JEE Mathematics Statistics MCQs Set 06

Answer: (b) \(\dfrac{n\cdot 2^{n-1}}{2^n-1}\)

Explanation:
1. The weight of the number \(k\) is \(\binom{n}{k}\).
2. Weighted mean = \(\dfrac{\sum k\binom{n}{k}}{\sum\binom{n}{k}}\).
3. \(\sum_{k=1}^{n} k\binom{n}{k}=n2^{n-1}\) and \(\sum_{k=1}^{n}\binom{n}{k}=2^n-1\).
4. So the mean is \(\dfrac{n2^{n-1}}{2^n-1}\).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) \(\dfrac{nM-a}{4}\)

Explanation:
1. Total of all observations = \(nM\).
2. Sum of the remaining 4 = nM \( - a \).
3. Their mean is (nM - a)/4.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (b) \(n_1:n_2=5:3\)

Explanation:
1. \(400n_1+480n_2=430(n_1+n_2)\).
2. \(30n_1=50n_2\), so \(n_1/n_2=5/3\).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) \(\bar X+\dfrac{n+1}{2}\)

Explanation:
1. Total increase = \(1+2+\dots+n=n(n+1)/2\).
2. Increase in mean = \((n+1)/2\).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (a) 10.1

Explanation:
1. There are 101 terms with mean \(1+50d\).
2. Mean deviation = \(\dfrac{d}{101}\sum_{k=0}^{100}|k-50|=\dfrac{d}{101}\times2\times\dfrac{50\cdot51}{2}=\dfrac{2550d}{101}\).
3. \(\dfrac{2550d}{101}=255\) gives \(d=10.1\).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.

Answer: (c) 2

Explanation:
1. The mean is 0, so each deviation squared is a squared.
2. Variance = \(a^2\), so SD = \(|a|\) = 2.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) \(\sqrt{\dfrac{22}{3}}\)

Explanation:
1. Combined variance = \(\dfrac{10\cdot4+20\cdot9}{30}=\dfrac{220}{30}=\dfrac{22}{3}\) (both deviations are about the same mean 5).
2. SD = \(\sqrt{22/3}\).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) 14

Explanation:
1. Since \(0<y<x<2y\), the order is \(x-y<y<x<2x+y\).
2. Median = \( (y + x)/2 = 10 \), so \( x + y = 20 \).
3. Range = \( (2x + y) - (x - y) = x + 2y = 28 \), so \( y = 8, x = 12 \).
4. Numbers: 4, 8, 12, 32 with mean \( 56/4 = 14 \).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (b) \(\dfrac{n+1}{2},\ \dfrac{n^2-1}{12}\)

Explanation:
1. Mean = \(\dfrac{n(n+1)/2}{n}=\dfrac{n+1}{2}\).
2. Variance = \(\dfrac{(n+1)(2n+1)}{6}-\dfrac{(n+1)^2}{4}=\dfrac{n^2-1}{12}\).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.

Answer: (d) \(2^2\times5\)

Explanation:
1. Multiplying each observation by \( k \) multiplies the variance by \( k \) squared.
2. New variance = \( 4 x 5 = 20 \).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (d) Variance

Explanation:
1. Adding 10 to every mark shifts the mean, median and mode by 10.
2. Deviations from the mean stay the same, so the variance is unchanged.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (a) 35 and 22.85

Explanation:
1. Mean = SD/CV \( x 100 \): \(\dfrac{21}{60}\times100=35\).
2. \(\dfrac{16}{70}\times100\approx22.85\).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (b) 93.33

Explanation:
1. The second set is twice the first, so the variance is multiplied by 4.
2. \( 4 x 23.33 = 93.33 \).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (d) 78

Explanation:
1. Corrected sum = \( 170 - 20 + 30 = 180 \); corrected sum of squares = \( 2830 - 400 + 900 = 3330 \).
2. Variance = \(\dfrac{3330}{15}-\left(\dfrac{180}{15}\right)^2=222-144=78\).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (a) 8.25

Explanation:
1. Variance of 1 to 10 = \((10^2-1)/12=8.25\).
2. Adding 1 to each number does not change the variance.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (b) 5

Explanation:
1. Mean = SD/CV \( x 100 \): \( 10/50 x 100 = 20 \) and \( 15/60 x 100 = 25 \).
2. Difference = 5.

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (d) 18

Explanation:
1. \((n^2-1)/12=10\) gives \(n=11\).
2. First \( m \) even numbers have variance \(4(m^2-1)/12=16\), so \(m=7\).
3. \( m + n = 18 \).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.

Answer: (a) 54

Explanation:
1. Sum = 80, so \( x + y = 80 - 64 = 16 \).
2. Sum of squares = \( 8(25 + 100) = 1000 \); the six known squares give 852, so \( x^2 + y^2 = 148 \).
3. xy = \( (16^2 - 148)/2 = 54 \).

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) 40

Explanation:
1. Adding 5 to every number leaves the deviations from the mean unchanged.
2. So the SD stays 40.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (d) 2

Explanation:
1. Variance is unchanged by shifting by 5: \(\dfrac{45}{9}-\left(\dfrac{9}{9}\right)^2=5-1=4\).
2. SD = 2.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (b) 5.2

Explanation:
1. Mean of \( U = (10 - 5)/2 = 2.5 \).
2. SD of \( X = 10 x 2.6/100 = 0.26 \), so SD of \( U = 0.13 \).
3. CV of \( U = 0.13/2.5 x 100 = 5.2 \).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (a) (a = 4, b = 3) and (a = 3, b = 4) only

Explanation:
1. Sum = 30, so \( a + b = 30 - 23 = 7 \).
2. Sum of squares = \( 5(6.8 + 36) = 214 \); so \( a^2 + b^2 = 214 - 189 = 25 \).
3. Then ab = \( (49 - 25)/2 = 12 \), so \( {a, b} = {3, 4} \); the pairs (0, 7) and (1, 6) fail the sum of squares.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (a) 18 and 20 only

Explanation:
1. Variance \(=\dfrac{400}{n}-\left(\dfrac{80}{n}\right)^2\ge0\).
2. This gives \(400n\ge6400\), so \(n\ge16\).
3. Among 15, 18, 20 and 12, only 18 and 20 qualify.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (b) \(6\text{ and }\tfrac{14}{3}\)

Explanation:
1. Variance = 12.25, so \(\dfrac{134+a^2}{4}-\left(\dfrac{16+a}{4}\right)^2=12.25\).
2. This gives \(3a^2-32a+84=0\).
3. \(a=\dfrac{32\pm4}{6}\), that is 6 or 14/3.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (c) 4

Explanation:
1. The 7 terms form an \( A.P \). with \( d = 7 \); SD = \(d\sqrt{(n^2-1)/12}=7\sqrt{4}=14\).
2. \( k + 10 = 14 \), so \( k = 4 \).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.

Answer: (d) 3

Explanation:
1. The new data are 3 times the old, so SD = \(3\sqrt{99/12}=3\cdot\dfrac{\sqrt{33}}{2}\).
2. So \(30+m=33\) and \( m = 3 \).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (b) 7

Explanation:
1. Ordered: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72; median = \( (46 + 49)/2 = 47.5 \).
2. Deviations: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5; their sum is 70.
3. Mean deviation = \( 70/10 = 7 \).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.

Answer: (c) \(\sigma^2\)

Explanation:
1. Multiplying by \( -1 \) does not change variance (it is squared).
2. Subtracting a shifts but does not change variance.

Teacher's Note:
1. Write the data in terms of the sum, the sum of squares and \( n \), then use variance = (sum of squares)/n \( - (mean)^2 \).
2. Adding a constant to every observation shifts the mean but leaves variance and standard deviation unchanged.

Answer: (d) \(\dfrac{n^2-1}{3}\)

Explanation:
1. The even numbers are 2 times 1, 2, ..., \( n \), so variance = \(4\times\dfrac{n^2-1}{12}=\dfrac{n^2-1}{3}\).

Teacher's Note:
1. Multiplying every observation by \( k \) multiplies the mean and SD by \( k \) and the variance by \( k \) squared.
2. Coefficient of variation is SD divided by mean, times 100, and has no units.

Answer: (a) A-s, B-r, C-q, D-p

Explanation:
1. A: sorted data 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21 has median 9; deviations sum to 58, so \( 58/11 = 5.27 (s) \).
2. \( B \): median 13.5; deviations sum to 28, so \( 28/12 = 2.33 (r) \).
3. \( C \) is the SD formula \( (q) \) and \( D \) is the CV formula \( (p) \).

Teacher's Note:
1. Mean deviation measures the average distance from the mean or the median, taken without sign.
2. Variance of a set of equally spaced numbers depends only on the common difference and the number of terms.