JEE Mathematics Statistics MCQs Set 10

Answer: (a) \( np \)

Explanation:
1. The frequencies are the terms of \( (q+p)^n \), so the total frequency is 1.
2. The mean is \( \sum r\binom nrq^{n-r}p^r=np(q+p)^{n-1}=np \).

Teacher's Note:
1. This is the binomial distribution: mean \( np \).

Answer: (c) 4675

Explanation:
1. Total \( N=43 \), so \( \frac N2=21.5 \). Cumulative frequencies: 3, 8, 28, ... so the median class is 4000-5000.
2. Median \( =4000+\frac{21.5-8}{20}\times1000=4675 \).

Teacher's Note:
1. Locate the class where the cumulative frequency first exceeds \( N/2 \).

Answer: (a) 75

Explanation:
1. The first five numbers without x add to \( 410 \), whose mean is 82. So \( 82+x=82 \) and \( x=0 \).
2. The second set without x adds to \( 375 \), with mean 75; the required mean is \( 75+x=75 \).

Teacher's Note:
1. The mean of numbers shifted by x is shifted by x.

Answer: (a) \( \frac{n(n+1)d}{2n+1} \)

Explanation:
1. There are \( 2n+1 \) terms with mean \( a+nd \) (the middle term).
2. The deviations are \( nd,(n-1)d,\dots,0,\dots,nd \) in absolute value, so the sum is \( 2d(1+2+\dots+n)=n(n+1)d \).
3. Mean deviation \( =\frac{n(n+1)d}{2n+1} \).

Teacher's Note:
1. The mean of an AP is its middle term.

Answer: (b) 8

Explanation:
1. Mean \( =10 \). Squared deviations: \( 16,4,0,4,16 \) with sum 40.
2. Variance \( =\frac{40}{5}=8 \).

Teacher's Note:
1. Variance is the mean of the squared deviations.

Answer: (c) 4 and 7

Explanation:
1. \( \sum x=20 \), so the other two add to \( 20-9=11 \).
2. \( \frac{\sum x^2}{5}-16=5.2 \) gives \( \sum x^2=106 \), so the other two have squares adding to \( 106-41=65 \).
3. With \( a+b=11 \) and \( a^2+b^2=65 \): \( ab=28 \), so the numbers are 4 and 7.

Teacher's Note:
1. Use the sum and the sum of squares to find the missing pair.

Answer: (a) 24, 12

Explanation:
1. Multiplying every observation by 3 multiplies both the mean and the standard deviation by 3.
2. New mean \( =24 \), new SD \( =12 \).

Teacher's Note:
1. Scaling affects mean and SD alike (variance by the square).

Answer: (a) weight

Explanation:
1. Length: mean \( 4.24 \), variance \( \frac{902.8}{50}-4.24^2=0.0784 \), SD \( =0.28 \), CV \( \approx6.6\% \).
2. Weight: mean \( 5.22 \), variance \( \frac{1457.6}{50}-5.22^2\approx1.90 \), SD \( \approx1.38 \), CV \( \approx26.4\% \).
3. The weight has the larger coefficient of variation, so it varies more.

Teacher's Note:
1. Compare coefficients of variation, not standard deviations, for different units.

Answer: (c) 46.67

Explanation:
1. The modal class is 40-50 with \( f_1=28 \), \( f_0=12 \), \( f_2=20 \) and \( h=10 \).
2. Mode \( =40+\frac{28-12}{56-12-20}\times10=40+\frac{160}{24}=46.67 \).

Teacher's Note:
1. Use the grouped data mode formula.

Answer: (b) 23

Explanation:
1. Using the empirical relation Mean \( - \) Mode \( =3( \)Mean \( - \) Median\( ) \): \( 20-\text{Mode}=3(20-21)=-3 \).
2. Mode \( =23 \).

Teacher's Note:
1. Mode = 3 Median - 2 Mean.

Answer: (a) Chemistry, Mathematics

Explanation:
1. Coefficient of variation \( =\frac{\sigma}{\bar x}\times100 \): Mathematics \( 28.57 \), Physics \( 46.87 \), Chemistry \( 48.89 \).
2. Chemistry has the highest variability and Mathematics the lowest.

Teacher's Note:
1. Higher CV means more variability.

Answer: (a) 11

Explanation:
1. The mean is \( \frac{(n+1)(2n+1)}{6} \), so \( \frac{(n+1)(2n+1)}{6}=\frac{46n}{11} \).
2. This gives \( 22n^2-243n+11=0 \), i.e. \( (n-11)(22n-1)=0 \), so \( n=11 \).

Teacher's Note:
1. Use \( \sum k^2=\frac{n(n+1)(2n+1)}{6} \).

Answer: (c) \( \frac13(2n+1) \)

Explanation:
1. Weighted mean \( =\frac{\sum k\cdot k}{\sum k}=\frac{n(n+1)(2n+1)/6}{n(n+1)/2} \).
2. This equals \( \frac{2n+1}{3} \).

Teacher's Note:
1. Divide the sum of \( k^2 \) by the sum of k.

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

Explanation:
1. There are \( n+1 \) terms and the sum of the GP is \( 2^{n+1}-1 \).
2. The mean is \( \frac{2^{n+1}-1}{n+1} \).

Teacher's Note:
1. Count the terms: the powers run from \( 2^0 \) to \( 2^n \).

Answer: (c) 63

Explanation:
1. The failed students scored \( 10\times28=280 \) in total.
2. The 40 who passed scored \( 2800-280=2520 \), so their average is \( \frac{2520}{40}=63 \).

Teacher's Note:
1. Subtract the failed students' total first.

Answer: (d) \( \bar x_1<\bar x<\bar x_2 \)

Explanation:
1. The combined mean \( \frac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2} \) is a weighted average of \( \bar x_1 \) and \( \bar x_2 \).
2. So it lies between them.

Teacher's Note:
1. A weighted average lies between the smallest and largest values.

Answer: (b) \( \bar x+n+1 \)

Explanation:
1. The mean of the added terms \( 2,4,\dots,2n \) is \( \frac{n(n+1)}{n}=n+1 \).
2. So the new mean is \( \bar x+n+1 \).

Teacher's Note:
1. Add the mean of the added terms.

Answer: (b) 10

Explanation:
1. Total \( =10\times12+20\times9=300 \).
2. Mean \( =\frac{300}{30}=10 \).

Teacher's Note:
1. Combined mean is the weighted mean.

Answer: (c) 44.46

Explanation:
1. Wrong total \( =4500 \). Correct total \( =4500-(91+13)+(19+31)=4446 \).
2. Correct mean \( =44.46 \).

Teacher's Note:
1. Replace the wrong readings in the total.

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

Explanation:
1. The new mean is \( \frac{\bar x}{\alpha}+10=\frac{\bar x+10\alpha}{\alpha} \).

Teacher's Note:
1. Apply the same operations to the mean.

Answer: (d) \( \bar x \)

Explanation:
1. The new mean is \( \frac{n\bar x+x_{n+1}}{n+1}=\bar x \).
2. So \( x_{n+1}=\bar x \).

Teacher's Note:
1. Adding the mean itself does not change the mean.