Class 11 Mathematics Statistics MCQs Set 08

Answer: (d) 2

Explanation:
1. Mean \( \bar{x} = \frac{-1 + 0 + 4}{3} = 1 \)
2. |xᵢ − x̄| = 2, 1, 3; total = 6
3. MD \( = \frac{6}{3} = 2 \)

Teacher's Note:
1. Mean deviation uses absolute deviations, so it can never be negative.
2. Find the mean first, then the distance of each value from it.

Answer: (a) 178

Explanation:
1. Mean \( = \frac{7230}{5} = 1446 \)
2. |xᵢ − 1446| = 89, 356, 220, 48, 177; total = 890
3. MD \( = \frac{890}{5} = 178 \) hours

Teacher's Note:
1. With large numbers, find each deviation carefully; one slip changes the answer.
2. The mean deviation has the same unit (hours) as the data.

Answer: (c) 5.27

Explanation:
1. Ascending order: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 (n = 11), so the median is the 6th value, M = 9.
2. |xᵢ − 9| = 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12; total = 58
3. MD(M) \( = \frac{58}{11} \approx 5.27 \)

Teacher's Note:
1. Always arrange the data in order before picking the median.
2. For odd n the median is the ((n + 1)/2)th observation.

Answer: (d) 5

Explanation:
1. n = 10, so M = mean of the 5th and 6th values \( = \frac{40 + 42}{2} = 41 \)
2. |xᵢ − 41| = 9, 7, 5, 3, 1, 1, 3, 5, 7, 9; total = 50
3. MD(M) \( = \frac{50}{10} = 5 \)

Teacher's Note:
1. For even n, the median is the average of the two middle observations.
2. For evenly spaced data the deviations form a symmetric pattern, which is a useful check.

Answer: (c) 5.1

Explanation:
1. N = 29; cumulative frequencies 3, 8, 14, 21, 29. The median is the 15th value, which lies in cf 21, so M = 30.
2. fᵢ|xᵢ − 30| = 3(15), 5(9), 6(3), 7(0), 8(5) = 45, 45, 18, 0, 40; total = 148
3. MD(M) \( = \frac{148}{29} \approx 5.1 \)

Teacher's Note:
1. For a discrete frequency table, use cumulative frequencies to locate the median.
2. Multiply each absolute deviation by its frequency before adding.

Answer: (b) 3.23

Explanation:
1. N = 26; cumulative frequencies 8, 14, 16, 18, 20, 26. The 13th and 14th values are both 7, so M = 7.
2. fᵢ|xᵢ − 7| = 8(2), 6(0), 2(2), 2(3), 2(5), 6(8) = 16, 0, 4, 6, 10, 48; total = 84
3. MD(M) \( = \frac{84}{26} \approx 3.23 \)

Teacher's Note:
1. When N is even, check both middle positions in the cumulative frequency column.
2. Here both the 13th and 14th observations fall in the class with cf 14.

Answer: (b) 2.57

Explanation:
1. Mean \( = \frac{49}{7} = 7 \)
2. |xᵢ − 7| = 4, 3, 3, 3, 0, 3, 2; total = 18
3. MD \( = \frac{18}{7} \approx 2.57 \)

Teacher's Note:
1. Repeated values must each be counted separately.
2. Divide by the number of observations, 7, not by the number of distinct values.

Answer: (c) 12.67

Explanation:
1. Ascending order: 20, 33, 39, 40, 50, 53, 59, 65, 69; median = 5th value = 50.
2. |xᵢ − 50| = 30, 17, 11, 10, 0, 3, 9, 15, 19; total = 114
3. MD(M) \( = \frac{114}{9} \approx 12.67 \)

Teacher's Note:
1. With 9 observations the median is the 5th value after sorting.
2. The deviation of the median itself is 0, a quick check on your list.

Answer: (b) 10

Explanation:
1. Mid-points xᵢ = 15, 25, 35, 45, 55, 65, 75; N = 40, Σfᵢxᵢ = 1800, so x̄ = 45.
2. fᵢ|xᵢ − 45| = 2(30), 3(20), 8(10), 14(0), 8(10), 3(20), 2(30) = 60, 60, 80, 0, 80, 60, 60; total = 400
3. MD \( = \frac{400}{40} = 10 \)

Teacher's Note:
1. For class intervals, use the mid-point of each class as xᵢ.
2. The frequencies are symmetric about 40-50, so the mean is the middle mid-point 45.

Answer: (b) 8.6

Explanation:
1. Ascending order: 44, 48, 52, 54, 56, 58, 64, 65, 73, 80; M \( = \frac{56 + 58}{2} = 57 \)
2. |xᵢ − 57| = 13, 9, 5, 3, 1, 1, 7, 8, 16, 23; total = 86
3. MD(M) \( = \frac{86}{10} = 8.6 \)

Teacher's Note:
1. For 10 values, the median is the average of the 5th and 6th values.
2. The median need not be one of the observations.

Answer: (a) 1.1

Explanation:
1. Sum = 54, n = 9, so \( \bar{x} = 6 \)
2. |xᵢ − 6| = 0, 1, 0.75, 0.5, 1.25, 1.5, 0.25, 1.75, 3; total = 10
3. MD \( = \frac{10}{9} \approx 1.1 \)

Teacher's Note:
1. Decimal data needs careful addition; pair numbers like 5.25 + 4.75 to simplify.
2. A mean deviation of about 1 suggests the data cluster closely round 6.

Answer: (a) 8.4

Explanation:
1. Sum = 500, n = 10, so \( \bar{x} = 50 \)
2. |xᵢ − 50| = 12, 20, 2, 10, 8, 5, 13, 4, 4, 6; total = 84
3. MD \( = \frac{84}{10} = 8.4 \)

Teacher's Note:
1. The mean does not require sorting, unlike the median.
2. Ignore signs of deviations when finding mean deviation.

Answer: (a) 3.24

Explanation:
1. N = 34; cumulative frequencies 3, 8, 12, 22, 30, 34. The 17th and 18th values are both 18, so M = 18.
2. fᵢ|xᵢ − 18| = 3(8), 5(6), 4(3), 10(0), 8(3), 4(5) = 24, 30, 12, 0, 24, 20; total = 110
3. MD(M) \( = \frac{110}{34} \approx 3.24 \) years

Teacher's Note:
1. Locate the median from the cumulative frequency column, not from the middle of the table.
2. Check that the frequencies add up to N before starting.

Answer: (c) 7

Explanation:
1. Ascending order: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72; M \( = \frac{46 + 49}{2} = 47.5 \)
2. |xᵢ − 47.5| = 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5; total = 70
3. MD(M) \( = \frac{70}{10} = 7 \)

Teacher's Note:
1. Write the sorted data in a row and the deviations below it to avoid missing any value.
2. Repeated values such as 46, 46 occupy separate positions.

Answer: (b) 2.33

Explanation:
1. Ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18; M \( = \frac{13 + 14}{2} = 13.5 \)
2. |xᵢ − 13.5| = 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5; total = 28
3. MD(M) \( = \frac{28}{12} \approx 2.33 \)

Teacher's Note:
1. For 12 values, average the 6th and 7th values to get the median.
2. Count the observations once more after sorting to make sure none is lost.

Answer: (d) 93.33

Explanation:
1. Each observation of the new data is 2 times the corresponding old observation.
2. If every observation is multiplied by k, the variance is multiplied by k².
3. New variance = 2² × 23.33 = 4 × 23.33 ≈ 93.33

Teacher's Note:
1. Multiplying data by k multiplies the standard deviation by |k| but the variance by k².
2. Doubling the variance (46.66) is a common mistake.

Answer: (d) 8.25

Explanation:
1. Adding a constant to every number does not change the variance.
2. Variance of 1, 2, …, 10: mean 5.5, \( \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 = \frac{385}{10} - 30.25 = 8.25 \)
3. Hence the new variance is also 8.25.

Teacher's Note:
1. Variance measures spread, and shifting all values by the same amount does not change the spread.
2. 2.87 is the standard deviation, not the variance.

Answer: (c) 19

Explanation:
1. Mid-points 6, 10, 14, 18; N = 20, Σfᵢxᵢ = 18 + 60 + 56 + 126 = 260, so x̄ = 13.
2. fᵢ(xᵢ − 13)² = 3(49), 6(9), 4(1), 7(25) = 147, 54, 4, 175; total = 380
3. \( \sigma^2 = \frac{380}{20} = 19 \)

Teacher's Note:
1. Use class mid-points as representative values.
2. Square each deviation before multiplying by the frequency.

Answer: (c) 9, 9.25

Explanation:
1. Sum = 72, n = 8, so \( \bar{x} = 9 \)
2. (xᵢ − 9)² = 9, 4, 1, 9, 16, 25, 1, 9; total = 74
3. \( \sigma^2 = \frac{74}{8} = 9.25 \)

Teacher's Note:
1. Variance is the mean of the squared deviations from the mean.
2. Check: the deviations themselves (−3, −2, 1, 3, 4, −5, −1, 3) add up to 0.

Answer: (a) \( \sqrt{\frac{52}{7}} \)

Explanation:
1. Sum = 63, n = 7, so \( \bar{x} = 9 \)
2. (xᵢ − 9)² = 9, 16, 0, 16, 9, 1, 1; total = 52
3. \( \sigma = \sqrt{\frac{52}{7}} \)

Teacher's Note:
1. Standard deviation is the positive square root of the variance.
2. 52/7 is the variance; do not stop before taking the root.

Answer: (c) \( \sqrt{9.25} \)

Explanation:
1. \( \bar{x} = \frac{72}{8} = 9 \); Σ(xᵢ − 9)² = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74
2. \( \sigma^2 = \frac{74}{8} = 9.25 \), so \( \sigma = \sqrt{9.25} \)

Teacher's Note:
1. Find the variance first and then take its square root.
2. √9.25 ≈ 3.04, which is a sensible spread for values from 4 to 13.

Answer: (a) 5000, 251600

Explanation:
1. Σx = n x̄ = 100 × 50 = 5000
2. \( \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 \Rightarrow 16 = \frac{\sum x^2}{100} - 2500 \)
3. Σx² = 100 × 2516 = 251600

Teacher's Note:
1. The form \( \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2 \) links the sum of squares to the mean and SD.
2. Remember to square the SD (4² = 16) before using it.

Answer: (b) 66.2

Explanation:
1. Sum = 560, n = 10, so \( \bar{x} = 56 \)
2. (xᵢ − 56)² = 121, 16, 36, 16, 36, 81, 4, 144, 144, 64; total = 662
3. \( \sigma^2 = \frac{662}{10} = 66.2 \)

Teacher's Note:
1. Write deviations with signs first, then square them.
2. 56 is the mean, which is one of the distractors written as 56.2.

Answer: (d) 17

Explanation:
1. N = 20, Σfᵢxᵢ = 30 + 30 + 90 + 160 + 50 = 360, so x̄ = 18.
2. fᵢ(xᵢ − 18)² = 3(64), 2(9), 5(0), 8(4), 2(49) = 192, 18, 0, 32, 98; total = 340
3. \( \sigma^2 = \frac{340}{20} = 17 \)

Teacher's Note:
1. For frequency data divide by N = Σfᵢ, not by the number of distinct values.
2. A table with columns xᵢ, fᵢ, fᵢxᵢ, (xᵢ − x̄)², fᵢ(xᵢ − x̄)² keeps the work organised.

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

Explanation:
1. The deviations xᵢ − x̄ have positive and negative values which cancel, so Σ(xᵢ − x̄) = 0; R is true.
2. That is why we take absolute values |xᵢ − x̄| (or squares) to measure dispersion; A is true.
3. R gives the reason for A, so R explains A.

Teacher's Note:
1. Mean deviation uses absolute values and variance uses squares, both to stop deviations cancelling.
2. The fact Σ(xᵢ − x̄) = 0 is a quick check on any table of deviations.

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

Explanation:
1. A: x̄ = 80/8 = 10; |xᵢ − 10| = 6, 3, 2, 1, 0, 2, 3, 7, total 24; MD = 24/8 = 3. A is true.
2. R: x̄ = 500/10 = 50; |xᵢ − 50| total = 84; MD = 8.4, not 8.5. R is false.

Teacher's Note:
1. Small arithmetic differences like 8.4 and 8.5 decide the truth of such statements, so compute exactly.
2. Each data set needs its own mean.

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

Explanation:
1. A: N = 25, Σfx = 350, x̄ = 14; Σf|x − 14| = 63 + 16 + 6 + 18 + 55 = 158; MD = 158/25 = 6.32. A is true.
2. R: N = 80, Σfx = 4000, x̄ = 50; Σf|x − 50| = 160 + 480 + 0 + 320 + 320 = 1280; MD = 1280/80 = 16, not 15. R is false.

Teacher's Note:
1. Mean deviation for a frequency table is Σfᵢ|xᵢ − x̄| / N.
2. Compute the mean of each table separately before the deviations.

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

Explanation:
1. When variability is very high, the median does not represent the data well.
2. So a mean deviation measured from such a median cannot be fully relied upon. Both are true and R explains A.

Teacher's Note:
1. Mean deviation is only as reliable as the central value it is measured from.
2. This is one of the stated limitations of mean deviation.

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

Explanation:
1. The sum of absolute deviations is least when taken about the median, so the sum about the mean is larger.
2. This makes mean deviation about the mean not very scientific for highly variable data, which is a limitation. Both are true and R explains A.

Teacher's Note:
1. Σ|xᵢ − a| is minimum when a is the median.
2. Such limitations are why variance and standard deviation are preferred.

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

Explanation:
1. A: let the fraction of boys be p; 52p + 42(1 − p) = 50 ⇒ 10p = 8 ⇒ p = 0.8, i.e. 80%. A is true.
2. R: let the fraction of girls be g; 55g + 50(1 − g) = 53 ⇒ 5g = 3 ⇒ g = 0.6, i.e. 60%, not 64%. R is false.

Teacher's Note:
1. The combined mean is a weighted average of the group means, weighted by the group sizes.
2. Working with fractions p and 1 − p avoids needing the actual class size.

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

Explanation:
1. A: sorted 27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45; median = 8th value = 35. The changes affect only the extreme values, so the new median is still 35. A is true.
2. R: sorted 40, 42, 45, 47, 50, 51, 54, 57; M \( = \frac{47 + 50}{2} = 48.5 \).
3. |xᵢ − 48.5| total = 8.5 + 6.5 + 3.5 + 1.5 + 1.5 + 2.5 + 5.5 + 8.5 = 38; MD = 38/8 = 4.75, not 4.78. R is false.

Teacher's Note:
1. The median depends only on the position of the middle value, so changing extreme values does not affect it.
2. Compute mean deviations to two decimal places to compare with the stated value.