More practice questions

5. Q5. Find the mean and variance for x_i = 92, 93, 97, 98, 102, 104, 109 with frequencies 3, 2, 3, 2, 6, 3, 3 (NCERT Exercise 13.2, Q5).

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   N = 22 and Σf_ix_i = 2200, so the mean is 100. The squared deviations 64, 49, 9, 4, 4, 16, 81 times the frequencies add up to 640, so the variance is 640/22 = 320/11 ≈ 29.09.

6. Q6. Find the mean and standard deviation by the short-cut method for x_i = 60, 61, …, 68 with frequencies 2, 1, 12, 29, 25, 12, 10, 4, 5 (NCERT Exercise 13.2, Q6).

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   Take A = 64 and h = 1, so y_i = −4, …, 4. N = 100, Σf_iy_i = 0 and Σf_iy_i² = 286. The mean is 64, the variance 286/100 = 2.86 and the standard deviation √2.86 ≈ 1.69.

7. Q7. Find the mean, variance and standard deviation of the heights 70–75, 75–80, …, 110–115 cm of 3, 4, 7, 7, 15, 9, 6, 6, 3 children by the short-cut method (NCERT Exercise 13.2, Q9).

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   The mid-points are 72.5, …, 112.5. With A = 92.5 and h = 5, y_i = −4, …, 4, N = 60, Σf_iy_i = 6 and Σf_iy_i² = 254. The mean is 92.5 + 5 × 6/60 = 93 cm, and

   σ² = 25 × (254/60 − (6/60)²) = 1267/12 ≈ 105.58,

   so σ ≈ 10.28 cm.

8. Q8. Find the mean and standard deviation of the diameters 33–36, 37–40, 41–44, 45–48, 49–52 mm of 15, 17, 21, 22, 25 circles (NCERT Exercise 13.2, Q10).

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   As NCERT’s hint says, make the classes continuous: 32.5–36.5, …, 48.5–52.5, with mid-points 34.5, 38.5, …, 50.5. With A = 42.5 and h = 4, y_i = −2, …, 2, N = 100, Σf_iy_i = 25 and Σf_iy_i² = 199. The mean is 42.5 + 4 × 0.25 = 43.5 mm, the variance 16 × (1.99 − 0.0625) = 30.84 and σ ≈ 5.55 mm.

9. Q9. 20 observations have mean 10 and standard deviation 2, but one observation 8 was wrong. Find the correct mean and standard deviation if the wrong item is (i) left out, (ii) replaced by 12 (NCERT Miscellaneous Exercise, Q5).

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   Σx = 200 and Σx² = 20(4 + 100) = 2080.

   (i) 19 values: Σx = 192 and Σx² = 2016. The mean is 192/19 ≈ 10.11, the variance 2016/19 − (192/19)² = 1440/361 and σ ≈ 2.00 (1.997).

   (ii) Σx = 204 and Σx² = 2160. The mean is 10.2, the variance 108 − 104.04 = 3.96 and σ ≈ 1.99.

10. Q10. 100 observations have mean 20 and standard deviation 3. Three of them, recorded as 21, 21 and 18, were wrong. Find the mean and standard deviation without them (NCERT Miscellaneous Exercise, Q6).

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    Σx = 2000 − 60 = 1940 for 97 values, so the mean is still 20. Σx² = 100(9 + 400) − (441 + 441 + 324) = 39 694, so the variance is 39 694/97 − 400 = 894/97 ≈ 9.22 and σ ≈ 3.04.

11. Q11. If x̄ and σ² are the mean and variance of x₁, …, x_n, prove that ax₁, …, ax_n (a ≠ 0) have mean ax̄ and variance a²σ² (NCERT Miscellaneous Exercise, Q4).

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    The new mean is (1/n) Σax_i = ax̄. Each new deviation, ax_i − ax̄, equals a(x_i − x̄), so its square is a²(x_i − x̄)², and the mean of these squares is a²σ².

12. Q12. Ten observations have mean 5 and variance 4. Two more observations, 3 and 7, are added. Find the new mean and variance.

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    Σx = 50 + 3 + 7 = 60 for 12 values, so the mean stays 5. Σx² = 10(4 + 25) + 9 + 49 = 348, so the variance is 348/12 − 25 = 4: unchanged, because the new values are each 2 from the mean, exactly σ.