Statistics - MCQs

STATISTICS 

This article contains few statistics MCQs related to mean, median, mode and standard deviation, asked in TCS Freshers placements.

Q 1) The mean of the median, the mode and the range of the following data: 15, 10, 17, 13, 25, 17, 11, 18, 14, 19, 12, 20

a. 15	b. 16(2/3)
c. 16	d. 15(1/3)

Solution:

c) 16
Arrange the data in increasing order-
10,11,12,13,14,15,17,17,18,19,20,25
Median = (Sum of the middle two terms)/2 = (15+17) /2 = 16
Mode = Number repeating the most number of times = 17
Range =10 to 25 = 15
Mean of median, mode and range = (16+17+15) /3 =16

Q 2) What is the mean of the mode and the median on the following data?
12, 28, 26, 27, 17, 16, 22, 25, 15, 16, 11:

a. 17	b. 16.5
c. 16	d. 16.3

Solution:

b) 16.5
Arrange the data in increasing order-
11,12,15,16,16,17,22,25,26,27,28
Median = Middle term = 17
Mode = Number repeating most number of times = 16
Mean of median and mode =16.5

Q 3) With what value should the highest quantity in the data 65,52,14,26,18,35,32,38 be replaced so that the mean and median become equal?

a. 66	b. 53
c. 64	d. 51

Solution:

b) 53
The mean of data = (65+52+14+26+18+35+32+38)/ 8 = (280/8) = 35
To find median we need to arrange data in ascending order.
The median of even data = (Sum of the middle terms) /2
= (32+35) /2 = 33.5
When we (280-65) = 215
Now, (215+A) /8= 33.5
Therefore A=53
Hence, 65 should be replaced with 53.

Q 4) What is the mean deviation of the data 8,9,12,15,16,20,24,30,32,34?

a. 10.2	b. 8
c. 0	d. 9.6

Solution:

b) 8
First find the mean of the data= (8+9+12+15+16+20+24+30+32+34) /10 = (200/10) = 20
Find the distance of each value from that mean

Value	Distance from mean i.e. 20
8	12
9	11
12	8
15	5
16	4
20	0
24	4
30	10
32	12
34	14

Now find the mean of those distances
= (12+11+8+5+4+0+4+10+12+14) /10 = (80/10) = 8

Q 5) The collection of numbers which comprise the data given below is arranged in ascending order. (3,7,9, N - 1,15,18,19,20)
If the median of the data is 12.5, what is the value of N?

a. 10.5	b. 11.5
c. 11	d. 12

Solution:

c) 11
Median of odd data= The value of the middle term
Median of even data= (Sum of middle terms)/2
12.5 = [(N-1) + 15] /2
Therefore, N=11

Q 6) The mean of a set of data is 5. What will be the mean if ten is subtracted from each data?

a. -5	b. 5
c. 10	d. -15

Solution:

a) -5
Let the size of data be N.
Mean= Sum of the data / Size of the data
Therefore, Sum of the data =5N
As 10 is subtracted from each data,
the new Sum= 5N-10N = -5N
the new Mean= -5N/N = -5
Hence the new Mean is -5

Q 7) The Range and the standard deviation of a data are R & S respectively. With the shift of the origin of the data, change(s) occur in the value(s) of:

a. Both R and S	b. R only
c. Neither R nor S	d. S only

Solution:

c) Neither R nor S

Q 8) In case of frequency distribution of ten continuous classes, the class width is 4 and the lower-class limit of the lowest class is 8. What is the upper-class limit for the highest class?

a. 48	b. 40
c. 43	d. 45

Solution:

a) 48
Let x and y be the upper and lower class limit of frequency distribution.
x – y = 4
But lower class limit is 8
Therefore, x – 8 = 4 i.e. x = 12 (Upper limit of lowest class)
Therefore the upper limit of the highest class
= Number of continuous class * Class width + Lower class limit of lowest class
= 10 * 4 + 8 = 48

Q 9) If mean of 29 observations is 33 and on adding one more observation the new mean becomes 34. What is the value of 30th observation?

a. 68	b. 63
c. 34	d. 55

Solution:

b) 63
Mean =Sum/Number
33= Sum / 29
Therefore, Sum= 33*29 = 957
New Mean =34 = (Sum + A) / 30
Therefore, (Sum + A) = 34*30= 1020
A=1020-957 = 63

Q 10) The index numbers of five commodities are 121,123,125,126,128 and the weight assigned to these are respectively 5,11,10,8,6. Then what is the weighted average index number?

a. 123.8	b. 125.2
c. 124.6	d. 124.2

Solution:

c) 124.6
Total Weight = 5+11+10+8+6 = 40
Commodity * Weight = (121*5) + (123*11) + (125*10) + (126*8) + (128*6) = 4984
Weighted Average = (Commodity * Weight) / Total Weight
= (4984 / 40) = 124.6

Q 11) If the mean of the data 5,7,10,14, x,25,32,19 is x, then what is the value of x?

a. 14.5	b. 15
c. 14	d. 16

Solution:

d) 16
Mean = x = (5+7+10+14+x+25+32+19) / 8 = (112+x) / 8
Therefore, 8x = 112 + x
Hence x = 16

Q 12)The mean and the standard deviation of a data which is comprised of a set of ten positive numbers are 8 and 2 respectively. If the sum of squares of 9 among the 10 members is 599, what is the 10th number?

a. 9	b. 8
c. 7	d. 11

Solution:

a) 9
Let the 10th number be x.
Mean of 10 numbers is 8. Therefore, the sum of those 10 numbers is 80.
The standard deviation is 2.
The square of 9 numbers is 599.
Therefore, square of all numbers= square of 9 numbers + square of 10th number= 599 + x^2
Substituting all these in the standard deviation formula,
2=square_root [ {(599+x^2)/10} – (8^2) ]
Solving this, we get x = -9 or x = 9
But as all the numbers are positive, x = 9

Q 13) If a and b represent the frequencies of the modal class and median class respectively of the data, the value of 3a+4b is _______.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	12	14	10	17	7

a. 98	b. 27
c. 90	d. 91

Solution:

d) 16
Here the modal class (class with the highest frequency) is 30-40.
Therefore, modal frequency(a) is 17.
There are (12+14+10+17+7) = 60 data items.
Therefore, the median class is 60/2 =30 i.e. 20-30.
Therefore, frequency of median class (b) = 10.
As per the question, 3(17) + 4(10) = 91

Q 14) The median of the frequency distribution table is _______.

Class	50-54	55-59	60-64	65-69	70-74	75-80
Frequency	12	26	30	8	13	11

a. 62	b. 61
c. 61.5	d. 62.5

Solution:

a) 62
To find the median group, we find the sum of frequency
= 12+26+30+8+13+11 = 100
Therefore, median group lies in 100/2 = 50
i.e. in 60-64 class (12+26+30 > 50)
L is the lower bound of median class. L = 60
N is the size of data. N = 100
B is the cumulative frequency of the groups before the median group. B= 12+26 = 38
G is the frequency of the median group. G=30
W is the width of the group. W = 5
As per the formula, median = L + [ { (N/2) – B} * (W/G) ]
Therefore, median = 62

Q 15)The median of the following data is _______.

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	8	12	10	14	16	10

a. 31.571	b. 33.571
c. 32.571	d. 34.571

Solution:

b) 33.571
To find the median group, we find the sum of frequency
= 8+12+10+14+16+10 = 70
Therefore, median group lies in 70/2 = 35
i.e. in 30-40 class (8+12+10+14 > 35)
L is the lower bound of median class. L = 30
N is the size of data. N = 70
B is the cumulative frequency of the groups before the median group. B= 8+12+10 = 30
G is the frequency of the median group. G=14
W is the width of the group. W = 10
As per the formula, median = L + [ { (N/2) – B} * (W/G) ]
Therefore, median = 33.571

Q 16) The median of the following data is _______.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	10	15	12	15	8

a. 24.366	b. 24.266
c. 24.123	d. 24.166

Solution:

d) 24.166
To find the median group, we find the sum of frequency
= 10+15+12+15+8 = 60
Therefore, median group lies in 60/2 = 30
i.e. in 20-30 class (10+15+12 >30)
L is the lower bound of median class. L = 30
N is the size of data. N = 60
B is the cumulative frequency of the groups before the median group. B= 10+15 =25
G is the frequency of the median group. G=12
W is the width of the group. W = 10
As per the formula, median = L + [ { (N/2) – B} * (W/G) ]
Therefore, median = 24.166

Q 17) Let x be the upper limit of the model class and y be the class mark of the median class of the following data:

Class	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	8	12	10	11	13	6

What is the value (x - y)?

a. 5.5	b. 9.5
c. 10.5	d. 7.5

Solution:

d) 7.5
The modal class is the one with highest frequency i.e. (30-35)
Therefore, the upper limit of modal class is 30
The size of data is 60
Therefore, the median class is 20-25 (60/2 =30 & CF of 30 lies in 20-25)
Therefore, the class mark of median class is 22.5
X-Y = 30-22.5 = 7.5