Practice Quantitative Aptitude questions on Statistics, covering mean, median, mode, range, averages, frequency distributions, and basic data analysis. Takes about 20–25 minutes and is useful for placement tests, TCS NQT, SSC, banking and MBA entrance exams.
Question 1
The heights of a group of players are normally distributed with a mean of 80 cm and a standard deviation of 8 cm. What is the probability that a randomly chosen player is taller than 92 cm?
Explanation
z-score for 92 = Value − MeanStandard deviation = 92 − 808 = 1.5.
From the standard normal table, P(z < 1.5) = 0.9332.
P(z > 1.5) = 1 − 0.9332 = 0.0668.
So there is about a 6.68% chance that a player is taller than 92 cm.
From the standard normal table, P(z < 1.5) = 0.9332.
P(z > 1.5) = 1 − 0.9332 = 0.0668.
So there is about a 6.68% chance that a player is taller than 92 cm.
Shortcut Method
z = 92 − 808 = 1.5
P(z < 1.5) = 0.9332
∴ P(z > 1.5) = 1 − 0.9332 = 6.68% ✓
P(z < 1.5) = 0.9332
∴ P(z > 1.5) = 1 − 0.9332 = 6.68% ✓
Question 2
The scores on a test are normally distributed with a mean of 60 and a standard deviation of 12. What is the probability that a randomly selected score is more than 72?
Explanation
z-score for 72 = Score − MeanStandard deviation = 72 − 6012 = 1.
From the standard normal table, P(z < 1) = 0.8413.
P(z > 1) = 1 − 0.8413 = 0.1587.
So there is about a 15.87% chance that a score is more than 72.
From the standard normal table, P(z < 1) = 0.8413.
P(z > 1) = 1 − 0.8413 = 0.1587.
So there is about a 15.87% chance that a score is more than 72.
Shortcut Method
z = 72 − 6012 = 1
P(z < 1) = 0.8413
∴ P(z > 1) = 1 − 0.8413 = 15.87% ✓
P(z < 1) = 0.8413
∴ P(z > 1) = 1 − 0.8413 = 15.87% ✓
Question 3
The mean of a data set is 120 and its standard deviation is 10. Using the empirical rule, approximately what percentage of the data lies between 100 and 140?
Explanation
The empirical (68-95-99.7) rule says about 95% of the data in a normal distribution lies within two standard deviations of the mean.
Mean − 2 SD = 120 − 20 = 100 and mean + 2 SD = 120 + 20 = 140.
So about 95% of the data lies between 100 and 140.
Mean − 2 SD = 120 − 20 = 100 and mean + 2 SD = 120 + 20 = 140.
So about 95% of the data lies between 100 and 140.
Shortcut Method
100 = 120 − 2(10) and 140 = 120 + 2(10) (±2 SD)
∴ 95% ✓
∴ 95% ✓
Question 4
The mean of a data set is 60 and its standard deviation is 8. Using the empirical rule, approximately what percentage of the data lies between 52 and 68?
Explanation
The empirical (68-95-99.7) rule says about 68% of the data in a normal distribution lies within one standard deviation of the mean.
Mean − SD = 60 − 8 = 52 and mean + SD = 60 + 8 = 68.
So about 68% of the data lies between 52 and 68.
Mean − SD = 60 − 8 = 52 and mean + SD = 60 + 8 = 68.
So about 68% of the data lies between 52 and 68.
Shortcut Method
52 = 60 − 8 and 68 = 60 + 8 (±1 SD)
∴ 68% ✓
∴ 68% ✓
Question 5
The median of the numbers 3, 5, 8, 12, 15, 20 and 25 is 12. If the number 30 is added to the set, what is the new median?
Explanation
After adding 30 the set is 3, 5, 8, 12, 15, 20, 25, 30, which has 8 values (an even number).
The median is the mean of the 4th and 5th values.
Median = 12 + 152 = 13.5.
The median is the mean of the 4th and 5th values.
Median = 12 + 152 = 13.5.
Shortcut Method
new set has 8 values
middle two = 12 and 15
∴ Median = 12 + 152 = 13.5 ✓
middle two = 12 and 15
∴ Median = 12 + 152 = 13.5 ✓
Question 6
A student scored 70, 80 and 90 in three class tests. The final exam counts as two tests, and she scored 75 in it. What is her average grade?
Explanation
The final exam counts twice, so its score is counted twice and the total weight is 3 + 2 = 5.
Weighted total = 70 + 80 + 90 + 75 + 75 = 390.
Average = 3905 = 78.
Weighted total = 70 + 80 + 90 + 75 + 75 = 390.
Average = 3905 = 78.
Shortcut Method
weights: 3 tests + 2 for the final = 5
total = 70 + 80 + 90 + 75 × 2 = 390
∴ Average = 3905 = 78 ✓
total = 70 + 80 + 90 + 75 × 2 = 390
∴ Average = 3905 = 78 ✓
Question 7
In a class of 40 students, 25 play cricket, 18 play badminton and 9 play both. How many students play neither game?
Explanation
Use n(C ∪ B) = n(C) + n(B) − n(C ∩ B), where C = cricket and B = badminton.
n(C ∪ B) = 25 + 18 − 9 = 34 students play at least one game.
Number playing neither = 40 − 34 = 6.
n(C ∪ B) = 25 + 18 − 9 = 34 students play at least one game.
Number playing neither = 40 − 34 = 6.
Shortcut Method
play at least one = 25 + 18 − 9 = 34
neither = 40 − 34
∴ 6 students ✓
neither = 40 − 34
∴ 6 students ✓
Question 8
The mean of five numbers is 24. If four of the numbers are 18, 30, 22 and 26, find the fifth number.
Explanation
Sum of the five numbers = 24 × 5 = 120.
Sum of the four known numbers = 18 + 30 + 22 + 26 = 96.
Fifth number = 120 − 96 = 24.
Sum of the four known numbers = 18 + 30 + 22 + 26 = 96.
Fifth number = 120 − 96 = 24.
Shortcut Method
total = 24 × 5 = 120
known sum = 96
∴ fifth = 120 − 96 = 24 ✓
known sum = 96
∴ fifth = 120 − 96 = 24 ✓
Question 9
The mean of 12, 7, 15, x, 9 and 10 is 11. Find the value of x.
Explanation
Mean = Sum of all numbersNumber of numbers.
11 = 12 + 7 + 15 + x + 9 + 106 = 53 + x6.
53 + x = 11 × 6 = 66, so x = 66 − 53 = 13.
11 = 12 + 7 + 15 + x + 9 + 106 = 53 + x6.
53 + x = 11 × 6 = 66, so x = 66 − 53 = 13.
Shortcut Method
total = 11 × 6 = 66
known sum = 12 + 7 + 15 + 9 + 10 = 53
∴ x = 66 − 53 = 13 ✓
known sum = 12 + 7 + 15 + 9 + 10 = 53
∴ x = 66 − 53 = 13 ✓
Question 10
A coin is tossed 80 times and heads comes up 36 times. What is the experimental probability of getting a head?
Explanation
Experimental probability = Number of times the event occursTotal number of trials.
Number of heads = 36 and total tosses = 80.
Probability = 3680 = 920 = 0.45 or 45%.
Number of heads = 36 and total tosses = 80.
Probability = 3680 = 920 = 0.45 or 45%.
Shortcut Method
headstosses = 3680
∴ 920 = 45% ✓
∴ 920 = 45% ✓
Question 11
The weights (in kg) of 25 students, already arranged in ascending order, are 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 52, 52, 53, 54, 55, 56, 57, 58. Find the median.
Explanation
Arrange the values in ascending order: 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 52, 52, 53, 54, 55, 56, 57, 58.
There are 25 values, which is odd, so the median is the middle value, that is the 13th value.
The 13th value is 48, so the median is 48.
There are 25 values, which is odd, so the median is the middle value, that is the 13th value.
The 13th value is 48, so the median is 48.
Shortcut Method
Sort: 40, 41, 41, 42, 43, 44, 44, 45, 46, 46, 47, 48, 48, 49, 50, 50, 51, 52, 52, 53, 54, 55, 56, 57, 58
middle value
∴ Median = 48 ✓
middle value
∴ Median = 48 ✓
Question 12
Find the median of these 8 numbers: 14, 22, 9, 31, 18, 27, 12, 25.
Explanation
Arrange the values in ascending order: 9, 12, 14, 18, 22, 25, 27, 31.
There are 8 values, which is even, so the median is the mean of the two middle values (the 4th and 5th).
Median = 18 + 222 = 20.
There are 8 values, which is even, so the median is the mean of the two middle values (the 4th and 5th).
Median = 18 + 222 = 20.
Shortcut Method
Sort: 9, 12, 14, 18, 22, 25, 27, 31
average of the two middle values
∴ Median = 20 ✓
average of the two middle values
∴ Median = 20 ✓
Question 13
Calculate the median of the following set of numbers: 23, 9, 17, 31, 12, 28, 20.
Explanation
Arrange the values in ascending order: 9, 12, 17, 20, 23, 28, 31.
There are 7 values, which is odd, so the median is the middle value, that is the 4th value.
The 4th value is 20, so the median is 20.
There are 7 values, which is odd, so the median is the middle value, that is the 4th value.
The 4th value is 20, so the median is 20.
Shortcut Method
Sort: 9, 12, 17, 20, 23, 28, 31
middle value
∴ Median = 20 ✓
middle value
∴ Median = 20 ✓
Question 14
The temperatures (in degrees Celsius) recorded on six days were 24, 31, 19, 27, 35, 22. Find the range.
Explanation
Range = Highest value − Lowest value.
Highest value = 35, lowest value = 19.
Range = 35 − 19 = 16.
Highest value = 35, lowest value = 19.
Range = 35 − 19 = 16.
Shortcut Method
Range = highest − lowest
35 − 19
∴ Range = 16 ✓
35 − 19
∴ Range = 16 ✓
Question 15
A student scored the following marks in five tests: 72, 58, 91, 64, 85. Find the range.
Explanation
Range = Highest value − Lowest value.
Highest value = 91, lowest value = 58.
Range = 91 − 58 = 33.
Highest value = 91, lowest value = 58.
Range = 91 − 58 = 33.
Shortcut Method
Range = highest − lowest
91 − 58
∴ Range = 33 ✓
91 − 58
∴ Range = 33 ✓
Question 16
A hockey club noted the number of goals scored in each of 30 matches: 1, 3, 0, 2, 1, 1, 4, 2, 1, 0, 3, 1, 2, 1, 5, 1, 2, 0, 1, 3, 1, 2, 1, 0, 2, 1, 4, 1, 2, 1. Find the mode.
Explanation
The mode is the value that occurs most often.
Counting each value: 0 occurs 4 times, 1 occurs 13 times, 2 occurs 7 times, 3 occurs 3 times, 4 occurs 2 times, 5 occurs 1 time.
1 occurs 13 times, more than any other value, so the mode is 1.
Counting each value: 0 occurs 4 times, 1 occurs 13 times, 2 occurs 7 times, 3 occurs 3 times, 4 occurs 2 times, 5 occurs 1 time.
1 occurs 13 times, more than any other value, so the mode is 1.
Shortcut Method
Count repeats: 0 ×4, 1 ×13, 2 ×7, 3 ×3, 4 ×2, 5 ×1
∴ Mode = 1 ✓
∴ Mode = 1 ✓
Question 17
In a survey of 30 families, the number of children in each family was recorded as follows: 2, 1, 3, 2, 0, 2, 1, 4, 2, 3, 1, 2, 2, 0, 1, 3, 2, 1, 2, 4, 2, 3, 1, 2, 0, 2, 1, 3, 2, 2. Find the mode.
Explanation
The mode is the value that occurs most often.
Counting each value: 0 occurs 3 times, 1 occurs 7 times, 2 occurs 13 times, 3 occurs 5 times, 4 occurs 2 times.
2 occurs 13 times, more than any other value, so the mode is 2.
Counting each value: 0 occurs 3 times, 1 occurs 7 times, 2 occurs 13 times, 3 occurs 5 times, 4 occurs 2 times.
2 occurs 13 times, more than any other value, so the mode is 2.
Shortcut Method
Count repeats: 0 ×3, 1 ×7, 2 ×13, 3 ×5, 4 ×2
∴ Mode = 2 ✓
∴ Mode = 2 ✓
Question 18
In a class of 40 students, 14 walk to school. What percentage of the students walk to school?
Explanation
Percentage = Number who walkTotal number of students × 100.
= 1440 × 100 = 0.35 × 100 = 35%.
= 1440 × 100 = 0.35 × 100 = 35%.
Shortcut Method
1440 × 100
∴ 35% ✓
∴ 35% ✓
Question 19
The number of units produced in 9 shifts was 6, 9, 7, 5, 8, 9, 6, 10, 9. Find the mean.
Explanation
Mean = Sum of all valuesNumber of values.
Sum = 6 + 9 + 7 + 5 + 8 + 9 + 6 + 10 + 9 = 69.
Number of values = 9.
Mean = 699 = 7.67.
Sum = 6 + 9 + 7 + 5 + 8 + 9 + 6 + 10 + 9 = 69.
Number of values = 9.
Mean = 699 = 7.67.
Shortcut Method
Mean = sumcount
699
∴ Mean = 7.67 ✓
699
∴ Mean = 7.67 ✓
Question 20
The points scored by a team in 8 matches were 15, 22, 30, 45, 48, 52, 60, 68. Find the mean.
Explanation
Mean = Sum of all valuesNumber of values.
Sum = 15 + 22 + 30 + 45 + 48 + 52 + 60 + 68 = 340.
Number of values = 8.
Mean = 3408 = 42.5.
Sum = 15 + 22 + 30 + 45 + 48 + 52 + 60 + 68 = 340.
Number of values = 8.
Mean = 3408 = 42.5.
Shortcut Method
Mean = sumcount
3408
∴ Mean = 42.5 ✓
3408
∴ Mean = 42.5 ✓
Question 21
A player scored the following runs in 10 matches: 24, 36, 40, 52, 58, 61, 70, 74, 82, 93. Find the mean.
Explanation
Mean = Sum of all valuesNumber of values.
Sum = 24 + 36 + 40 + 52 + 58 + 61 + 70 + 74 + 82 + 93 = 590.
Number of values = 10.
Mean = 59010 = 59.
Sum = 24 + 36 + 40 + 52 + 58 + 61 + 70 + 74 + 82 + 93 = 590.
Number of values = 10.
Mean = 59010 = 59.
Shortcut Method
Mean = sumcount
59010
∴ Mean = 59 ✓
59010
∴ Mean = 59 ✓
Question 22
The daily wages (in Rs.) of 10 workers are 320, 340, 300, 360, 340, 320, 380, 340, 300, 300. Find the mean.
Explanation
Mean = Sum of all valuesNumber of values.
Sum = 320 + 340 + 300 + 360 + 340 + 320 + 380 + 340 + 300 + 300 = 3300.
Number of values = 10.
Mean = 330010 = 330.
Sum = 320 + 340 + 300 + 360 + 340 + 320 + 380 + 340 + 300 + 300 = 3300.
Number of values = 10.
Mean = 330010 = 330.
Shortcut Method
Mean = sumcount
330010
∴ Mean = 330 ✓
330010
∴ Mean = 330 ✓
Question 23
Find the mode of the following set of numbers: 5, 8, 5, 2, 9, 5, 8, 1, 5, 3.
Explanation
The mode is the value that occurs most often.
Counting each value: 5 occurs 4 times, 8 occurs 2 times.
5 occurs 4 times, more than any other value, so the mode is 5.
Counting each value: 5 occurs 4 times, 8 occurs 2 times.
5 occurs 4 times, more than any other value, so the mode is 5.
Shortcut Method
Count repeats: 5 ×4, 8 ×2
∴ Mode = 5 ✓
∴ Mode = 5 ✓
Question 24
The marks of 15 students are 41, 43, 44, 44, 46, 47, 47, 48, 50, 51, 52, 53, 54, 55, 57. What are the modes of this data?
Explanation
The mode is the value that occurs most often.
44 occurs 2 times and 47 occurs 2 times, and no other value occurs as often.
Two values share the highest frequency, so the data is bimodal with modes 44 and 47.
44 occurs 2 times and 47 occurs 2 times, and no other value occurs as often.
Two values share the highest frequency, so the data is bimodal with modes 44 and 47.
Shortcut Method
44 ×2, 47 ×2, others fewer
∴ Modes = 44 and 47 ✓
∴ Modes = 44 and 47 ✓
Question 25
The quiz scores of 12 students are 6, 8, 9, 9, 7, 8, 5, 9, 8, 4, 9, 8. What are the modes of this data?
Explanation
The mode is the value that occurs most often.
8 occurs 4 times and 9 occurs 4 times, and no other value occurs as often.
Two values share the highest frequency, so the data is bimodal with modes 8 and 9.
8 occurs 4 times and 9 occurs 4 times, and no other value occurs as often.
Two values share the highest frequency, so the data is bimodal with modes 8 and 9.
Shortcut Method
8 ×4, 9 ×4, others fewer
∴ Modes = 8 and 9 ✓
∴ Modes = 8 and 9 ✓
