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UPSC CDS  Statistics- Question papers Selection Exam 2012   Objective Questions  Statistics

1. The age distribution of workers in a factory is as follows :

 Age in years No. of workers 20-28 28-36 36-44 44-52 52-60 45 95 100 42 18

If 15% of the total strength starting from lowest age group is retrenched and 20% of the total strength from the highest age groups is given premature retirement, then the age limit of workers retained in the factory is

(a) 20-36

(b) 28-44 (Ans)

(c) 28-52

(d) 36-52

Hints & Solutions  :  Total number pf workers = 300

Retrenched = 15% of 300 = 45

These are all from age group 20-28.

Premature retired = 20% of 300 = 60

= 18 from age group 52-60

and 42 from age group (44-52)

∴ Age limit of workers retained is 28-44.

2. The following age groups are included in the proportion indicated

 Age group Relative proportion in population 12-17 18-23 24-29 30-35 36 + 0.17 0.31 0.27 0.21 0.04

How many of each age-group should be included in a sample of 3000 people to mark the sample representative ?

(a) 850, 155, 135, 905, 955

(b) 510, 930, 810, 630, 120 (Ans)

(c) 600, 600, 600, 600, 600

(d) 510, 630, 950, 100, 810

3.  A data has highest value 120 and lowest value 71. A frequency distribution in descending order with seven classes is to be constructed. The limits of the second class-interval shall be

(a) 71 and 78

(b) 78 and 85 (Ans)

(c) 113 and 120

(d) 106 and 113

4.  If the sum of 11 consecutive natural numbers is 2761, the the middle number is

(a) 249

(b) 250

(c) 251 (Ans)

(d) 252

Hints & Solutions  :  According to question

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

⇒    11x + 55 =  2761

⇒               x =  2761 - 55   = 246
11

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

5.  Mean of 100 items is 49. It was discovered that three items which should have been 60, 70, 80 were wrongly read as 40, 20, 50 respectively. The correct mean is

(a) 48

(b) 82

(c) 50 (Ans)

(d) 80

Hints & Solutions  :  The sum of items = 49 * 100 - (40 + 20 + 50) + (60 + 70 + 80) = 5000

∴  Mean of 100 items =  5000   = 50
100

6.  Consider the following statements :

1.   In a bar graph not only height but also width of each rectangle matters

2.  In a bar graph height of each rectangle matters and not its width

3.  In a histogram the height as well as the width of each rectangle matters.

4.  A bar graph is two-dimensional of these statements

(a) (1) alone is correct

(b) (3) alone is correct

(c) (2) and (3) are correct  (Ans)

(d) (1) and (4) are correct

7.  A pie chart is to be drawn for representing the following data

 Items of expenditure Number of families Education Food and clothing House rent Electricity Miscellaneous 150 400 40 250 160

The value of the central angle for food and clothing would be

(a) 900

(b) 2080

(c) 1500

(d) 1440 (Ans)

Hints & Solutions  :  Central angle for food and clothing

=  400  * 360 = 1440
1000

8. The frequency distribution of some given numbers is

 Value Frequency 1 2 3 4 5 4 6 f

If the mean is known to be 3, then the value of f is

(a) 3

(b) 7

(c) 10

(d) 14 (Ans)

Hints & Solutions  :

 x f fx 1 2 3 4 5 4 6 f 5 8 18 4f ∑f = 15 + f ∑fx = 31 + 4f

∴    Mean  = ∑fx /∑f   = 31 + 4f /15 + f

⇒           3 = 31 + 4f /15 + f

⇒   45 - 31 = f

⇒            f = 14

9. The average of the squares of the numbers 0, 1, 2, 3, 4, ......, n is

(a) 1/2 n (n + 1)

(b) 1/6 n (2n + 1) (Ans)

(c) 1/6 n (n + 1)(2n + 1)

(d) 1/6 n (n + 1)

Hints & Solutions  : Mean =  12 + 22 + 32 + ............ + n2/ (n + 1)

= n (n + 1)(2n + 1)  =  1  n (2n + 1)
6 (n + 1)           6

10. If the average of the numbers 148, 146, 144, 142, ..... in AP be 125, then the total numbers in the series will be

(a) 18

(b) 24 (Ans)

(c) 30

(d) 48

Hints & Solutions  : a =148, d = (146 - 148) = -2

Sn = n/2 [2a + (n - 1)d] = 125n

⇒                     125n = n/2 [2 * 148 + (n - 1) * (-2)]

⇒                 n2 - 24n = 0

⇒               n (n - 24) = 0

⇒                           n = 24                                        [n≠0]

11. If the values 1, 1/2, 1/3, 1/4, 1/5, .........., 1/n occur at frequencies 1, 2, 3, 4, 5, ......, n respectively, in a  frequency distribution, then the mean is

(a) 1

(b) n

(c) 1/n

(d) 2/n + 1 (Ans)

12. The mean of the values of 1,2,3, ....., n with respectively frequencies  x, 2x, 3x, ..... nx is

(a) n/2

(b) 1/3(2n + 1) (Ans)

(c) 1/6(2n + 1)

(d) n/2

Hints & Solutions  :f = x + 2x + 3x + ..... + nx

= x (1 + 2 + 3 + .... + n)

= 1/2 nx (n + 1)

∑(* d) = 1 * x + 2 * 2x + 3 * 3x + ...... + n * nx

=  x(12 + 22 + 32 + .... + n2)

= x 1/6 n (n + 1)(2n + 1)

Mean = ∑(* d)/∑f

= 1/3(2n + 1)

13. The arithmetic mean of data with observations a, a + d, a + 2d, ...., a + 2md is

(a) a + md (Ans)

(b) a + (m - 1)d

(c) a + 1/2 md

(d) a + 1/2 (m + 1)d

Hints & Solutions  : S = a + (a + d) + (a + 2d).... + (a + 2md)

= 2m + 1 [2a  + 2md]
2

= (2m + 1) (a + md)  =   (a + md)
(2m + 1)

14. In a class of 50 students, 10 have failed and their average marks are 28. The total marks obtained by the entire class are 2800. The average marks of those who have passed are

(a) 43

(b) 53

(c) 63 (Ans)

(d) 70

Hints & Solutions  : Total marks of 10 failed students = 28 * 10 = 280

Total marks of 50 students = 2800

∴                                 Total of 40 passed students = 2800 - 280 = 2520

∴                    Average marks of 40 passed students = 2520/40 = 63

15. Consider the frequency distribution given below

 Class-interval Frequency 0-10 10-20 20-30 30-40 40-50 4 6 10 16 14

The mean of the above data is

(a) 25

(b) 35

(c) 30

(d) 31 (Ans)

Hints & Solutions  :

 CI x f fx 0-10 10-20 20-30 30-40 40-50 5 15 25 35 45 4 6 10 16 14 20 90 250 560 630 ∑f = 50 ∑fx = 1550

∴  Mean = ∑fx/∑= 1550/50 = 31

16. The mean of 30 given numbers, when it is given that the mean of 10 of them is 12 and the mean of the remaining 20 is 9, is equal to

(a) 11

(b) 10 (Ans)

(c) 9

(d) 5
_                          _
Hints & Solutions  :
Given that, n1 = 10, x1= 12, n2 = 20, x2 = 9