JEE Mathematics Statistics MCQs Set 07

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

Explanation:
1. The mean gives a + b + 23 = 30, so a + b = 7.
2. Variance = Σx²/5 − 36 = 6.8, so Σx² = 214, i.e. a² + b² + 189 = 214 and a² + b² = 25.
3. a = 3, b = 4 satisfies both (3 + 4 = 7, 9 + 16 = 25).

Teacher's Note:
1. Use both the mean condition and the variance condition.

Answer: (a) 1

Explanation:
1. B is A shifted by 50 units.
2. Variance is unchanged by a shift, so V_A = V_B and the ratio is 1.

Teacher's Note:
1. Shifting every value by the same amount does not change the spread.

Answer: (c) \( \bar x+\frac{n+1}{2} \)

Explanation:
1. The total increase is 1 + 2 + ... + n = n(n + 1)/2.
2. Dividing by n gives an increase of (n + 1)/2 in the mean.

Teacher's Note:
1. Add the increments and divide by n.

Answer: (c) 57 kg

Explanation:
1. The total weight is 7 × 55 = 385 kg.
2. The six known weights add to 328 kg.
3. The seventh weight is 385 − 328 = 57 kg.

Teacher's Note:
1. Use total = mean × number of items.

Answer: (d) x = y = z

Explanation:
1. The sum is 50, so x = 5.
2. Sorted: 1, 2, 3, 5, 5, 5, 6, 6, 8, 9, so the median y = 5.
3. The value 5 occurs most often (three times), so z = 5. Hence x = y = z.

Teacher's Note:
1. Sort the data to find the median.

Answer: (c) \( \sqrt{10} \)

Explanation:
1. The second set is the first set shifted by 20.
2. Standard deviation does not change under a shift, so it remains √10.

Teacher's Note:
1. A shift leaves the spread unchanged.

Answer: (b) \( 2\bar x+3 \)

Explanation:
1. The mean of ax + b is a x̄ + b.
2. With a = 2 and b = 3, the mean is 2x̄ + 3.

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

Answer: (b) 10.1

Explanation:
1. There are 101 terms with mean 1 + 50d; the deviations are |kd − 50d| for k = 0, ..., 100.
2. The sum of the deviations is 2d(1 + 2 + ... + 50) = 2550d, so the mean deviation is 2550d/101.
3. 2550d/101 = 255 gives d = 10.1.

Teacher's Note:
1. The deviations are symmetric about the middle term.

Answer: (a) 10

Explanation:
1. Mean = 2/3 + 3/2 + 11/6 = 4.
2. E(x²) = 4/3 + 9/2 + 121/6 = 26.
3. Variance = 26 − 16 = 10.

Teacher's Note:
1. Use Var = E(x²) − (E x)².

Answer: (d) 52.25

Explanation:
1. Total marks = 40(50) + 35(60) + 45(55) + 42(45) = 2000 + 2100 + 2475 + 1890 = 8465.
2. Number of students = 162, so the average is 8465/162 = 52.25.

Teacher's Note:
1. Take the weighted mean.

Answer: (d) Group A is less variable than group B because group A's coefficient of variation is smaller

Explanation:
1. Compare variability with the coefficient of variation CV = (SD/mean) × 100.
2. CV of A = 10/22 × 100 ≈ 45.5% and CV of B = 12/23 × 100 ≈ 52.2%.
3. A has the smaller CV, so it is less variable (the proper reason is the CV).

Teacher's Note:
1. Use the coefficient of variation when the means differ.

Answer: (d) 38

Explanation:
1. Total of five = 150; total of four = 112.
2. The excluded number is 150 − 112 = 38.

Teacher's Note:
1. Total = mean × count.

Answer: (a) 78.00

Explanation:
1. Corrected Σx = 170 − 20 + 30 = 180 and corrected Σx² = 2830 − 400 + 900 = 3330.
2. Variance = 3330/15 − (180/15)² = 222 − 144 = 78.

Teacher's Note:
1. Correct both sums before computing the variance.

Answer: (b) \( \frac{nM-x_n+x'}{n} \)

Explanation:
1. The old total is nM. The new total is nM − x_n + x′.
2. Dividing by n gives (nM − x_n + x′)/n.

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

Answer: (d) 14

Explanation:
1. These form an AP with common difference d = 7 and n = 7 terms.
2. For an AP, SD = d √((n² − 1)/12) = 7 √(48/12) = 7 × 2 = 14.

Teacher's Note:
1. Use the AP formula for standard deviation.

Answer: (b) y = 1.02x + 8.74

Explanation:
1. b_yx = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²) = (34670 − 28600)/(22880 − 16900) = 6070/5980 ≈ 1.02.
2. The means are x̄ = 13 and ȳ = 22, so the intercept is 22 − 1.02(13) ≈ 8.74.
3. The line is y = 1.02x + 8.74.

Teacher's Note:
1. Compute the slope from the sums, then use the means.

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

Explanation:
1. Weighted mean = Σr(r² + r)/Σ(r² + r) = (Σr³ + Σr²)/(Σr² + Σr).
2. Numerator = n(n + 1)(3n + 1)(n + 2)/12 and denominator = n(n + 1)(n + 2)/3.
3. The ratio is (3n + 1)/4.

Teacher's Note:
1. Write the sums with standard formulas and cancel common factors.

Answer: (b) 39.95, 14.98

Explanation:
1. Corrected mean = (8000 − 10)/200 = 39.95.
2. Original Σx² = 200(225 + 1600) = 365000; corrected Σx² = 365000 − 2500 + 1600 = 364100.
3. Variance = 364100/200 − 39.95² ≈ 224.5, so SD ≈ 14.98.

Teacher's Note:
1. Correct Σx and Σx² separately.

Answer: (d) 169.18

Explanation:
1. Original total = 50 × 169 = 8450.
2. Corrected total = 8450 − 134 + 143 = 8459.
3. Corrected mean = 8459/50 = 169.18.

Teacher's Note:
1. Add the difference 9 to the total.