Class 11 Mathematics Statistics MCQs Set 09

Answer: (a) 7991

Explanation:
1. Incorrect sum = 200 × 40 = 8000.
2. Correct sum = 8000 − (34 + 53) + (43 + 35) = 8000 − 87 + 78 = 7991

Teacher's Note:
1. Remove the wrong values and add the correct ones; do not recompute from scratch.
2. Incorrect total = n × incorrect mean.

Answer: (b) 39.955

Explanation:
1. Correct mean \( = \frac{7991}{200} = 39.955 \)

Teacher's Note:
1. The number of candidates stays 200; only the total changes.
2. The new mean is slightly less than 40 because the true values add up to less than the misread ones.

Answer: (a) \( \frac{\sum_{i=1}^{n}(x_i - \bar{x})^2}{n} \)

Explanation:
1. Variance is the mean of the squared deviations: \( \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \bar{x})^2 \)

Teacher's Note:
1. Without dividing by n, the expression is only the sum of squares.
2. The sum of deviations without squaring is always 0, so it cannot measure spread.

Answer: (c) 224.143

Explanation:
1. Incorrect Σx² = n(σ² + x̄²) = 200(225 + 1600) = 365000
2. Correct Σx² = 365000 − 34² − 53² + 43² + 35² = 365000 − 1156 − 2809 + 1849 + 1225 = 364109
3. Correct \( \sigma^2 = \frac{364109}{200} - (39.955)^2 = 1820.545 - 1596.402 = 224.143 \)

Teacher's Note:
1. Use \( \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 \) to recover Σx² from the incorrect mean and SD.
2. Fix Σx² the same way as Σx: subtract the squares of the wrong values and add the squares of the right ones.

Answer: (a) 14.971

Explanation:
1. \( \sigma = \sqrt{224.143} \approx 14.971 \)

Teacher's Note:
1. SD is the positive square root of the new variance.
2. It is close to the original 15, as expected for a small correction.

Answer: (c) 45.5

Explanation:
1. Sum = 455, n = 10, so \( \bar{x} = 45.5 \)

Teacher's Note:
1. Add carefully and divide by the number of observations.
2. The mean need not be one of the observations.

Answer: (d) 9.0

Explanation:
1. |xᵢ − 45.5| = 11.5, 20.5, 15.5, 7.5, 1.5, 4.5, 5.5, 14.5, 3.5, 5.5; total = 90
2. MD \( = \frac{90}{10} = 9.0 \)

Teacher's Note:
1. Use the mean 45.5 found in part (i).
2. Absolute deviations are all taken as positive.

Answer: (c) 43

Explanation:
1. Ascending order: 30, 34, 38, 40, 42, 44, 50, 51, 60, 66; M \( = \frac{42 + 44}{2} = 43 \)

Teacher's Note:
1. For 10 values, average the 5th and 6th values.
2. Sorting first is essential for the median.

Answer: (d) 8.7

Explanation:
1. |xᵢ − 43| = 13, 9, 5, 3, 1, 1, 7, 8, 17, 23 (sorted order); total = 87
2. MD(M) \( = \frac{87}{10} = 8.7 \)

Teacher's Note:
1. Mean deviation about the median is never more than that about the mean.
2. Here 8.7 < 9.0, as expected.

Answer: (c) 0.3

Explanation:
1. MD about mean = 9.0 and MD about median = 8.7.
2. Difference = 9.0 − 8.7 = 0.3

Teacher's Note:
1. The sum of absolute deviations is least about the median.
2. So this difference is always zero or positive.

Answer: (b) 7.5

Explanation:
1. N = 40, Σfᵢxᵢ = 4 + 40 + 60 + 56 + 80 + 60 = 300
2. \( \bar{x} = \frac{300}{40} = 7.5 \)

Teacher's Note:
1. Multiply each value by its frequency before adding.
2. Divide by N = Σfᵢ = 40.

Answer: (c) 2.3

Explanation:
1. fᵢ|xᵢ − 7.5| = 2(5.5), 8(2.5), 10(1.5), 7(0.5), 8(2.5), 5(4.5) = 11, 20, 15, 3.5, 20, 22.5; total = 92
2. MD \( = \frac{92}{40} = 2.3 \)

Teacher's Note:
1. Multiply each absolute deviation by its frequency.
2. Keep decimals exact; the total 92 divides neatly by 40.

Answer: (c) 7

Explanation:
1. N = 40; cumulative frequencies 2, 10, 20, 27, 35, 40.
2. 20th value = 6 and 21st value = 8, so M \( = \frac{6 + 8}{2} = 7 \)

Teacher's Note:
1. For even N take the average of the (N/2)th and (N/2 + 1)th values.
2. Here the two middle values fall in different rows of the table.

Answer: (d) 2.3

Explanation:
1. fᵢ|xᵢ − 7| = 2(5), 8(2), 10(1), 7(1), 8(3), 5(5) = 10, 16, 10, 7, 24, 25; total = 92
2. MD(M) \( = \frac{92}{40} = 2.3 \)

Teacher's Note:
1. Here both mean deviations happen to be equal, 2.3.
2. Still, MD about the median can never exceed MD about the mean.

Answer: (c) 0.5

Explanation:
1. Mean = 7.5 and median = 7.
2. Difference = 0.5

Teacher's Note:
1. Use the values found in the earlier parts.
2. A mean greater than the median hints at a slight right skew in the data.

Answer: (a) \( \bar{x} = \frac{\Sigma f_i x_i}{N} \)

Explanation:
1. For a frequency distribution, \( \bar{x} = \frac{\Sigma f_i x_i}{N} \), where N = Σfᵢ.

Teacher's Note:
1. Each value is weighted by its frequency.
2. Σfᵢxᵢ² is used for variance, not for the mean.

Answer: (a) \( \sigma^2 = \frac{1}{N}\Sigma f_i(x_i - \bar{x})^2 \)

Explanation:
1. Variance of a discrete frequency distribution: \( \sigma^2 = \frac{1}{N}\Sigma f_i(x_i - \bar{x})^2 \)

Teacher's Note:
1. Variance uses squared deviations from the mean, weighted by frequency.
2. The plain sum of deviations is always 0, so option (c) is useless.

Answer: (c) 14

Explanation:
1. N = 30, Σfᵢxᵢ = 12 + 40 + 99 + 85 + 80 + 72 + 32 = 420
2. \( \bar{x} = \frac{420}{30} = 14 \)

Teacher's Note:
1. Add the frequencies to get N = 30.
2. Tabulate fᵢxᵢ in a column to avoid slips.

Answer: (b) 45.8

Explanation:
1. (xᵢ − 14)² = 100, 36, 9, 9, 36, 100, 324
2. Σfᵢ(xᵢ − 14)² = 300 + 180 + 81 + 45 + 144 + 300 + 324 = 1374
3. \( \sigma^2 = \frac{1374}{30} = 45.8 \)

Teacher's Note:
1. Square the deviations before multiplying by the frequencies.
2. Divide by N = 30.

Answer: (a) 6.77

Explanation:
1. \( \sigma = \sqrt{45.8} \approx 6.77 \)

Teacher's Note:
1. SD is the positive square root of the variance.
2. Check: 6.77² ≈ 45.8.

Answer: (a) \( M = l + \frac{\frac{N}{2} - cf}{f} \times h \)

Explanation:
1. For grouped data, \( M = l + \frac{\frac{N}{2} - cf}{f} \times h \), where l is the lower limit of the median class, cf the cumulative frequency of the preceding class, f the frequency of the median class and h the class width.

Teacher's Note:
1. The median class is the class whose cumulative frequency first exceeds N/2.
2. cf is the cumulative frequency of the class before the median class, not of the median class itself.

Answer: (b) \( MD = \frac{\Sigma f_i|x_i - M|}{N} \)

Explanation:
1. \( MD(M) = \frac{1}{N}\Sigma f_i|x_i - M| \), where xᵢ are the class mid-points.

Teacher's Note:
1. Deviations are measured from the median, so the sign inside the modulus is minus.
2. Each deviation must be weighted by its frequency.

Answer: (c) 50

Explanation:
1. N = 6 + 7 + 15 + 16 + 4 + 2 = 50

Teacher's Note:
1. Total frequency is the sum of all frequencies, not the number of classes.
2. 60 is the upper limit of the last class, a tempting wrong choice.

Answer: (a) 28

Explanation:
1. N/2 = 25; cumulative frequencies 6, 13, 28, 44, 48, 50. The median class is 20-30.
2. l = 20, cf = 13, f = 15, h = 10: \( M = 20 + \frac{25 - 13}{15} \times 10 = 20 + 8 = 28 \)

Teacher's Note:
1. The median class is the first class whose cumulative frequency is at least N/2.
2. The median must lie inside the median class, here between 20 and 30.

Answer: (a) 10.16

Explanation:
1. Mid-points 5, 15, 25, 35, 45, 55; |xᵢ − 28| = 23, 13, 3, 7, 17, 27.
2. fᵢ|xᵢ − 28| = 138, 91, 45, 112, 68, 54; total = 508
3. MD(M) \( = \frac{508}{50} = 10.16 \)

Teacher's Note:
1. Use class mid-points as xᵢ for continuous distributions.
2. Keep the median 28 from the previous part.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Variance is in squared units (e.g. cm²) while data and mean are in cm; R is true.
2. To get a measure in the same unit as the data we take the positive square root of variance, the standard deviation; A is true and R explains it.

Teacher's Note:
1. Standard deviation has the same unit as the data, which is why it is preferred for reporting spread.
2. σ is always taken as the positive root of σ².

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. N = 40, Σfᵢxᵢ = 12 + 40 + 98 + 216 + 192 + 112 + 90 = 760, so x̄ = 19. A is true.
2. Σfᵢ(xᵢ − 19)² = 338 + 324 + 175 + 12 + 200 + 324 + 363 = 1736; σ² = 1736/40 = 43.4. R is true.
3. The variance does not explain the value of the mean, so R is not the explanation of A.

Teacher's Note:
1. The mean must be found before the variance, but the variance does not justify the mean.
2. Organise the work in columns of fᵢxᵢ and fᵢ(xᵢ − x̄)².

Answer: (d) A is false, but R is true.

Explanation:
1. Take assumed mean a = 64, dᵢ = xᵢ − 64 = −4, …, 4; N = 100.
2. Σfᵢdᵢ = 0, so x̄ = 64 + 0 = 64, not 32. A is false.
3. Σfᵢdᵢ² = 286, so \( \sigma = \sqrt{\frac{286}{100} - 0^2} = \sqrt{2.86} \approx 1.69 \). R is true.

Teacher's Note:
1. In the shortcut method, x̄ = a + Σfᵢdᵢ / N.
2. The mean must lie between the smallest and largest values (60 to 68), so 32 is impossible.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. New mean = x̄ + a, so each new deviation (xᵢ + a) − (x̄ + a) = xᵢ − x̄ is unchanged.
2. Hence the variance is unchanged. A is true and R states the general property behind it.

Teacher's Note:
1. Shifting all data changes the mean by the same amount but leaves the spread unchanged.
2. Multiplying the data, by contrast, changes the variance.

Answer: (d) A is false, but R is true.

Explanation:
1. Mean \( = 5 + \frac{\Sigma(x - 5)}{n} = 5 + \frac{3}{18} = 5.1666 \), not 4.1666. A is false.
2. \( \sigma^2 = \frac{43}{18} - \left(\frac{3}{18}\right)^2 = 2.3889 - 0.0278 = 2.3611 \), so σ ≈ 1.54. R is true.

Teacher's Note:
1. Deviations from an assumed value a give x̄ = a + Σd/n.
2. Variance from such deviations is Σd²/n − (Σd/n)².