JEE Mathematics Statistics MCQs Set 05

Answer: (c) the 10th man weighs three times as much as the 11th man

Explanation:
1. Let the weights be w1, ..., w11. Then \( w_1+\dots+w_9=9x \).
2. With the 10th man: \( 9x+w_{10}=10\times1.05x\Rightarrow w_{10}=1.5x \).
3. With the 11th man the average is x again: \( 9x+1.5x+w_{11}=11x\Rightarrow w_{11}=0.5x \).
4. So \( w_{10}=3w_{11} \).

Teacher's Note:
1. Write the total weight after each man joins.
2. Subtract successive totals to isolate each new man's weight.

Answer: (c) 3

Explanation:
1. Coefficient of variation \( =\frac\sigma{\bar x}\times100=50\Rightarrow\sigma=0.5\times6=3 \).

Teacher's Note:
1. CV is the standard deviation expressed as a percentage of the mean.
2. Do not confuse the 50% with 50.

Answer: (b) 1

Explanation:
1. Mode = 3 Median − 2 Mean, so Mean − Mode = 3(Mean − Median).
2. Hence Mean − Median \( =\frac33=1 \).

Teacher's Note:
1. Use the empirical relation among mean, median and mode.
2. Rewrite it as Mean − Mode = 3(Mean − Median).

Answer: (d) \(\frac23\)

Explanation:
1. Coefficient of range \( =\frac{\text{max}-\text{min}}{\text{max}+\text{min}}=\frac{10-2}{10+2}=\frac23 \).

Teacher's Note:
1. The coefficient of range is a unit-free ratio.
2. Divide the range by the sum of the extreme values.

Answer: (c) \(\frac{33}4\)

Explanation:
1. \( \sum n^2=385,\ \sum n=55 \).
2. Variance \( =\frac{385}{10}-\left(\frac{55}{10}\right)^2=38.5-30.25=\frac{33}4 \).

Teacher's Note:
1. Use the formula (n² − 1)/12 as a check: (100 − 1)/12 = 33/4.
2. Variance = mean of squares − square of mean.

Answer: (d) The median is severely affected by fluctuations in extreme values.

Explanation:
1. (a) is true: Σ(xi − x̄) = 0.
2. (b) is false: the median depends only on the middle position and is not affected by extreme values.
3. (c) is a standard result and (d) holds because a change of origin does not change the standard deviation.

Teacher's Note:
1. The median is a positional average, which is why it is preferred when there are outliers.
2. Check every statement against the properties of the mean, median and standard deviation.

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

Explanation:
1. Mean \( =\frac{n+1}2 \) and \( \frac{\sum k^2}{n}=\frac{(n+1)(2n+1)}6 \).
2. Variance \( =\frac{(n+1)(2n+1)}6-\frac{(n+1)^2}4=\frac{(n+1)(n-1)}{12}=\frac{n^2-1}{12} \).

Teacher's Note:
1. Variance = mean of squares − square of the mean.
2. Factor (n + 1) out of both terms.

Answer: (b) 20

Explanation:
1. \( \sum x=750,\ \sum x^2=5(18+22500)=112590 \).
2. With the new student: \( \sum=906,\ \sum x^2=112590+24336=136926 \).
3. Mean 151; variance \( =\frac{136926}{6}-151^2=22821-22801=20 \).

Teacher's Note:
1. Rebuild the totals Σx and Σx² for the five students.
2. Add the new student's contribution and recompute.

Answer: (b) 4.34

Explanation:
1. Combined mean \( =\frac{25\cdot40+35\cdot45}{60}=42.917 \).
2. Combined variance \( =\frac{25(25)+35(4)}{60}+\frac{25\cdot35}{60^2}(40-45)^2=\frac{765}{60}+6.076=18.83 \).
3. Standard deviation \( \approx4.34 \).

Teacher's Note:
1. Use the combined variance formula that includes the term for the difference of the sample means.
2. Take the square root of the combined variance.

Answer: (b) \(\frac{43}5\)

Explanation:
1. Ordered: 34, 38, 42, 44, 46, 48, 54, 55, 63, 70. The median is \( \frac{46+48}2=47 \).
2. Absolute deviations from 47: 13, 9, 5, 3, 1, 1, 7, 8, 16, 23, with sum 86.
3. Mean deviation \( =\frac{86}{10}=8.6=\frac{43}5 \).

Teacher's Note:
1. For an even number of observations, average the two middle scores to get the median.
2. Add the absolute deviations and divide by the number of observations.

Answer: (c) 0

Explanation:
1. \( \sum(x_i-\bar X)=\sum x_i-n\bar X=n\bar X-n\bar X=0 \).

Teacher's Note:
1. This is the defining balance property of the arithmetic mean.
2. It is why mean deviation is computed using absolute values.

Answer: (c) 3 only

Explanation:
1. For the geometric mean: \( \frac{(x_1\cdots x_n)^{1/n}}{(y_1\cdots y_n)^{1/n}}=\left(\frac{x_1}{y_1}\cdots\frac{x_n}{y_n}\right)^{1/n} \), so the property always holds.
2. Arithmetic mean fails: for x = (2, 4) and y = (1, 4), 3/2.5 = 1.2 but the mean of 2 and 1 is 1.5.
3. Median fails: for x = (1, 5, 6) and y = (3, 1, 2), 5/2 = 2.5 but the median of 1/3, 5, 3 is 3. Only (3) satisfies the property.

Teacher's Note:
1. A property of this type works only for a measure built from products and roots.
2. One counter-example is enough to reject the A.M. and the median.

Answer: (b) 16, 14

Explanation:
1. Adding a constant to every element adds the same constant to the mean and to the median.
2. New mean 14 + 2 = 16; new median 12 + 2 = 14.

Teacher's Note:
1. Both the mean and the median shift with a change of origin.
2. The spread measures do not change.

Answer: (b) \(\frac{2^{2n}}{n+1}\)

Explanation:
1. By symmetry, the sum of the first n + 1 binomial coefficients of order 2n + 1 is half of the total: \( \frac12\cdot2^{2n+1}=2^{2n} \).
2. There are n + 1 terms, so the mean is \( \frac{2^{2n}}{n+1} \).

Teacher's Note:
1. The coefficients ⁿCr and ⁿC(n−r) are equal, so exactly half of the 2n + 2 coefficients sum to half the total.
2. Divide the half-total by the number of terms, n + 1.

Answer: (a) \(3a^2-32a+84=0\)

Explanation:
1. \( \frac{4+9+a^2+121}{4}-\left(\frac{16+a}{4}\right)^2=\frac{49}4 \).
2. Multiplying by 16: \( 4(134+a^2)-(16+a)^2=196 \), i.e. \( 3a^2-32a+84=0 \).

Teacher's Note:
1. Use variance = mean of squares − square of the mean, with variance 3.5² = 12.25.
2. Multiply out and collect terms to get a quadratic in a.

Answer: (d) 10

Explanation:
1. Arranged in order: 6, 8, 9, 10, 11, 12, 14.
2. There are 7 values, so the median is the 4th value, 10.

Teacher's Note:
1. Always arrange the data in order before reading the median.
2. For an odd number of values the median is the middle one.

Answer: (d) 20

Explanation:
1. Subtracting a constant from every observation subtracts it from the mean: 25 − 5 = 20.

Teacher's Note:
1. The mean shifts by the same constant as the data.
2. The number of observations is irrelevant here.

Answer: (d) The standard deviation of 1001, 1002, 1003, 1004, 1005 is √2

Explanation:
1. The values 1001, ..., 1005 are the values 1, ..., 5 shifted by 1000.
2. A shift of origin does not change the standard deviation, so it is still √2.
3. The other sets are not simple shifts or scalings of 1, 2, 3, 4, 5.

Teacher's Note:
1. Standard deviation is independent of the origin.
2. Only a shift by a constant leaves it unchanged.

Answer: (d) 18.5

Explanation:
1. The failed students occupy the lowest 8 positions. The passing marks in order are 17, 18, 19, 20, 21, 23, 24, 26, 27, 27, 28, 29.
2. The median of 20 values is the mean of the 10th and 11th. The 9th value is 17, the 10th is 18 and the 11th is 19.
3. Median \( =\frac{18+19}2=18.5 \).

Teacher's Note:
1. Because the eight failures lie below every passing mark, their exact marks do not matter.
2. Count positions from the bottom: failures take ranks 1 to 8.

Answer: (d) \(n\)

Explanation:
1. The sum of the first n odd numbers is \( n^2 \).
2. Mean \( =\frac{n^2}{n}=n \).

Teacher's Note:
1. The sum of the first n odd numbers is a perfect square, n².
2. Divide by the number of terms.

Answer: (d) \(n\bar X-k\)

Explanation:
1. Total of all n numbers \( =n\bar X \).
2. The nth number \( =n\bar X-k \).

Teacher's Note:
1. The total is the mean times the number of terms.
2. Subtract the known part of the total.

Answer: (a) \(\frac n2\)

Explanation:
1. Mean \( =\frac{\sum r\,{}^nC_r}{\sum{}^nC_r}=\frac{n2^{n-1}}{2^n}=\frac n2 \).

Teacher's Note:
1. Use r·ⁿCr = n·ⁿ⁻¹C(r−1), so the numerator sums to n·2ⁿ⁻¹.
2. The denominator is the total, 2ⁿ.

Answer: (c) 3, 12

Explanation:
1. Sum \( =49\Rightarrow a+b=49-34=15 \).
2. Variance: \( \sum(x-7)^2=100\Rightarrow25+9+0+16+9+(a-7)^2+(b-7)^2=100\Rightarrow(a-7)^2+(b-7)^2=41 \).
3. With a + b = 15 this gives a = 3, b = 12.

Teacher's Note:
1. The mean gives a + b; the variance gives the sum of squared deviations.
2. Test the answer pairs against both conditions.