CBSE Class 11 Mathematics Statistics Competitive Exam MCQs Set 01

Answer: (c) \( \frac{\bar{x} + 10\alpha}{\alpha} \)

Explanation:
1. Each new observation is \( y_i = \frac{x_i}{\alpha} + 10 \).
2. Mean of \( y_i \) = \( \frac{\bar{x}}{\alpha} + 10 \) because the mean follows every linear change.
3. \( \frac{\bar{x}}{\alpha} + 10 = \frac{\bar{x} + 10\alpha}{\alpha} \), which is option (c).

Teacher's Note:
1. The mean is affected by both change of origin (adding) and change of scale (multiplying or dividing).
2. Apply the operations to the mean in the same order in which they are applied to the observations.

Answer: (d) \( a\bar{x} + b \)

Explanation:
1. Sum of new numbers = \( a(x_1 + x_2 + \ldots + x_n) + nb = an\bar{x} + nb \).
2. New mean = \( \frac{an\bar{x} + nb}{n} = a\bar{x} + b \).

Teacher's Note:
1. If y = ax + b for every observation, then the mean of y is a x̄ + b.
2. Option (b) is the sum of the new numbers, not their mean; do not forget to divide by n.

Answer: (b) \( \frac{15}{4} \)

Explanation:
1. Sum of first 10 = 10 × 4.5 = 45; sum of remaining 30 = 30 × 3.5 = 105.
2. Total = 150 for 40 observations.
3. Average = \( \frac{150}{40} = \frac{15}{4} \).

Teacher's Note:
1. To combine groups, first convert each average back into a total, add the totals, then divide by the total count.
2. The combined mean always lies between the group means, here between 3.5 and 4.5.

Answer: (b) 44.46

Explanation:
1. Wrong total = 100 × 45 = 4500.
2. Correct total = 4500 − (91 + 13) + (19 + 31) = 4500 − 104 + 50 = 4446.
3. Correct mean = \( \frac{4446}{100} = 44.46 \).

Teacher's Note:
1. For wrongly recorded values, subtract the wrong values from the total and add the right ones.
2. Work with totals, not with the mean directly, to avoid errors.

Answer: (d) 58 kg

Explanation:
1. Total weight of students = 35 × 40 = 1400 kg.
2. With the teacher, 36 persons have average 40.5 kg, total = 36 × 40.5 = 1458 kg.
3. Teacher's weight = 1458 − 1400 = 58 kg.

Teacher's Note:
1. The teacher must supply the extra half kg for all 36 persons plus his own 40 kg: 40 + 36 × 0.5 = 58.
2. Always multiply the new average by the new number of persons.

Answer: (a) \( \frac{nM - a}{4} \)

Explanation:
1. Sum of all n observations = nM.
2. Sum of remaining 4 observations = nM − a.
3. Their mean = \( \frac{nM - a}{4} \).

Teacher's Note:
1. The product of the mean and the number of observations gives their total.
2. Divide the leftover total by the number of leftover observations, here 4.

Answer: (d) measure of dispersion

Explanation:
1. Mean and median are measures of central tendency; they locate the centre of the data.
2. A number that describes the spread or variability of the data is a measure of dispersion.

Teacher's Note:
1. Range, mean deviation, variance and standard deviation are all measures of dispersion.
2. Two data sets can have the same mean but very different spreads, which is why dispersion is needed.

Answer: (c) 5.27

Explanation:
1. Arranged: 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21; n = 11, median = 6th value = 9.
2. |x − 9|: 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12; sum = 58.
3. M.D. = \( \frac{58}{11} \) ≈ 5.27.

Teacher's Note:
1. Always arrange the data in order before picking the median.
2. Take absolute deviations, so every term is non-negative before adding.

Answer: (b) 2.57

Explanation:
1. Mean = \( \frac{3 + 10 + 10 + 4 + 7 + 10 + 5}{7} = \frac{49}{7} = 7 \).
2. |x − 7|: 4, 3, 3, 3, 0, 3, 2; sum = 18.
3. M.D. = \( \frac{18}{7} \) ≈ 2.57.

Teacher's Note:
1. Find the mean first, then the absolute deviations from it.
2. Check that the signed deviations add to zero; that confirms the mean is right.

Answer: (a) 178

Explanation:
1. Mean = \( \frac{1357 + 1090 + 1666 + 1494 + 1623}{5} = \frac{7230}{5} = 1446 \).
2. |x − 1446|: 89, 356, 220, 48, 177; sum = 890.
3. M.D. = \( \frac{890}{5} = 178 \) hours.

Teacher's Note:
1. With large numbers, compute the deviations carefully and add them in pairs to reduce slips.
2. The mean deviation has the same unit as the data, here hours.

Answer: (c) 12.67

Explanation:
1. Arranged: 20, 33, 39, 40, 50, 53, 59, 65, 69; median = 5th value = 50.
2. |x − 50|: 30, 17, 11, 10, 0, 3, 9, 15, 19; sum = 114.
3. M.D. = \( \frac{114}{9} \) ≈ 12.67.

Teacher's Note:
1. For an odd number n of observations, the median is the \( \left(\frac{n+1}{2}\right)^{th} \) value after ordering.
2. Mean deviation about the median is the smallest possible mean deviation for a data set.

Answer: (b) 2.57

Explanation:
1. Mean = \( \frac{2 + 9 + 9 + 3 + 6 + 9 + 4}{7} = \frac{42}{7} = 6 \).
2. |x − 6|: 4, 3, 3, 3, 0, 3, 2; sum = 18.
3. M.D. = \( \frac{18}{7} \) ≈ 2.57.

Teacher's Note:
1. Repeated values must each be counted; 9 appears three times here.
2. Dividing by the number of observations, not by the number of distinct values, is essential.

Answer: (c) 8

Explanation:
1. Mean = \( \frac{30}{5} = 6 \).
2. Squared deviations: 16, 4, 0, 4, 16; sum = 40.
3. Variance = \( \frac{40}{5} = 8 \).

Teacher's Note:
1. Variance is the mean of the squared deviations from the mean.
2. Shortcut: 2, 4, …, 10 is 2 × (1, 2, …, 5), so variance = 4 × \( \frac{5^2 - 1}{12} \) = 8.

Answer: (a) K

Explanation:
1. Each number of the second set is obtained by adding 10 to the corresponding number of the first set.
2. Adding a constant does not change the spread, so the S.D. stays K.

Teacher's Note:
1. Standard deviation is independent of change of origin but depends on change of scale.
2. Shifting all values moves the mean by the same amount, so the deviations from the mean are unchanged.

Answer: (a) Only Statement I is true

Explanation:
1. I: let fraction of boys be p: 52p + 42(1 − p) = 50 ⇒ 10p = 8 ⇒ p = 0.8, i.e. 80% boys. True.
2. II: let fraction of girls be g: 55g + 50(1 − g) = 53 ⇒ 5g = 3 ⇒ g = 0.6, i.e. 60% girls, not 64%. False.
3. Hence only Statement I is true.

Teacher's Note:
1. Write the combined mean as a weighted average with the fractions as weights.
2. A quick check: the combined mean divides the gap between the two group means in the inverse ratio of the group sizes.

Answer: (a) \( \frac{n^2 - 1}{12} \)

Explanation:
1. Mean = \( \frac{n + 1}{2} \); \( \frac{\Sigma x^2}{n} = \frac{(n + 1)(2n + 1)}{6} \).
2. Variance = \( \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} = \frac{(n + 1)[2(2n + 1) - 3(n + 1)]}{12} \).
3. \( = \frac{(n + 1)(n - 1)}{12} = \frac{n^2 - 1}{12} \).

Teacher's Note:
1. This result is worth memorising; check it with n = 3 (data 1, 2, 3 has variance 2/3).
2. Use variance = mean of squares − square of mean for such formula-based data.

Answer: (b) \( v = \sigma^2 \)

Explanation:
1. Standard deviation is defined as the positive square root of the variance.
2. So σ = √v, i.e. v = σ².

Teacher's Note:
1. Variance has squared units, while standard deviation has the same unit as the data.
2. Since σ ≥ 0, a negative value can never be a standard deviation.

Answer: (a) σ²

Explanation:
1. New observations \( x_i + \lambda \) have mean \( \bar{x} + \lambda \).
2. Deviations \( (x_i + \lambda) - (\bar{x} + \lambda) = x_i - \bar{x} \) are unchanged.
3. So the variance remains σ².

Teacher's Note:
1. Variance is independent of change of origin.
2. Only multiplying or dividing all observations changes the variance.

Answer: (b) λ²σ²

Explanation:
1. New observations \( \lambda x_i \) have mean \( \lambda\bar{x} \).
2. Each deviation becomes \( \lambda(x_i - \bar{x}) \), so each squared deviation is multiplied by λ².
3. New variance = λ²σ².

Teacher's Note:
1. Multiplying data by λ multiplies the variance by λ² and the S.D. by |λ|.
2. Do not confuse this with adding λ, which leaves variance unchanged.

Answer: (c) a²σ²

Explanation:
1. Mean of \( ax_i \) is \( a\bar{x} \); each deviation becomes \( a(x_i - \bar{x}) \).
2. Squared deviations are multiplied by a², so the variance becomes a²σ².

Teacher's Note:
1. Variance is affected by change of scale by the square of the scale factor.
2. The standard deviation would become |a|σ.

Answer: (a) 40

Explanation:
1. Increasing each number by 5 increases the mean by 5.
2. Deviations from the mean are unchanged, so S.D. remains 40.

Teacher's Note:
1. Standard deviation does not depend on change of origin.
2. The number of observations, 25, is not needed in this question.

Answer: (b) 4 : 1

Explanation:
1. Let there be m men and w women: 26m + 21w = 25(m + w).
2. m = 4w, so m : w = 4 : 1.

Teacher's Note:
1. The combined mean 25 is closer to 26, so men must be in the majority.
2. Group sizes are in the inverse ratio of the distances of the group means from the combined mean: (25 − 21) : (26 − 25) = 4 : 1.

Answer: (c) 4 and 7

Explanation:
1. Let the other two be x, y: 1 + 2 + 6 + x + y = 20 ⇒ x + y = 11.
2. Σx² = 5(5.2 + 4²) = 106 ⇒ 1 + 4 + 36 + x² + y² = 106 ⇒ x² + y² = 65.
3. 2xy = 121 − 65 = 56 ⇒ xy = 28, so x, y = 4 and 7.

Teacher's Note:
1. Use Σx = n × mean and Σx² = n(variance + mean²) to set up equations.
2. Knowing x + y and xy, the two numbers are the roots of t² − 11t + 28 = 0.

Answer: (b) \( x_{51} \)

Explanation:
1. The sum of absolute deviations Σ|x_i − k| is least when k is the median.
2. For 101 ordered observations the median is the \( \frac{101 + 1}{2} \) = 51st value, i.e. \( x_{51} \).

Teacher's Note:
1. Mean deviation is minimum about the median, while the sum of squared deviations is minimum about the mean.
2. Option (c) is the mean, which is a common trap.

Answer: (b) Only Statement II is true

Explanation:
1. The variance of the first n natural numbers is \( \frac{n^2 - 1}{12} \), not \( \frac{n^2 + 1}{12} \), so Statement I is false.
2. Multiples of 3 are 3 × (1, 2, …, 10): mean = 3 × 5.5 = 16.5, variance = 9 × \( \frac{100 - 1}{12} \) = 9 × 8.25 = 74.25.
3. So only Statement II is true.

Teacher's Note:
1. Multiplying data by 3 multiplies the mean by 3 and the variance by 9.
2. Check the formula (n² − 1)/12 with n = 1: a single number must have variance 0.

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

Explanation:
1. Mean = \( \frac{63}{7} = 9 \).
2. Squared deviations: 9, 16, 0, 16, 9, 1, 1; sum = 52.
3. σ = \( \sqrt{\frac{52}{7}} \).

Teacher's Note:
1. Option (b) is the variance; the S.D. is its square root.
2. Keep the answer in surd form when the options are in surd form.

Answer: (c) 93.32

Explanation:
1. Each value of the second set is twice the corresponding value of the first set.
2. Variance is multiplied by 2² = 4: 4 × 23.33 = 93.32.

Teacher's Note:
1. Look for a simple relation (multiple or shift) between two data sets before recalculating.
2. Doubling data doubles the S.D. but quadruples the variance.

Answer: (c) 252500

Explanation:
1. \( \sigma^2 = \frac{\Sigma x^2}{n} - \bar{x}^2 \) ⇒ \( 25 = \frac{\Sigma x^2}{100} - 2500 \).
2. Σx² = 100 × 2525 = 252500.

Teacher's Note:
1. The relation Σx² = n(σ² + x̄²) is very useful in questions on wrongly recorded or combined data.
2. Never forget to square the S.D. before using it.

Answer: (a) 8.25

Explanation:
1. Variance of 1, 2, …, 10 = \( \frac{10^2 - 1}{12} = \frac{99}{12} = 8.25 \).
2. Multiplying by −1 multiplies variance by (−1)² = 1; adding 1 does not change it.
3. So the variance remains 8.25.

Teacher's Note:
1. Multiplying by a negative number reverses the order of the data but does not change its spread.
2. Variance changes by the square of the multiplier, so the sign of the multiplier never matters.

Answer: (c) 3.87

Explanation:
1. First set: Σx = 25 × 18.2 = 455; Σx² = 25(3.25² + 18.2²) = 25 × 341.8025 = 8545.06.
2. All 40: Σx = 455 + 279 = 734; Σx² = 8545.06 + 5524 = 14069.06.
3. \( \sigma^2 = \frac{14069.06}{40} - \left(\frac{734}{40}\right)^2 = 351.73 - 336.72 = 15.00 \) (approx.)
4. σ ≈ √15.00 ≈ 3.87.

Teacher's Note:
1. To combine data sets, convert each set to Σx and Σx², add them, and then use variance = Σx²/n − (Σx/n)².
2. Carry enough decimal places in the middle steps because the final answer is a small difference of large numbers.

Answer: (b) 2² × 5

Explanation:
1. Multiplying each observation by 2 multiplies the variance by 2².
2. New variance = 2² × 5 = 20.

Teacher's Note:
1. Variance scales with the square of the multiplier.
2. The number of observations, 20, does not affect the answer.

Answer: (d) 20, 3.03

Explanation:
1. Σx = 100 × 20 = 2000; Σx² = 100(9 + 400) = 40900.
2. Omitting 21, 21, 18: Σx = 2000 − 60 = 1940, Σx² = 40900 − (441 + 441 + 324) = 39694, n = 97.
3. Mean = 1940/97 = 20; \( \sigma^2 = \frac{39694}{97} - 400 \) ≈ 9.216.
4. σ ≈ 3.03 (more precisely 3.036), so the answer is 20, 3.03.

Teacher's Note:
1. When observations are removed, reduce n as well as Σx and Σx².
2. The mean stays 20 here because the omitted values themselves average 20.

Answer: (a) s

Explanation:
1. Adding k to every observation increases the mean by k.
2. Deviations from the mean are unchanged, so the S.D. remains s.

Teacher's Note:
1. S.D. is independent of change of origin.
2. Only the mean changes, from m to m + k.

Answer: (c) ks

Explanation:
1. Multiplying every observation by k multiplies every deviation from the mean by k.
2. So the S.D. becomes |k|s, which is ks for k > 0.

Teacher's Note:
1. S.D. depends on change of scale: it is multiplied by the absolute value of the factor.
2. Variance would be multiplied by k².

Answer: (a) l = 1.25, k = −5

Explanation:
1. S.D. of \( w_i \) = |l| × 12 = 15 ⇒ |l| = 1.25.
2. Mean: 55 = 48l + k.
3. With l = 1.25: k = 55 − 60 = −5 (option a). With l = −1.25, k = 115, which is not offered.

Teacher's Note:
1. Use the S.D. to find the scale factor first, since S.D. ignores the constant k.
2. Then use the mean, which depends on both l and k.