JEE Mathematics Statistics MCQs Set 04

Answer: (a) \(\frac{625}9\)

Explanation:
1. Each mark is multiplied by \( \frac{100}{60}=\frac53 \).
2. New SD \( =\frac53\times5=\frac{25}3 \), so the variance is \( \left(\frac{25}3\right)^2=\frac{625}9 \).

Teacher's Note:
1. Rescaling multiplies the SD by the factor and the variance by its square.
2. Convert SD to variance by squaring at the end.

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 from the mean are \( \pm d,\pm2d,\dots,\pm nd \), so the sum of absolute deviations is \( 2|d|(1+2+\dots+n)=n(n+1)|d| \).
3. Mean deviation \( =\frac{n(n+1)|d|}{2n+1} \).

Teacher's Note:
1. In an A.P. with an odd number of terms the mean equals the middle term.
2. Pair the deviations on either side of the mean.

Answer: (d) 7

Explanation:
1. \( x_1+x_2=20-11=9 \).
2. \( \frac{9+16+16+x_1^2+x_2^2}{5}-16=5.2\Rightarrow x_1^2+x_2^2=65 \).
3. So \( 2x_1x_2=81-65=16 \) and \( (x_1-x_2)^2=65-16=49 \), giving |x1 − x2| = 7.

Teacher's Note:
1. The mean gives the sum of the other two, the variance gives their sum of squares.
2. (x1 − x2)² = (x1² + x2²) − 2x1x2.

Answer: (b) \(\sqrt5\)

Explanation:
1. Expanding: \( \sum x^2+2\sum x+n=9n \) and \( \sum x^2-2\sum x+n=5n \).
2. Adding and subtracting: \( \sum x^2=6n \) and \( \sum x=n \).
3. Variance \( =6-1=5 \), so SD \( =\sqrt5 \).

Teacher's Note:
1. Expand both squares and add and subtract the two equations.
2. Variance = mean of squares − (mean)².

Answer: (c) 21

Explanation:
1. Empirical relation: Mode = 3 Median − 2 Mean.
2. So Mean − Mode = 3(Mean − Median) = 63.
3. Mean − Median = 21.

Teacher's Note:
1. The empirical relation for moderately skewed data connects the three averages.
2. Rewrite Mode = 3 Median − 2 Mean as Mean − Mode = 3 (Mean − Median).

Answer: (d) 5.25

Explanation:
1. Total days \( =2(1)+14(2)+8(5)+32(7)=2+28+40+224=294 \).
2. Total weeks \( =2+14+8+32=56 \).
3. Mean \( =\frac{294}{56}=5.25 \).

Teacher's Note:
1. Use the weighted mean with the number of weeks as frequencies.
2. Divide total days by total weeks.

Answer: (d) TTF

Explanation:
1. Variance is an average of squares, so it cannot be negative: (I) true.
2. The standard deviation is the non-negative square root of the variance: (II) true.
3. The median depends only on the middle value(s), not on extreme values: (III) false. So TTF.

Teacher's Note:
1. Variance is an average of squares and so is non-negative.
2. The median is a positional average and is not influenced by extremes.

Answer: (c) 12

Explanation:
1. SD of ax is |a| times SD of x: \( |-4|\times3=12 \).

Teacher's Note:
1. A standard deviation is never negative.
2. Scaling by −4 scales the SD by 4.

Answer: (b) 4 : 9

Explanation:
1. \( x+y=25-12=13 \).
2. \( \frac{1+9+64+x^2+y^2}{5}-25=9.2\Rightarrow x^2+y^2=97 \).
3. So \( 2xy=169-97=72\Rightarrow xy=36 \) and the numbers are 4 and 9, ratio 4 : 9.

Teacher's Note:
1. Use the mean for x + y and the variance for x² + y².
2. The product follows from (x + y)² − (x² + y²).

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

Explanation:
1. Variance of 2x is \( 2^2\sigma^2=4\sigma^2 \), so A is true.
2. The mean of 2xi is \( 2\bar x \), not 4 x̄, so R is false.

Teacher's Note:
1. Multiplying by k multiplies the mean by k and the variance by k².
2. Compare the claimed mean with 2 x̄.

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

Explanation:
1. There are n + 1 terms forming a G.P. with first term 1 and ratio 2.
2. Sum \( =\frac{2^{n+1}-1}{2-1}=2^{n+1}-1 \).
3. Mean \( =\frac{2^{n+1}-1}{n+1} \).

Teacher's Note:
1. Count the terms carefully: 1 = 2⁰ up to 2ⁿ gives n + 1 terms.
2. Use the G.P. sum formula before dividing by the number of terms.

Answer: (d) \(\sqrt2\)

Explanation:
1. The mean is 1/2 (the shifts −d and +d cancel).
2. Variance \( =\frac{10d^2+0+10d^2}{30}=\frac{2d^2}3 \).
3. \( \frac{2d^2}{3}=\frac43\Rightarrow d^2=2 \), so \( |d|=\sqrt2 \).

Teacher's Note:
1. Only the deviations from the mean matter for variance.
2. Equal numbers above and below the central value keep the mean at 1/2.

Answer: (a) 51

Explanation:
1. Increasing every observation by 2% multiplies each observation, the mean and every deviation by 1.02.
2. New mean deviation \( =1.02\times50=51 \).

Teacher's Note:
1. A percentage increase is a change of scale, which changes the mean deviation in the same ratio.
2. A change of origin (adding a constant) would leave it unchanged.

Answer: (d) variance

Explanation:
1. Adding 10 to every mark shifts the mean, the median and the mode by 10.
2. The deviations from the mean are unchanged, so the variance stays the same.

Teacher's Note:
1. Grace marks are a change of origin.
2. Measures of position shift; measures of spread do not.

Answer: (d) 76

Explanation:
1. Each observation becomes 0.95 times itself, so each deviation also becomes 0.95 times itself.
2. New mean deviation \( =0.95\times80=76 \).

Teacher's Note:
1. A decrease of 5% multiplies all values by 0.95.
2. Mean deviation scales exactly like the data.

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

Explanation:
1. \( \frac1n\sum(x_i+2i)=\bar x+\frac2n\cdot\frac{n(n+1)}2=\bar x+(n+1) \).

Teacher's Note:
1. Split the sum into the sum of the xi and the sum of 2i.
2. Use 1 + 2 + ... + n = n(n + 1)/2.

Answer: (a) variance decreases

Explanation:
1. Original \( \sum x=1100,\ \sum x^2=100(2+121)=12300 \).
2. Corrected: \( \sum x=1104,\ \sum x^2=12300-25+81=12356 \), mean 11.04.
3. Corrected variance \( =123.56-121.8816=1.6784<2 \), so the variance decreases.

Teacher's Note:
1. Recompute both sums after replacing 5 by 9.
2. The corrected value is closer to the mean, so the spread falls.

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

Explanation:
1. \( a+b+23=30\Rightarrow a+b=7 \).
2. Variance: \( \frac{(a-6)^2+(b-6)^2+4+1+16}{5}=6.8\Rightarrow(a-6)^2+(b-6)^2=13 \).
3. With b = 7 − a this gives \( a^2-7a+12=0 \), so a = 3, b = 4 (or a = 4, b = 3). Only option (a) matches.

Teacher's Note:
1. The mean gives a + b and the variance gives (a − 6)² + (b − 6)².
2. Check the options against both conditions.

Answer: (c) 24

Explanation:
1. Let the two remaining values be x and y. Mean: \( x+y+42=56\Rightarrow x+y=14 \).
2. Variance: \( \frac{4+16+100+144+196+x^2+y^2}{7}-64=16\Rightarrow x^2+y^2=560-460=100 \).
3. So \( xy=\frac{196-100}2=48 \) and the numbers are 6 and 8, with LCM 24.

Teacher's Note:
1. Use the mean to get x + y and the variance to get x² + y².
2. The numbers then follow from their sum and product.

Answer: (c) 4

Explanation:
1. First group: \( \sum x^2=100(9+225)=23400 \). Whole group: \( \sum z=250\times15.6=3900 \) and \( \sum z^2=250(13.44+243.36)=64200 \).
2. Second group (150 items): \( \sum y=3900-1500=2400 \), mean 16, and \( \sum y^2=64200-23400=40800 \).
3. Variance \( =\frac{40800}{150}-256=16 \), so the SD is 4.

Teacher's Note:
1. Work with sums and sums of squares of the groups.
2. The second group's data are the whole group's data minus the first group's.

Answer: (b) 5.5

Explanation:
1. First set: \( \sum x^2=5(4+4)=40 \), \( \sum x=10 \). Second set: \( \sum y^2=5(5+16)=105 \), \( \sum y=20 \).
2. Combined: \( \frac{145}{10}-\left(\frac{30}{10}\right)^2=14.5-9=5.5 \).

Teacher's Note:
1. Rebuild the sums of squares from variance and mean, then combine.
2. Subtract the square of the combined mean (3).

Answer: (c) change of origin but not of scale

Explanation:
1. Adding a constant to every observation shifts the mean by the same amount, so deviations are unchanged.
2. Multiplying by a constant k multiplies the standard deviation by |k|.
3. Hence it is independent of a change of origin but not of a change of scale.

Teacher's Note:
1. Think of what happens to deviations from the mean when you add or multiply.
2. Variance transforms as Var(ax + b) = a² Var(x).