Previous year questions: Statistics

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each one uses the formulas from the first page, and every answer is worked out step by step so that the method behind the correct option is clear. The set covers grouped means, mean deviation, variance, standard deviation, combined data and shifts of data.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The mean of the grouped data with classes 5-10, 10-15, 15-20, 20-25, 25-30 and 30-35 and frequencies 2, k, 28, 54, k + 1 and 5 is 21. Find k, and state which of the given quadratic equations has k as a root.

Answer: The class marks are 7.5, 12.5, 17.5, 22.5, 27.5 and 32.5. The total frequency is 90 + 2k. The weighted sum is 15 + 12.5k + 490 + 1215 + 27.5(k + 1) + 162.5 = 1910 + 40k. Setting the mean equal to 21 gives 1910 + 40k = 21(90 + 2k), so 1910 + 40k = 1890 + 42k and k = 10. Substituting k = 10 in the options, the equation 2x squared - 19x - 10 = 0 gives 200 - 190 - 10 = 0, so k is a root of that equation. The correct option is 2x squared - 19x - 10 = 0. A quicker check is to substitute k = 10 into each option; only one option gives zero.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

The mean and variance of n observations are 8 and 16. The sum of the first n - 1 observations is 48 and the sum of their squares is 496. Find n.

Answer: The total sum is 8n, so the last observation is 8n - 48. The variance relation gives (sum of squares)/n - 64 = 16, so the total sum of squares is 80n. The square of the last observation is then 80n - 496. Hence (8n - 48) squared = 80n - 496, which simplifies to 64n squared - 768n + 2304 = 80n - 496, that is 64n squared - 848n + 2800 = 0. Dividing by 16 gives 4n squared - 53n + 175 = 0, whose roots are n = 7 and n = 25/4. Since n is a count, n = 7. The correct option is 7. The other root 25/4 is rejected because the number of observations must be a positive integer.

Q3 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

For 10 observations, the sum of (xi + 2) squared is 180 and the sum of (xi - 1) squared is 90. Find the standard deviation.

Answer: Expand both sums. The first gives sum of xi squared plus 4 times sum of xi plus 40 equal to 180. The second gives sum of xi squared minus 2 times sum of xi plus 10 equal to 90. Subtracting the second from the first gives 6 times sum of xi plus 30 equal to 90, so the sum of xi is 10 and the mean is 1. Substituting back, the sum of xi squared is 100. The variance is 100/10 - 1 = 9, so the standard deviation is 3. The correct option is 3. The two expanded sums differ only in the linear and constant terms, so subtracting them isolates the sum of the observations without knowing the individual values.

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

The values 5, 7, 9, 10, 12 and 15 occur with frequencies 8, 6, 2, 2, 2 and 6. Work out the mean deviation about the mean.

Answer: The total frequency is 26 and the weighted sum is 40 + 42 + 18 + 20 + 24 + 90 = 234, so the mean is 9. The weighted absolute deviations are 8 times 4, 6 times 2, 2 times 0, 2 times 1, 2 times 3 and 6 times 6, that is 32, 12, 0, 2, 6 and 36. Their sum is 88, so the mean deviation is 88/26 = 44/13. The correct option is 44/13.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The mean and variance of the seven observations 2, 4, alpha, 8, beta, 12 and 14 are 8 and 16, with alpha less than beta. Find the quadratic equation whose roots are 3 alpha + 2 and 2 beta + 1.

Answer: The sum of the observations is 40 + alpha + beta, so 40 + alpha + beta = 56 and alpha + beta = 16. The sum of squares is 424 + alpha squared + beta squared, and the variance relation gives (424 + alpha squared + beta squared)/7 - 64 = 16, so alpha squared + beta squared = 136. Then alpha beta = ((alpha + beta) squared - (alpha squared + beta squared))/2 = (256 - 136)/2 = 60. The pair is 6 and 10, with alpha = 6 and beta = 10. The roots are 20 and 21, whose sum is 41 and product is 420, so the equation is x squared - 41x + 420 = 0. Check the roots: 20 and 21 add to 41 and multiply to 420, as required by the sum and product relations.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Find the variance of all ten observations together.

Answer: For the first set, the sum of squares is 4 times (13 + 1) = 56. For the second set, the sum of squares is 6 times (1 + 4) = 30. The ten observations have total sum 4 + 12 = 16 and total sum of squares 86. The combined mean is 16/10 = 1.6, and the combined variance is 86/10 - (1.6) squared = 8.6 - 2.56 = 6.04. The correct option is 6.04. A common mistake is to average 13 and 1, which ignores both the unequal group sizes and the gap between the group means.

Q7 (NCERT Exemplar, 1 mark)

Find the mean deviation of the data 2, 9, 9, 3, 6, 9 and 4 from the mean.

Answer: The sum is 42 and the mean is 6. The absolute deviations are 4, 3, 3, 3, 0, 3 and 2, whose sum is 18. With seven observations the mean deviation is 18/7, which is about 2.57. The correct option is 2.57. The deviations are taken about the mean and their absolute values are averaged, so the order of the observations does not matter.

Q8 (NCERT Exemplar, 1 mark)

A set of n values has standard deviation 6. Find the standard deviation of the n values obtained by adding the same constant k to every value.

Answer: Adding a constant to every observation adds the same constant to the mean, so each deviation from the mean is unchanged. The squared deviations are unchanged, so the variance and the standard deviation are unchanged. The standard deviation of the shifted values is therefore 6, the same as before. The correct option is that the standard deviation stays 6. Only the mean moves with the shift, and the spread about the mean is untouched.

Q9 (NCERT Exemplar, 2 marks)

Find the variance and standard deviation of the observations 57, 64, 43, 67, 49, 59, 44, 47, 61 and 59.

Answer: The sum is 550 and the mean is 55. The deviations are 2, 9, minus 12, 12, minus 6, 4, minus 11, minus 8, 6 and 4, whose squares add to 4 + 81 + 144 + 144 + 36 + 16 + 121 + 64 + 36 + 16 = 662. The variance is 662/10 = 66.2 and the standard deviation is the square root of 66.2, about 8.13.

Q10 (NCERT Exemplar, 2 marks)

The distribution has values 1, 3, 5, 7, 9, 11, 13 and 15, occurring with frequencies 3, 3, 4, 14, 7, 4, 3 and 4. Compute the mean deviation about the mean.

Answer: The total frequency is 42 and the weighted sum is 3 + 9 + 20 + 98 + 63 + 44 + 39 + 60 = 336, so the mean is 8. The weighted absolute deviations are 3 times 7, 3 times 5, 4 times 3, 14 times 1, 7 times 1, 4 times 3, 3 times 5 and 4 times 7, that is 21, 15, 12, 14, 7, 12, 15 and 28. Their sum is 124, so the mean deviation is 124/42, which is about 2.95. The value with the largest frequency, 7, sits close to the mean, which is why the mean deviation is small.