JEE Mathematics Statistics MCQs Set 03

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

Explanation:
1. \( \sum(A+iB)=nA+B\frac{n(n+1)}2 \).
2. Dividing by n: \( A+B\frac{n+1}2 \).

Teacher's Note:
1. Sum the constant term and the term in i separately.
2. The sum of the first n natural numbers is n(n + 1)/2.

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

Explanation:
1. The first n even natural numbers are 2, 4, ..., 2n. Their variance is \( 4\cdot\frac{n^2-1}{12}=\frac{n^2-1}3 \), not (n² − 1)/4, so A is false.
2. R states the two standard summation formulas, which are true.

Teacher's Note:
1. The variance of the first n natural numbers is (n² − 1)/12 and doubling the numbers multiplies it by 4.
2. The summation formulas are standard and correct.

Answer: (a) 12

Explanation:
1. There are 48 values; the median is \( \frac{76+77}2=76.5 \).
2. Absolute deviations: 23.5, 22.5, ..., 0.5 on each side, summing to \( 2\times\frac{24^2}2=576 \).
3. Mean deviation \( =\frac{576}{48}=12 \).

Teacher's Note:
1. The deviations are symmetric about the median, so double the one-sided sum.
2. The sum 0.5 + 1.5 + ... + 23.5 equals 24²/2.

Answer: (c) Rs. 12000

Explanation:
1. There are 17 staff, and two groups earn 90000 and 95000, which are extreme values that distort the mean.
2. So the median is the appropriate measure. Arranged in order the salaries are 7000 (4 staff), 10000 (2), 12000 (5), 15000 (1), 90000 (3), 95000 (2).
3. The 9th value is 12000, so the median is Rs. 12000.

Teacher's Note:
1. Extreme values make the mean unrepresentative; the median is not affected by them.
2. Locate the middle position, here the 9th of 17 values.

Answer: (d) 31

Explanation:
1. \( \sum(x_i-30)=20\Rightarrow\sum x_i=20+20\times30=620 \).
2. Mean \( =\frac{620}{20}=31 \).

Teacher's Note:
1. Deviation from 30 means xi − 30.
2. Sum of deviations equals Σxi minus 20 × 30.

Answer: (c) \(\alpha-\frac54\)

Explanation:
1. Order the values: \( \alpha-\frac72,\ \alpha-3,\ \alpha-\frac52,\ \alpha-2,\ \alpha-\frac12,\ \alpha+\frac12,\ \alpha+4,\ \alpha+5 \).
2. There are 8 values, so the median is the mean of the 4th and 5th: \( \frac{(\alpha-2)+(\alpha-\frac12)}{2}=\alpha-\frac54 \).

Teacher's Note:
1. With an even number of values the median is the average of the two middle values.
2. Compare the numbers by their constant offsets from α when ordering.

Answer: (a) 507.5

Explanation:
1. \( \sum x=50 \) and \( \sum x^2=5(9+100)=545 \).
2. Adding −50: total of 6 observations is 0 (mean 0), and \( \sum x^2=545+2500=3045 \).
3. Variance \( =\frac{3045}6-0=507.5 \).

Teacher's Note:
1. Because the new mean is 0, the variance is just the mean of squares.
2. Rebuild Σx² from the original variance and mean.

Answer: (d) 81

Explanation:
1. \( \sum x=3150,\ \sum y=2160,\ \sum x^2=50(81+3969)=202500,\ \sum y^2=40(36+2916)=118080 \).
2. Combined mean \( =\frac{5310}{90}=59 \).
3. Variance \( =\frac{320580}{90}-59^2=3562-3481=81 \).

Teacher's Note:
1. Find the sum and sum of squares of each sample separately.
2. Combine and subtract the square of the combined mean.

Answer: (b) 32, 2

Explanation:
1. Every true weight is the reported weight plus 2, a change of origin.
2. Correct mean \( =30+2=32 \) and the standard deviation is unchanged at 2.

Teacher's Note:
1. A constant error shifts the mean by the same amount.
2. Spread is unaffected by a constant shift.

Answer: (d) 10%

Explanation:
1. \( \sigma=\sqrt{\frac{250}{10}}=5 \).
2. Coefficient of variation \( =\frac{5}{50}\times100=10\% \).

Teacher's Note:
1. The variance is the sum of squared deviations from the mean divided by n.
2. The coefficient of variation is σ/mean × 100.

Answer: (a) 17

Explanation:
1. Ordered: 4, 8, 12, 17, 19, 23, 27.
2. With 7 values, the median is the 4th value, 17.

Teacher's Note:
1. Sort the values before locating the middle one.
2. Do not average values when the count is odd.

Answer: (a) 3

Explanation:
1. Variance is unchanged by shifting the origin, so use the values xi − 4.
2. Variance \( =\frac{44}{11}-\left(\frac{11}{11}\right)^2=4-1=3 \).

Teacher's Note:
1. A shift of origin does not change the variance.
2. Apply variance = mean of squares − square of the mean to the shifted values.

Answer: (c) 2

Explanation:
1. Mean \( =\frac{-1+0+4}3=1 \).
2. Absolute deviations: 2, 1, 3, whose sum is 6.
3. Mean deviation \( =\frac63=2 \).

Teacher's Note:
1. Mean deviation uses absolute deviations, not squares.
2. The square-root option is the standard deviation, a different measure.

Answer: (b) 12

Explanation:
1. \( \sigma^2=\frac{\sum x^2}n-\bar x^2=156-12=144 \), so \( \sigma=12 \).

Teacher's Note:
1. Variance is the mean of squares minus the square of the mean.
2. Take the square root at the end.

Answer: (a) 10.1

Explanation:
1. There are 101 terms with mean \( 1+50d \).
2. The deviations are \( \pm d,\pm2d,\dots,\pm50d \), so the sum of absolute deviations is \( 2d(1+2+\dots+50)=2550d \).
3. Mean deviation \( =\frac{2550d}{101}=255\Rightarrow d=10.1 \).

Teacher's Note:
1. In an A.P. with 101 terms the mean is the 51st term.
2. Pair the deviations on the two sides of the mean.

Answer: (c) \(\sqrt{\frac{n^2-1}3}\)

Explanation:
1. The standard deviation of 1, 2, ..., n is \( \sqrt{\frac{n^2-1}{12}} \).
2. Multiplying every value by 2 doubles the standard deviation: \( 2\sqrt{\frac{n^2-1}{12}}=\sqrt{\frac{n^2-1}3} \).

Teacher's Note:
1. The variance of the first n natural numbers is (n² − 1)/12.
2. A change of scale by 2 multiplies the SD by 2.

Answer: (c) \(\frac{(n+1)(n+2)(n+3)}4\)

Explanation:
1. Write \( r(r+1)(r+2)=\frac14\left[r(r+1)(r+2)(r+3)-(r-1)r(r+1)(r+2)\right] \).
2. Summing, the terms telescope: \( \sum_{r=1}^nr(r+1)(r+2)=\frac{n(n+1)(n+2)(n+3)}4 \).
3. Dividing by n gives \( \frac{(n+1)(n+2)(n+3)}4 \).

Teacher's Note:
1. The product of three consecutive integers can be written as a difference of products of four consecutive integers.
2. The sum then collapses to a single term.

Answer: (b) 833

Explanation:
1. For 2, 4, ..., 100: mean \( =51 \) and \( \frac{\sum(2n)^2}{50}=\frac{4\cdot50\cdot51\cdot101/6}{50}=3434 \).
2. Variance \( =3434-51^2=3434-2601=833 \).

Teacher's Note:
1. The variance of the first n natural numbers is (n² − 1)/12 and the even numbers are twice those.
2. So the variance is 4 × (2500 − 1)/12 = 833.

Answer: (b) 2

Explanation:
1. Ordered data: 6, 8, 9, 10, 11, 12, 14. Q1 is the 2nd term, 8, and Q3 is the 6th term, 12.
2. Quartile deviation \( =\frac12(12-8)=2 \).

Teacher's Note:
1. Order the data first.
2. Quartile deviation is half the inter-quartile range.

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

Explanation:
1. Shift the origin to 8: \( \sigma=\sqrt{\frac{\sum(x_i-8)^2}{9}-\left(\frac{\sum(x_i-8)}9\right)^2}=\sqrt{5-1}=2 \). So A is true.
2. R is true, and it is exactly why the shift to 8 is allowed in A.

Teacher's Note:
1. Shifting every observation by the same constant leaves the spread unchanged.
2. So the deviations from 8 give the same variance as the original data.

Answer: (b) 67.2

Explanation:
1. Combined mean \( =\frac{200\cdot25+300\cdot10}{500}=16 \).
2. Sum of squares: \( 200(9+625)+300(16+100)=126800+34800=161600 \).
3. Variance \( =\frac{161600}{500}-16^2=323.2-256=67.2 \).

Teacher's Note:
1. Find Σx² for each sample from its variance and mean, then combine.
2. Subtract the square of the combined mean at the end.

Answer: (d) 17

Explanation:
1. Range = largest value − smallest value = 18 − 1 = 17.

Teacher's Note:
1. The range uses only the two extreme values.
2. It is the simplest measure of dispersion but is sensitive to outliers.