}x_{i} {/tex}

Therefore, {tex} var \left(\overline{x}\right) = \frac{1}{n^{2}}\left[ \sum^{n}_{i=1} var \left(x_{j}\right)+2\sum^{n}_{i \neq j} cov \left(x_{i},\ x_{j}\right) \right] {/tex}

Q 9.

Correct2

Incorrect-0.5

Coefficient of skewness for the values median {tex} = 18.8,\ Q_{1} = 14.6,\ Q_{3} = 25.2 {/tex} is

0.2

B

0.5

C

0.7

D

None of these

Explanation

Coefficient of skewness = {tex} \frac{Q_{3}-Q_{1}-2 \left(median\right) }{Q_{3}-Q_{1}} {/tex}

= {tex} \frac{25.2+14.6-2 \left(18.8\right) }{25.2-14.6} = \frac{2.2}{10.6} = 0.20 {/tex}

Q 10.

Correct2

Incorrect-0.5

The 7th percentile is equal to

7th decile

B

{tex} Q_{3} {/tex}

C

6th decile

D

None of these

Explanation

7th decile {tex} D_{7} = \frac{7n}{10} {/tex} .. (i)

and 7th percentile, {tex} P_{10} = \frac{7n}{100} {/tex} .. (ii)

From equations (i) and (ii), we get

{tex} D_{7}\neq P_{70} {/tex}

Q 11.

Correct2

Incorrect-0.5

Consider the following statements:
1. The values of median and mods can be determined graphically.
2. Mean, median and mode have the same unit.
3. Range is the best measure of dispersion
which of these is/are correct?

(1) alone

B

(2) alone

C

Both (2) and (3)

D

None of these

Explanation

It is true that median and mode can be determined graphically.

Q 12.

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Incorrect-0.5

In a factory daily wages of 7 employees are Rs.50, Rs.60, Rs.70, Rs.80, Rs.90, Rs.100 and Rs.110. Calculate the arithmetic mean (AM)

80

B

70

C

90

D

60

Explanation

Arithmetic mean (AM) = {tex} \frac{50+60+70+80+90+100+110}{7} {/tex}

={tex} \frac{560}{7} {/tex} = 80

Q 13.

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Calculate the mean of the following marks secured by 40 students

Marks: 354045506570
Number of students: 4321687

A

74

54

C

84

D

64

Explanation

Marks(x) Frequency(f)fx
35 4140
40 3120
45 290
50 16800
65 8520
40 7490
{tex} \sum x=40 {/tex}{tex} \sum fx =2160{/tex}


Arithmetic mean (A.M.) {tex} \sum f=40 \sum fx =2160{/tex}

{tex} \overline{x} = \frac{\sum fx}{\sum f} {/tex}

={tex} \frac{216}{40} {/tex} = 54

Q 14.

Correct2

Incorrect-0.5

Calculate the mean for the following frequency distribution.

x: 10-2020-3030-4040-5050-6060-7070-80
f: 689107515

A

54.32

B

62.24

48.16

D

42.14

Explanation

x fm{tex} x' = \frac{m-A}{i} {/tex}fx'
10-20 615-3-18
20-30 825-2-16
30-40 935-1-9
40-50 104500
50-60 75517
60-70 565210
70-80 1575345
{tex} \sum =60 {/tex}{tex} \sum fx'=19 {/tex}


m - middle point of the interval.

i. -differene of the class interval.

A = 45

Arithmetic mean {tex} \overline{x} = A+ \frac{\sum f'x}{\sum f}i {/tex}

= {tex} 45+\frac{19}{60} \times 10 {/tex}

={tex} 45+3.16 = 48.16 {/tex}

Q 15.

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Incorrect-0.5

Find the median for the following data 15, 24, 18, 9, 23, 6, 19

A

24

B

6

C

19

18

Explanation

Arrange the given data in ascending order. The data can be written as

6. 9, 15, 18, 19, 23, 24

The middle No. is 18.

Hence 18 is the median.

Q 16.

Correct2

Incorrect-0.5

Find the median for the following data 42, 24, 51, 33, 15, 60

37.5

B

38.5

C

40

D

40.5

Explanation

Arrange the given data in ascending order. The data can be written as

15, 24, 33, 42, 51, 60

Here the number of data is even number. So take the two middle data and find the average.

Median = {tex} \frac{33+42}{2}=\frac{75}{2} = 37.5{/tex}

Q 17.

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Find the standard deviation for the given data.
The weight odf 5 boys (in kg.) are given below:
45, 40, 36, 34, 65

A

13.25

11.15

C

14.35

D

12.25

Explanation

x {tex} d=x-\overline{x} {/tex}{tex} d^{2} {/tex}
45 45-44=11
40 40-44=-416
36 36-44=-864
34 34-44=-10100
65 65-44=21441
{tex} \sum d^{2} {/tex}622


Variance = {tex} \frac{ \sum d^{2}}{n} = \frac{622}{5} =124 {/tex} {tex} \frac{2}{5} {/tex}=124.4

Standard deviation ={tex} \sqrt{124.4} = 11.15 {/tex}

Q 18.

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When dice is thrown, find the probability to get the sum of the numbers is 5.

A

{tex} \frac{1}{4} {/tex}

B

{tex} \frac{1}{3} {/tex}

{tex} \frac{1}{2} {/tex}

D

{tex} \frac{1}{5} {/tex}

Explanation

Number of occurrence =

{tex} \{ \left(1, 1\right) \left(1, 2\right) \left(1, 3\right) \left(1, 4\right) \left(1, 5\right) \left(1, 6\right) {/tex} {tex} \left(2, 1\right) \left(2, 2\right) \left(2, 3\right) \left(2,4\right) \left(2, 5\right) \left(2, 6\right) {/tex} {tex} \left(3, 1\right) \left(3, 2\right) \left(3, 3\right) \left(3, 4\right) \left(3, 5\right) \left(3, 6\right) {/tex} {tex} \left(4, 1\right) \left(4, 2\right) \left(4, 3\right) \left(4, 4\right) \left(4, 5\right) \left(4, 6\right) {/tex} {tex} \left(5, 1\right) \left(5, 2\right) \left(5, 3\right) \left(5, 4\right) \left(5, 5\right) \left(5, 6\right) {/tex} {tex} \left(6, 1\right) \left(6, 2\right) \left(6, 3\right) \left(6, 4\right) \left(6, 5\right) \left(6, 6\right) \} {/tex}

n(S) = 36

Number of events to get total of 5 is {tex} \{ \left(1, 4\right) \left(2, 3\right) \left(3, 2\right) \left(4, 1\right) \} {/tex}

n(E) =4

p(E) = {tex} \frac{n \left(E\right) }{n \left(S\right) }=\frac{4}{36}=\frac{1}{9} {/tex}

Note: Probability to get either head or tail when the coin is tossed is {tex} \frac{1}{2} {/tex}

Q 19.

Correct2

Incorrect-0.5

When three coins are tossed. Find the probability to get at most one head.

A

{tex} \frac{2}{8} {/tex}

B

{tex} \frac{5}{8} {/tex}

C

{tex} \frac{7}{8} {/tex}

{tex} \frac{3}{8} {/tex}

Explanation

When the three coins are tossed.

S = {tex} \{ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT \} {/tex}

n(S) = 8

n(E) = {tex} \{ HTT, THT,TTH \} = 3 {/tex}

p(E) = {tex} \frac{n \left(E\right) }{n \left(S\right) }=\frac{3}{8} {/tex}