Chapter Exam

Chapter 8: Statistics and Probability — Online MCQ Test

Q: The range of a set of observations is defined as ______.

Answer: A. highest value - lowest value — Range measures the spread of data and is found by subtracting the smallest observation from the largest observation.

Q: The median of an arranged data set with an odd number of observations is the ______.

Answer: B. middle value — For an odd number of observations, the median is the middle observation after arranging the data in ascending or descending order.

Q: The probability of an event always lies between ______.

Answer: B. 0 and 1 — The probability of any event cannot be negative and cannot exceed 1.

Q: When a single coin is tossed once, the sample space is ______.

Answer: C. {H, T} — A single coin toss has two possible outcomes: Head and Tail. Hence the sample space is {H, T}.

Q: Find the mean of the observations 5, 7, 8, 10, 10.

Answer: B. 8 — Mean = (5 + 7 + 8 + 10 + 10) / 5 = 40 / 5 = 8.

Q: Find the median of the observations 12, 5, 9, 20.

Answer: C. 10.5 — Arranged data: 5, 9, 12, 20. Median = (9 + 12) / 2 = 10.5.

Q: Find the mean for the grouped data: class intervals 0-10, 10-20, 20-30 with frequencies 2, 3, 5.

Answer: C. 18 — Class marks are 5, 15, and 25. Mean = (2×5 + 3×15 + 5×25) / (2 + 3 + 5) = 180/10 = 18.

Q: Find the variance of the observations 2, 4, 6, 8.

Answer: B. 5 — The mean is 5. The squared deviations are 9, 1, 1, 9, so variance = (9 + 1 + 1 + 9) / 4 = 5.

Q: For the grouped data with class intervals 0-10, 10-20, 20-30 and frequencies 3, x, 5, the mean is 16. Find x.

Answer: C. 12 — Class marks are 5, 15, 25. So (3×5 + x×15 + 5×25)/(3 + x + 5) = 16, giving (140 + 15x)/(8 + x) = 16, hence x = 12.

Q: Two fair dice are rolled together. What is the probability of getting at least one six?

Answer: C. 11/36 — Use the complement: probability of no six on both dice = (5/6)×(5/6) = 25/36. Therefore, probability of at least one six = 1 - 25/36 = 11/36.

MATHS · CLASS 10th · Tamil Nadu State Board

Practice Chapter 8: Statistics and Probability with a free chapter-wise online MCQ test for Tamil Nadu State Board CLASS 10th MATHS. This chapter covers: Mean Median Mode Range Variance Standard deviation Frequency distribution Probability Random experiment Sample space Event Addition theorem Complementary events. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 8: Statistics and Probability — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 10th MATHS — Chapter 8: Statistics and Probability, with answers and explanations. These are sample questions; the exam has its own separate question set.

The range of a set of observations is defined as ______.

A. highest value - lowest value ✓
B. highest value + lowest value
C. sum of all values / number of values
D. middle value after arranging

Answer: A. highest value - lowest value
Range measures the spread of data and is found by subtracting the smallest observation from the largest observation.

The median of an arranged data set with an odd number of observations is the ______.

A. most repeated value
B. middle value ✓
C. average of all values
D. difference between two extreme values

Answer: B. middle value
For an odd number of observations, the median is the middle observation after arranging the data in ascending or descending order.

The probability of an event always lies between ______.

A. -1 and 1
B. 0 and 1 ✓
C. 1 and 2
D. 0 and 100 only

Answer: B. 0 and 1
The probability of any event cannot be negative and cannot exceed 1.

When a single coin is tossed once, the sample space is ______.

A. {H}
B. {T}
C. {H, T} ✓
D. {HH, TT}

Answer: C. {H, T}
A single coin toss has two possible outcomes: Head and Tail. Hence the sample space is {H, T}.

Find the mean of the observations 5, 7, 8, 10, 10.

A. 7
B. 8 ✓
C. 9
D. 10

Answer: B. 8
Mean = (5 + 7 + 8 + 10 + 10) / 5 = 40 / 5 = 8.

Find the median of the observations 12, 5, 9, 20.

A. 9
B. 10
C. 10.5 ✓
D. 12

Answer: C. 10.5
Arranged data: 5, 9, 12, 20. Median = (9 + 12) / 2 = 10.5.

Find the mean for the grouped data: class intervals 0-10, 10-20, 20-30 with frequencies 2, 3, 5.

A. 15
B. 16
C. 18 ✓
D. 20

Answer: C. 18
Class marks are 5, 15, and 25. Mean = (2×5 + 3×15 + 5×25) / (2 + 3 + 5) = 180/10 = 18.

Find the variance of the observations 2, 4, 6, 8.

A. 4
B. 5 ✓
C. 6
D. 10

Answer: B. 5
The mean is 5. The squared deviations are 9, 1, 1, 9, so variance = (9 + 1 + 1 + 9) / 4 = 5.

For the grouped data with class intervals 0-10, 10-20, 20-30 and frequencies 3, x, 5, the mean is 16. Find x.

A. 8
B. 10
C. 12 ✓
D. 15

Answer: C. 12
Class marks are 5, 15, 25. So (3×5 + x×15 + 5×25)/(3 + x + 5) = 16, giving (140 + 15x)/(8 + x) = 16, hence x = 12.

Two fair dice are rolled together. What is the probability of getting at least one six?

A. 1/6
B. 5/36
C. 11/36 ✓
D. 25/36

Answer: C. 11/36
Use the complement: probability of no six on both dice = (5/6)×(5/6) = 25/36. Therefore, probability of at least one six = 1 - 25/36 = 11/36.

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