Aptitude Questions on Statistics for Placements

Welcome to our practice page dedicated to helping you ace Statistics questions for placement aptitude tests! This page is packed with carefully crafted questions designed to build your skills in solving a range of questions. Practice, review solutions, and keep challenging yourself. With consistent practice, you’ll develop a solid foundation to handle any Statistics problem that comes your way!

Q. 1 The average marks of 50 students in a class is 72. If two students scoring 64 and 58 are excluded, what is the new average of the remaining students?

A) 73.2
B) 72.5
C) 71.8
D) 70.9

Check Solution

Ans: B) 72.5

Original total marks = $50 \times 72 = 3600$
Marks of two excluded students = 64+58=122
New total marks = 3600−122=3478
New average = $\frac{3478}{48} = 72.458 \approx 72.5$

Q. 2 The marks scored by 11 students are: 48, 55, 63, 45, 52, 65, 70, 40, 50, 60, 58. What is the median score?

A) 55
B) 58
C) 60
D) 63

Check Solution

Ans: A) 55

Arranging the scores in ascending order: 40, 45, 48, 50, 52, 55, 58, 60, 63, 65, 70
Median = Middle value = 55

Q. 3 The scores of a group are: 5, 7, 9, 11, and 13. What is the standard deviation?

A) 2.58
B) 2.82
C) 3.16
D) 3.45

Check Solution

Ans: B) 2.82

Mean $\mu = \frac{5+7+9+11+13}{5}$
Variance $\sigma^2 = \frac{(5-9)^2 + (7-9)^2 + (9-9)^2 + (11-9)^2 + (13-9)^2}{5}$
Standard deviation $\sigma = \sqrt{8} \approx 2.82$

Q. 4 In a probability distribution, the mean is 25, and the standard deviation is 5. Approximately what percentage of values lie between 20 and 30 (assuming a normal distribution)?

A) 68%
B) 50%
C) 95%
D) 99%

Check Solution

Ans: A) 15 and 25.

In a normal distribution, approximately 68% of data lies within 1 standard deviation of the mean.
Here, 25−5=20 and 25+5=30.

Q. 5 The scores of a class are: 56, 62, 56, 64, 70, 56, 68. What is the mode?

A) 56
B) 62
C) 64
D) 70

Check Solution

Ans: A) 56

The number 56 appears 3 times, more than any other score.

Q. 6 Two groups of students have averages of 75 and 85, with 30 and 20 students, respectively. What is the combined average?

A) 79
B) 80
C) 81
D) 82

Check Solution

Ans: A) 79

Combined total marks = $75 \times 30 + 85 \times 20 = 2250 + 1700 = 3950$
Combined average = $\frac{3950}{50} = 79$

Q. 7 If the minimum and maximum values of a dataset are 15 and 95, respectively, what is the range?

A) 70
B) 75
C) 80
D) 85

Check Solution

Ans: C) 80

Range = 95−15=80

Q. 8 A dataset has a mean of 50 and a standard deviation of 5. What is the coefficient of variation (CV)?

A) 5%
B) 10%
C) 20%
D) 50%

Check Solution

Ans: B) 10%

Coefficient of Variation CV=$ \frac{\sigma}{\mu} \times 100 = \frac{5}{50} \times 100 = 10\%$

Q. 9 In a dataset, the mean is greater than the median. What type of skewness does it have?

A) No skewness
B) Positive skewness
C) Negative skewness
D) Cannot be determined

Check Solution

Ans: B) Positive skewness

If the mean is greater than the median, the dataset is positively skewed

Q. 10 The marks obtained by 8 students are: 12, 18, 24, 30, 36, 42, 48, 54. What is the third quartile (Q3)?

A) 36
B) 42
C) 48
D) 54

Check Solution

Ans: C) 48

Q3=Value at (3/4)(n+1)th position=(3/4)(8+1)=6.75
Interpolating between the 6th and 7th values (42 and 48):
$Q3 = 42 + 0.75(48 – 42) = 42 + 4.5 = 46.5 \approx 48$

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