Class 11 Maths Chapter 13: Statistics worksheet with answer key
A 20-mark worksheet on Statistics for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (b) 2
Multiplying each observation by k = 1/2 multiplies the variance by k² = 1/4: new variance 4, so the new standard deviation is √4 = 2.
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Answer: (c) 19 and 9
Adding a number to each observation shifts the mean by that number (12 + 7 = 19) and leaves the variance unchanged (9).
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Answer: (a) 0
The mean is 36/4 = 9; the deviations are −6, −1, 1, 6, and their sum is 0, as it is for every data about its mean.
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Answer: (a) 4.83
In order: 3, 6, 8, 11, 15, 20. n is even, so M = (8 + 11)/2 = 9.5. The absolute deviations are 6.5, 3.5, 1.5, 1.5, 5.5, 10.5, sum 29; M.D.(M) = 29/6 = 4.83. Dividing by 5 gives the slip 5.80.
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Answer: (d) 2.5
The standard deviation is the positive square root of the variance: √6.25 = 2.5.
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x̄ = 40/4 = 10; the deviations are −5, −1, 2, 4, and their sum is 0. Since this happens for every data set about its mean, the mean of the deviations is always 0/n = 0 and tells nothing about scatter.
- mean 10 and the deviations summing to 0 (1)
- the mean of deviations about the mean is always 0, so it is of no use (1)
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x̄ = 50/5 = 10; |xi − 10| = 7, 3, 1, 3, 8, sum 22, so M.D.(x̄) = 22/5 = 4.4. The median is the 3rd observation, 9; |xi − 9| = 6, 2, 0, 4, 9, sum 21, so M.D.(M) = 21/5 = 4.2 < 4.4.
- M.D.(x̄) = 4.4 (1)
- M = 9, M.D.(M) = 4.2 and the comparison (1)
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Range = 31 − 9 = 22. The range depends only on the two extreme values; it does not tell how the observations are spread about a central value such as the mean, for which a measure based on the deviations from it is needed.
- range 22 (1)
- says the range ignores how the values lie about a central value (1)
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(i) P: deviations −12, 0, 12, so Σ = 144 + 0 + 144 = 288. Q: deviations −m, …, m, so Σ = 2(1² + 2² + … + m²) = m(m + 1)(2m + 1)/3. For m = 7 this is 7 × 8 × 15/3 = 280 < 288; for m = 8 it is 8 × 9 × 17/3 = 408 > 288. The least m is 8 (Q is 22, 23, …, 38, seventeen integers). (ii) Variance of P = 288/3 = 96; variance of Q = 408/17 = 24. Q has the larger sum although P's values lie far wider of the mean: the sum grows with the number of observations, so the mean of the squared deviations, the variance, is the proper measure.
- P's sum 288 and Q's sum m(m + 1)(2m + 1)/3 (1)
- m = 8 (m = 7 gives 280), Q's sum 408 (1)
- variances 96 and 24, and the sum rejected as a measure, with reason (1)
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N = 16, Σ fi xi = 2 + 16 + 36 + 32 + 10 = 96, x̄ = 6. (xi − 6)² = 16, 4, 0, 4, 16; Σ fi (xi − x̄)² = 16 + 16 + 0 + 16 + 16 = 64; σ² = 64/16 = 4, σ = 2. Doubling every frequency doubles both Σ fi (xi − x̄)² and N, and leaves the mean at 6, so σ is still 2.
- x̄ = 6 (1)
- σ² = 64/16 = 4, σ = 2 (1)
- doubling frequencies leaves σ = 2, with reason (1)
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|xi − 30| = 20, 10, 0, 10, 20, so Σ fi |xi − M| = 80 + 70 + 0 + 90 + 20k = 240 + 20k and N = 32 + k. (240 + 20k)/(32 + k) = 10 gives 240 + 20k = 320 + 10k, so k = 8. Check: N = 40, cumulative frequencies 4, 11, 23, 32, 40; the 20th and 21st observations are 30, so the median is indeed 30.
- Σ fi |xi − 30| = 240 + 20k and N = 32 + k (1)
- equation (240 + 20k)/(32 + k) = 10 (1)
- k = 8, with the median checked (1)
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