CBSE Class 8 Mathematics Statistics and Probability MCQs Set 25

Answer: (b) 3

Explanation:
1. Σf = 41 + p and Σfx = 18 + 40 + 105 + 9p + 88 + 52 = 303 + 9p.
2. Mean = (303 + 9p) / (41 + p) = 7.5, so 303 + 9p = 307.5 + 7.5p.
3. 1.5p = 4.5, so p = 3.

Teacher's Note:
1. Mean = Σfx / Σf for a frequency table.
2. Form the equation and solve for p.

Answer: (a) 52

Explanation:
1. Let n be the number from school III: 4500 + 3840 + 55n + 2000 = 66(148 + n).
2. 10340 + 55n = 9768 + 66n, so 11n = 572 and n = 52.

Teacher's Note:
1. Total score = sum of (candidates × average).
2. Overall average = total score / total candidates.

Answer: (b) 10

Explanation:
1. Let there be b blue balls: b / (b + 5) = 2 × 5 / (b + 5), so b = 10.

Teacher's Note:
1. Both probabilities have the same denominator.
2. Blue balls = 2 × red balls.

Answer: (a) \( \frac{2}{3} \)

Explanation:
1. Favourable outcomes: 2, 3, 4, 6, which is 4 outcomes.
2. Probability = 4 / 6 = 2/3.

Teacher's Note:
1. List the favourable outcomes; do not count 6 twice.
2. Total outcomes = 6.

Answer: (a) \( \frac{5}{36} \)

Explanation:
1. Pairs with sum 8: (2,6), (3,5), (4,4), (5,3), (6,2), which is 5 pairs.
2. Probability = 5 / 36.

Teacher's Note:
1. Total outcomes with two dice = 36.
2. Count ordered pairs.

Answer: (c) \( \frac{11}{36} \)

Explanation:
1. Count pairs where (first even and second multiple of 3) or (second even and first multiple of 3).
2. First even (2, 4, 6) and second in {3, 6}: 3 × 2 = 6 pairs. Second even and first in {3, 6}: another 6 pairs.
3. The pair (6, 6) is in both lists, so total = 6 + 6 − 1 = 11 pairs.
4. Probability = 11 / 36.

Teacher's Note:
1. Use inclusion-exclusion to avoid double counting.
2. The pair (6, 6) satisfies both conditions.

Answer: (b) \( \frac{1}{7} \)

Explanation:
1. An ordinary year has 52 weeks and 1 extra day.
2. That extra day can be any one of the 7 weekdays, so P(Sunday) = 1/7.

Teacher's Note:
1. Extra days decide whether there is a 53rd occurrence.
2. Ordinary year: 1 extra day; leap year: 2.

Answer: (c) \( \frac{1}{7} \)

Explanation:
1. A leap year has 52 weeks and 2 extra days.
2. The pairs of consecutive extra days are 7 equally likely: (Sun, Mon), (Mon, Tue), ..., (Sat, Sun).
3. Only (Sun, Mon) gives both, so the probability is 1/7.

Teacher's Note:
1. List the 7 possible pairs of extra days.
2. Exactly one pair contains both Sunday and Monday in that order.

Answer: (d) \( \frac{1}{26} \)

Explanation:
1. Black kings: king of spades and king of clubs, so 2 cards.
2. Probability = 2 / 52 = 1/26.

Teacher's Note:
1. Count cards satisfying both conditions.
2. Reduce 2/52.

Answer: (b) 29.9

Explanation:
1. Wrong total = 100 × 30 = 3000.
2. Correct total = 3000 − 32 − 12 + 23 + 11 = 2990.
3. Correct mean = 2990 / 100 = 29.9.

Teacher's Note:
1. Correct the total, not the mean.
2. Subtract the wrong values and add the right ones.

Answer: (d) Rs. 1450

Explanation:
1. Total of 10 = 14450. New total = 14450 + 1500 = 15950.
2. Mean = 15950 / 11 = Rs. 1450.

Teacher's Note:
1. Find the total, update it, then divide by the new count.
2. The new salary is only slightly above the old mean, so the mean rises slightly.

Answer: (a) 14

Explanation:
1. The numbers are already in order and n = 5.
2. Median = the 3rd value = 14.

Teacher's Note:
1. For odd n the median is the middle value.
2. Arrange in order first.

Answer: (d) 22.5

Explanation:
1. Arranged: 13, 15, 16, 19, 20, 22, 23, 25, 27, 30, 32, 35.
2. For n = 12, the median is the average of the 6th and 7th values: (22 + 23) / 2 = 22.5.

Teacher's Note:
1. For even n average the two middle values.
2. Always sort the data first.

Answer: (d) 3

Explanation:
1. Frequencies: 3 occurs 4 times, 2 and 1 twice each, 4 and 5 once.
2. Mode = 3.

Teacher's Note:
1. Mode is the most frequent value.
2. Count each value.

Answer: (b) 5

Explanation:
1. Sum = 10 + 12 + x + 8 + 20 = 50 + x = 5 × 11 = 55, so x = 5.

Teacher's Note:
1. Total = mean × number of observations.
2. Check: 55 / 5 = 11.

Answer: (d) 2 and 3 only

Explanation:
1. First five natural numbers: mean = 15 / 5 = 3, so 1 is false.
2. First ten odd numbers sum to 100, mean 10, so 2 is true.
3. (3 − 1 + 7 + 0 + 11) / 5 = 20 / 5 = 4, so 3 is true.
4. Primes 7, 11, 13, 17, 19, 23 sum to 90, mean 15, so 4 is false.

Teacher's Note:
1. Check each statement by direct calculation.
2. Sum of the first n odd numbers is n².