Q5. Find M.D.(M) for xi = 15, 21, 27, 30, 35 with frequencies 3, 5, 6, 7, 8 (NCERT Exercise 13.1, Q8 data).
N = 29, so the median is the 15th observation. The cumulative frequencies are 3, 8, 14, 21, 29, so M = 30. The distances 15, 9, 3, 0, 5 times the frequencies give 45, 45, 18, 0, 40, a total of 148. M.D.(M) = 148/29 ≈ 5.10.
Q6. Find the mean deviation about the mean of the daily incomes ₹0–100, 100–200, …, 700–800 of 4, 8, 9, 10, 7, 5, 4, 3 persons (NCERT Exercise 13.1, Q9 data).
The mid-points are 50, 150, …, 750, N = 50 and Σfixi = 17 900, so x̄ = ₹358. The distances 308, 208, 108, 8, 92, 192, 292, 392 times the frequencies add up to 7896, so M.D.(x̄) = 7896/50 = ₹157.92.
Q7. Find the mean deviation about the mean of the heights 95–105, 105–115, …, 145–155 cm of 9, 13, 26, 30, 12, 10 boys (NCERT Exercise 13.1, Q10 data).
The mid-points are 100, 110, …, 150, N = 100 and Σfixi = 12 530, so x̄ = 125.3 cm. The distances 25.3, 15.3, 5.3, 4.7, 14.7, 24.7 times the frequencies give 227.7, 198.9, 137.8, 141, 176.4, 247, a total of 1128.8, so M.D.(x̄) = 11.288 ≈ 11.29 cm.
Q8. Find the mean deviation about the median age for ages 16–20, 21–25, …, 51–55 of 5, 6, 12, 14, 26, 12, 16, 9 persons (NCERT Exercise 13.1, Q12 data).
Make the classes continuous: 15.5–20.5, 20.5–25.5, …, 50.5–55.5, with mid-points 18, 23, …, 53. N = 100 and the cumulative frequencies are 5, 11, 23, 37, 63, …, so the median class is 35.5–40.5:
M = 35.5 + ((50 − 37)/26) × 5 = 38.
The distances 20, 15, 10, 5, 0, 5, 10, 15 times the frequencies add up to 735, so M.D.(M) = 7.35 years.
Q9. Find the mean height in Q7 by step deviations, with A = 130 and h = 10.
di = −3, −2, −1, 0, 1, 2 and Σfidi = −27 − 26 − 26 + 0 + 12 + 20 = −47. So x̄ = 130 + 10 × (−47/100) = 125.3 cm.
Q10. The mean deviation about the median of a data set is 5. Find it when every value x is replaced by −x/5 + 7.
Every distance is multiplied by |−1/5| = 1/5, and adding 7 changes nothing, so the new mean deviation is 5/5 = 1.
Q11. Find the mean deviation of 1, 4, 6, 13 about the mean and about the median.
The mean is 6 and the median is (4 + 6)/2 = 5. The distances from 6 are 5, 2, 0, 7 and from 5 are 4, 1, 1, 8; both add up to 14, so both mean deviations are 3.5. They are equal although the mean and the median differ, because the mean lies between the two middle values.
Q12. Show that the 2m + 1 values a, a + d, …, a + 2md (d > 0) have mean deviation m(m + 1)d/(2m + 1) about the mean.
The mean is the middle value a + md. The distances from it are d, 2d, …, md on each side. Since 1 + 2 + … + m = m(m + 1)/2, they add up to m(m + 1)d. Dividing by 2m + 1 gives the result; for five values (m = 2) it is 6d/5.