JEE Mathematics Statistics MCQs Set 09

Answer: (c) 5

Explanation:
1. \( \bar x=\frac{\sigma\times100}{CV} \): the means are \( \frac{1000}{50}=20 \) and \( \frac{1500}{60}=25 \).
2. The difference is 5.

Teacher's Note:
1. CV = SD / mean x 100.

Answer: (a) 5

Explanation:
1. The sum is 22, so the other two sum to \( 22-9=13 \).
2. \( \sum x^2=5(8.24+19.36)=138 \); the other two have squares summing to \( 138-41=97 \). Then \( xy=\frac{169-97}{2}=36 \), so they are 4 and 9, with difference 5.

Teacher's Note:
1. Use the sum and the sum of squares.

Answer: (c) 7

Explanation:
1. Ordered: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72. The median is \( \frac{46+49}{2}=47.5 \).
2. Absolute deviations: 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5; total 70. Mean deviation \( =7 \).

Teacher's Note:
1. Find the median of ten observations as the average of the middle two.

Answer: (c) 8

Explanation:
1. Variance \( =\frac{1000}{10}-\left(\frac{60}{10}\right)^2=100-36=64 \).
2. SD \( =8 \).

Teacher's Note:
1. Use variance = mean of squares - square of mean.

Answer: (a) 2

Explanation:
1. Mean 3; squared deviations 4, 1, 0, 1, 4 sum to 10.
2. Variance \( =2 \), so \( a=2 \).

Teacher's Note:
1. The standard deviation is the square root of the variance.

Answer: (a) 8

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

Teacher's Note:
1. Mean first, then squared deviations.

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

Explanation:
1. A is the definition of mean deviation, so it is true.
2. The sum of the deviations from the mean is always 0, not 1, so R is false.

Teacher's Note:
1. The deviations from the mean cancel out.

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

Explanation:
1. The mean deviation is the sum of absolute deviations divided by n, not the product, so A is false.
2. R is a true statement about how dispersion is measured.

Teacher's Note:
1. Check the definition.

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

Explanation:
1. Scaling by 2 multiplies the variance by 4, so A is true.
2. The mean is multiplied by 2, giving \( 2\bar x \), so R is false.

Teacher's Note:
1. Scaling by k scales the mean by k and the variance by \( k^2 \).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are standard definitions and are true.
2. R defines variance, which has nothing to do with the range, so it does not explain A.

Teacher's Note:
1. Check whether R is relevant to A.

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

Explanation:
1. Mean \( =\frac{42}{7}=6 \); absolute deviations 4, 3, 3, 3, 0, 3, 2 sum to 18, and \( \frac{18}{7}\approx2.57 \), so A is true.
2. R is the formula used to obtain A.

Teacher's Note:
1. Apply the formula.

Answer: (c) 35

Explanation:
1. Total of six: 180; total of the remaining five: 145.
2. Excluded number \( =180-145=35 \).

Teacher's Note:
1. Work with totals.

Answer: (d) 14.6

Explanation:
1. The wrong total is \( 20\times15.5=310 \).
2. Corrected total \( =310-42+24=292 \), so the mean is \( \frac{292}{20}=14.6 \).

Teacher's Note:
1. Correct the total, then divide.

Answer: (b) 8.8

Explanation:
1. Old total \( =386 \). New mean \( =18.8 \) for 21 observations, so the new total is \( 394.8 \).
2. The new observation is \( 394.8-386=8.8 \).

Teacher's Note:
1. Compare the totals before and after.

Answer: (b) 62

Explanation:
1. With 10 observations the median is the average of the 5th and 6th: \( \frac{x+(x+2)}{2}=x+1 \).
2. \( x+1=63 \) gives \( x=62 \).

Teacher's Note:
1. The median of 10 values is the mean of the two middle values.

Answer: (c) 14

Explanation:
1. Total of all 13: 182. First 7 total 84, last 7 total 112.
2. The 7th observation is counted in both groups: \( 84+112-182=14 \).

Teacher's Note:
1. The overlap is the 7th observation.

Answer: (a) mean \( =\dfrac{n+1}{2} \), variance \( =\dfrac{n^2-1}{12} \)

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

Teacher's Note:
1. Use \( \sum k \) and \( \sum k^2 \).

Answer: (c) mean = 19, S.D = 6.59

Explanation:
1. N = 40 and mean \( =\frac{760}{40}=19 \).
2. \( \sum fx^2=16176 \), so variance \( =\frac{16176}{40}-19^2=404.4-361=43.4 \) and the S.D. is \( \sqrt{43.4}\approx6.59 \).

Teacher's Note:
1. Compute the mean, then the variance.

Answer: (d) 2

Explanation:
1. Mean \( =1 \).
2. Absolute deviations 2, 1, 3; mean deviation \( =\frac63=2 \).

Teacher's Note:
1. Mean deviation = average of |x - mean|.

Answer: (c) 93.32

Explanation:
1. The second data are twice the first.
2. Doubling the observations multiplies the variance by 4: \( 4\times23.33=93.32 \).

Teacher's Note:
1. Variance scales by the square of the multiplier.

Answer: (c) 252500

Explanation:
1. \( \sigma^2=\frac{\sum x^2}{n}-\bar x^2 \), so \( \frac{\sum x^2}{100}=25+2500=2525 \).
2. \( \sum x^2=252500 \).

Teacher's Note:
1. Rearrange the variance formula.

Answer: (d) 8.25

Explanation:
1. Adding a constant does not change the variance.
2. Variance of 1 to 10 is \( \frac{10^2-1}{12}=8.25 \).

Teacher's Note:
1. Variance is independent of origin.

Answer: (a) 0

Explanation:
1. \( \sigma_1=\frac{50\times30}{100}=15 \) and \( \sigma_2=\frac{60\times25}{100}=15 \).
2. The difference is 0.

Teacher's Note:
1. SD = CV x mean / 100.

Answer: (a) 10%

Explanation:
1. Variance \( =\frac{250}{10}=25 \), SD \( =5 \).
2. CV \( =\frac{5}{50}\times100=10\% \).

Teacher's Note:
1. CV = SD / mean x 100.

Answer: (d) 25%

Explanation:
1. Variance \( =\frac{1530}{10}-144=9 \), SD \( =3 \).
2. CV \( =\frac{3}{12}\times100=25\% \).

Teacher's Note:
1. Use variance = mean of squares - square of mean.