Statistics Questions

Statistics Questions are provided to give you a basic idea of the concepts. It is important to learn Statistics for the students of Class 9, 10, and 11 according to the CBSE syllabus. The questions listed here cover a range of topics from mean, median, and mode to other complex formulas in probability. Practising them will help you be better prepared for your exams.

Statistics Questions with Solution

Question 1: The ages of 10 students in a class are: 15, 16, 14, 15, 16, 15, 17, 16, 14, and 15. Find the mean age of the students.

Solution:

To find the mean age, add up all the ages and divide by the number of students.
Total sum of ages = 15 + 16 + 14 + 15 + 16 + 15 + 17 + 16 + 14 + 15 = 153.
Number of students = 10.
Mean age = Total sum of ages ÷ Number of students = 153 ÷ 10 = 15.3 years.

Question 2: In a survey of 20 people, 8 people said they liked chocolate ice cream, and 12 people said they liked vanilla. What is the percentage of people who like chocolate ice cream?

Solution:

Percentage = (Number who like chocolate ice cream ÷ Total number of people surveyed) × 100.
= (8 ÷ 20) × 100 = 0.4 × 100 = 40%.
So, 40% of people surveyed like chocolate ice cream.

Question 3: Find the mode of the following set of numbers: 3, 7, 3, 2, 9, 10, 3, 4, 5, 6.

Solution:

The mode is the number that appears most frequently in a set.
In the given set, the number 3 appears three times, more than any other number.
Therefore, the mode of this set is 3.

Question 4: The test scores of 7 students are: 85, 90, 75, 88, 92, 80, and 78. Find the median score.

Solution:

To find the median, first arrange the scores in ascending order: 75, 78, 80, 85, 88, 90, 92.
The median is the middle number in a sorted list.
Since there are 7 scores, the median is the 4th score, which is 85.

Question 5: A student scored the following marks in five subjects: 85, 76, 90, 65, and 88. Calculate the range of the marks.

Solution:

The range is the difference between the highest and lowest values.
Highest mark = 90, Lowest mark = 65.
Range = Highest mark – Lowest mark = 90 – 65 = 25.

Question 6: In a class of 25 students, the heights (in cm) are: 150, 152, 153, 154, 155, 155, 156, 156, 157, 158, 158, 159, 160, 160, 161, 162, 162, 163, 164, 165, 166, 167, 168, 170, 172. Find the median height.

Solution:

First, arrange the heights in ascending order (already arranged).
Since there are 25 heights, the median is the 13th value in the sorted list.
The median height is 160 cm.

Question 7: A dice is rolled 60 times, and the number 6 comes up 15 times. What is the experimental probability of rolling a 6?

Solution:

Experimental probability = (Number of times event occurs) / (Total number of trials).
Number of times a 6 comes up = 15.
Total number of trials = 60.
Probability of rolling a 6 = 15 / 60 = 0.25 or 25%.

Question 8: The scores of a batsman in 10 cricket matches are as follows: 38, 45, 50, 60, 65, 70, 75, 80, 85, 90. Calculate the mean score.

Solution:

To find the mean score, add up all the scores and divide by the number of matches.
Total sum of scores = 38 + 45 + 50 + 60 + 65 + 70 + 75 + 80 + 85 + 90 = 658.
Number of matches = 10.
Mean score = Total sum of scores / Number of matches = 658 / 10 = 65.8.

Question 9: The weights (in kg) of 15 students are: 42, 45, 48, 50, 50, 52, 53, 54, 54, 55, 56, 57, 58, 60, 62. Find the mode of these weights.

Solution:

The mode is the number that appears most frequently in a set.
In this set, the numbers 50 and 54 appear twice, more frequently than any other numbers.
Therefore, the set has two modes: 50 and 54 (bimodal).

Question 10: A bag contains 3 red balls, 4 blue balls, and 5 green balls. If a ball is randomly selected, what is the probability that it is not blue?

Solution:

Total number of balls = 3 (red) + 4 (blue) + 5 (green) = 12.
Number of balls that are not blue = 3 (red) + 5 (green) = 8.
Probability of not selecting a blue ball = Number of non-blue balls / Total number of balls = 8 / 12 = 2 / 3.

Related Links:

• What is Mean?
• What is Mode?
• What is Median?

Question 11: The marks obtained by a group of students in a test are 10, 20, 15, 25, 20, 30, 20, 15, 10, and 35. Find the range of marks.

Solution:

The range is the difference between the highest and the lowest marks.
Highest mark = 35, Lowest mark = 10.
Range = Highest mark – Lowest mark = 35 – 10 = 25 marks.

Question 12: In a survey, 40 people were asked about the number of books they read in a month. The results are as follows: 1, 0, 2, 3, 1, 4, 2, 1, 0, 2, 1, 5, 1, 1, 3, 2, 0, 1, 4, 1, 2, 1, 3, 1, 0, 2, 4, 1, 2, 0, 1, 1, 3, 2, 1, 4, 1, 2, 1, and 0. Calculate the mode of the data.

Solution:

The mode is the value that appears most frequently.
In this data, the number 1 appears most frequently.
Therefore, the mode of this data is 1 book.

Question 13: Calculate the median of the following set of numbers: 11, 18, 14, 20, 12, 16, 17.

Solution:

First, arrange the numbers in ascending order: 11, 12, 14, 16, 17, 18, 20.
Since there are 7 numbers, the median is the 4th number.
Therefore, the median is 16.

Question 14: If the mean of five numbers is 18, and four of these numbers are 15, 20, 25, and 10, find the fifth number.

Solution:

The mean of five numbers is the sum of the numbers divided by 5.
Let the fifth number be x.
Mean = (15 + 20 + 25 + 10 + x) / 5 = 18.
Solving for x: 70 + x = 18 × 5.
x = 90 – 70 = 20.
Therefore, the fifth number is 20.

Question 15: A set of data is 4, 8, 6, 5, 9, 7, 8, 8, 10. What is the mean of this data set?

Solution:

The mean is the sum of the numbers divided by the count of numbers.
Sum of numbers = 4 + 8 + 6 + 5 + 9 + 7 + 8 + 8 + 10 = 65.
Number of data points = 9.
Mean = Sum of numbers / Number of data points = 65 / 9 ≈ 7.22.

Question 16: A student’s grades on five tests are 70, 75, 80, 85, and 90. If the final exam, worth two test grades, is 85, what is the student’s average grade?

Solution:

First, calculate the total of the five test grades and the final exam (counted twice).
Total grades = 70 + 75 + 80 + 85 + 90 + 85 + 85.
Total = 570.
Divide the total by the number of tests (6, since the final counts as two).
Average = 570 / 6 = 95.

Question 17: The mean of 8, 11, 6, 14, x, and 10 is 12. Find the value of x.

Solution:

Mean = (Sum of all numbers) / (Number of numbers).
12 = (8 + 11 + 6 + 14 + x + 10) / 6.
72 = 49 + x.
x = 72 – 49 = 23.

Question 18: In a class of 30 students, 18 play football, 15 play basketball, and 7 play both. How many play neither?

Solution:

Use the formula: n(F ∪ B) = n(F) + n(B) – n(F ∩ B).
n(F ∪ B) is the number of students playing either football or basketball.
n(F) = 18, n(B) = 15, n(F ∩ B) = 7.
n(F ∪ B) = 18 + 15 – 7 = 26.
Number playing neither = Total students – n(F ∪ B) = 30 – 26 = 4.

Question 19: The weights (in kg) of 20 students are normally distributed with a mean of 55 kg and a standard deviation of 5 kg. Find the probability that a randomly selected student weighs less than 60 kg.

Solution:

First, find the z-score: z = (X – mean) / standard deviation.
z = (60 – 55) / 5 = 1.
Use the z-table to find the probability for z = 1.
The probability (approximately) is 0.8413.
So, there is an 84.13% chance a student weighs less than 60 kg.

Question 20: A set of numbers has a median of 21 and a mode of 19. If one more number, 24, is added to the set, what is the new median?

Solution:

Since the median is 21, adding 24 (which is greater than 21) will shift the median upwards.
However, without knowing the number of elements or their specific values, the exact new median cannot be determined.
The new median will be greater than or equal to 21.

Question 21: In a batch of 400 manufactured parts, 20 are found to be defective. What is the probability that a randomly chosen part is not defective?

Solution:

The probability of not being defective = (Total parts – Defective parts) / Total parts.
= (400 – 20) / 400 = 380 / 400.
= 0.95 or 95%.

Question 22: The mean of a data set is 50, and its standard deviation is 5. Using the empirical rule, what percentage of the data falls between 45 and 55?

Solution:

According to the empirical rule (68-95-99.7 rule), approximately 68% of the data in a normal distribution falls within one standard deviation of the mean.
Since 45 and 55 are one standard deviation away from the mean (50), approximately 68% of the data falls between 45 and 55.

Question 23: A set of scores is normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that a randomly selected score is more than 130?

Solution:

First, calculate the z-score for 130: z = (Score – Mean) / Standard deviation.
z = (130 – 100) / 15 = 2.
Using the z-table, find the probability of z being less than 2 and subtract it from 1.
Probability (z < 2) ≈ 0.9772.
Probability (z > 2) = 1 – 0.9772 = 0.0228.
So, there is a 2.28% chance of a score being more than 130.

Question 24: In a correlation study between variables X and Y, the correlation coefficient is found to be -0.8. Interpret this result.

Solution:

A correlation coefficient of -0.8 indicates a strong negative correlation between X and Y.
This means that as X increases, Y tends to decrease, and vice versa.
The value -0.8 is close to -1, indicating a strong linear relationship.

Question 25: The lifetimes of a certain type of component are normally distributed with a mean of 4 years and a standard deviation of 0.5 years. What proportion of these components will last less than 3 years?

Solution:

Calculate the z-score for 3 years: z = (Lifetime – Mean) / Standard deviation.
z = (3 – 4) / 0.5 = -2.
Using the z-table, find the probability of z being less than -2.
Probability (z < -2) ≈ 0.0228.
Thus, about 2.28% of the components are expected to last less than 3 years.

Conclusion

From analyzing population growth to calculating the mean, median, and mode of different datasets, statistics helps us make a lot of decisions. The solutions to the questions we discussed above will improve your problem-solving skills statistics.