Statistics & Probability — Practice

Question 1

Find the mean of the following frequency distribution using the step-deviation method: Class: 0–10, 10–20, 20–30, 30–40, 40–50. Frequency: 6, 8, 20, 10, 6.

Accepted Answer

Midpoints xi = 5, 15, 25, 35, 45. Take assumed mean A = 25, h = 10, so ui = (xi−25)/10 = −2, −1, 0, 1, 2.

fiui = 6(−2)+8(−1)+20(0)+10(1)+6(2) = −12−8+0+10+12 = 2. Σfi = 50.

Mean = A + (Σfiui/Σfi)×h = 25 + (2/50)×10 = 25 + 0.4 = 25.4.

(Cross-check via direct method: Σfixi = 6(5)+8(15)+20(25)+10(35)+6(45) = 30+120+500+350+270 = 1270; mean = 1270/50 = 25.4 ✓.)

Answer: 25.4. Trap: 25.04 comes from adding (Σfiui/Σfi) directly to A without multiplying by h first — using 2/50=0.04 instead of 0.4. Trap: 0.4 is only the correction term (Σfiui/Σfi)×h with the assumed mean A never added back. Trap: 24.6 flips the sign of the correction term, subtracting instead of adding it to A.

Question 2

Find the median of the following frequency distribution: Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60. Frequency: 3, 7, 12, 15, 8, 5.

Accepted Answer

N = Σfi = 3+7+12+15+8+5 = 50, so N/2 = 25. Cumulative frequencies: 3, 10, 22, 37, 45, 50.

The class 30–40 is the first with cumulative frequency (37) ≥ 25, so it is the median class: l = 30, cf = 22 (cumulative frequency of the class BEFORE it, i.e. up to 20–30), f = 15, h = 10.

Median = l + [(N/2 − cf)/f] × h = 30 + [(25−22)/15]×10 = 30 + (3/15)×10 = 30 + 2 = 32.

Answer: 32. Trap: 22 comes from using the median class's OWN cumulative frequency (37) as cf instead of the preceding class's (22): 30 + [(25−37)/15]×10 = 30−8 = 22. Trap: 46.67 forgets to subtract cf entirely, computing 30 + (25/15)×10. Trap: 32.5 misidentifies 20–30 as the median class (using l=20, cf=10, f=12 instead of the correct class): 20 + [(25−10)/12]×10 = 32.5.

Question 3

Find the mode of the following frequency distribution: Class: 0–10, 10–20, 20–30, 30–40, 40–50, 50–60. Frequency: 6, 12, 12, 20, 8, 5.

Accepted Answer

The highest frequency is 20, in class 30–40 — this is the modal class: l = 30, f1 = 20, f0 = 12 (frequency of the preceding class 20–30), f2 = 8 (frequency of the succeeding class 40–50), h = 10.

Mode = l + [(f1−f0)/(2f1−f0−f2)] × h = 30 + [(20−12)/(40−12−8)]×10 = 30 + [8/20]×10 = 30 + 4 = 34.

Answer: 34. Trap: 36 swaps f0 and f2 in the numerator, computing (f1−f2) = (20−8) = 12 instead of (f1−f0): 30+[12/20]×10 = 36. Trap: 35 simply reports the modal class's midpoint (30+40)/2 = 35, confusing it with the actual mode. Trap: 32 uses only 2f1 as the denominator (forgetting to subtract f0 and f2): 30 + [8/40]×10 = 32.

Question 4

Find the mean deviation about the mean for the data: 2, 4, 6, 8, 10, 12.

Accepted Answer

Mean x̄ = (2+4+6+8+10+12)/6 = 42/6 = 7.

Absolute deviations |xi−7|: 5, 3, 1, 1, 3, 5. Sum = 18.

Mean deviation = 18/6 = 3.

Answer: 3. Trap: 0 comes from forgetting the absolute value — the signed deviations (−5,−3,−1,1,3,5) sum to exactly zero by definition of the mean, which is why the modulus is essential. Trap: 3.6 divides the sum of absolute deviations by (n−1)=5 instead of n=6. Trap: 11.67 confuses mean deviation with variance, squaring the deviations instead of taking absolute values: Σ(xi−7)² = 25+9+1+1+9+25 = 70, and 70/6 ≈ 11.67.

Question 5

The variance of the data 3, 5, 7, 9, 11 is:
(a) 8  (b) 10  (c) 57  (d) 2.4

Accepted Answer

Mean x̄ = (3+5+7+9+11)/5 = 35/5 = 7. Σxi² = 9+25+49+81+121 = 285.

Variance σ² = Σxi²/n − x̄² = 285/5 − 49 = 57 − 49 = 8.

Answer: (a) 8. Trap (b) 10 uses the (n−1)=4 'sample variance' divisor instead of the correct population divisor n=5: Σ(xi−x̄)²/(n−1) = 40/4 = 10 — NDA/CBSE variance always uses n. Trap (c) 57 is Σxi²/n alone, with the −x̄² term forgotten entirely. Trap (d) 2.4 computes the mean deviation (Σ|xi−x̄|/n = 12/5 = 2.4) instead of the variance, confusing the two different dispersion measures.

Question 6

Batsman A has mean score 60 with SD 15; batsman B has mean score 45 with SD 12. Which batsman is more consistent?
(a) A is more consistent (CV_A = 25%, CV_B ≈ 26.67%)  (b) B is more consistent (CV_A = 25%, CV_B ≈ 26.67%)  (c) A is more consistent (CV_A = 15%, CV_B = 12%)  (d) A is more consistent (CV_A = 900%, CV_B = 540%)

Accepted Answer

CV = (σ/x̄)×100. CV_A = (15/60)×100 = 25%. CV_B = (12/45)×100 ≈ 26.67%.

The SMALLER coefficient of variation indicates more consistent (less variable) performance, so since 25% < 26.67%, batsman A is more consistent.

Answer: (a). Trap (b) uses the correct CV values but reverses the rule, picking the HIGHER-CV batsman as 'more consistent' — the rule is smaller CV = more consistent, not the other way round. Trap (c) reports the raw SD values as if they were already the CV, skipping the ÷mean×100 normalisation entirely — this is meaningless when the two means differ. Trap (d) inverts the formula, multiplying σ by x̄ instead of dividing (15×60=900, 12×45=540).

Question 7

Two dice are thrown together. What is the probability of getting a sum of 8?
(a) 5/36  (b) 1/6  (c) 1/36  (d) 1/11

Accepted Answer

Total equally-likely outcomes = 36. Pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2) — 5 outcomes.

P(sum = 8) = 5/36.

Answer: (a). Trap (b) 1/6 = 6/36 miscounts the favourable outcomes as six instead of five. Trap (c) 1/36 counts only a single qualifying pair, missing the other four ways to make 8. Trap (d) 1/11 wrongly treats the 11 possible sums (2 through 12) as equally likely outcomes and assigns each a probability of 1/11 — but the sums are NOT equally likely; only the 36 elementary (die1, die2) outcomes are.

Question 8

A card is drawn at random from a well-shuffled pack of 52 playing cards. What is the probability that it is either a king or a heart?
(a) 4/13  (b) 17/52  (c) 1/52  (d) 1/4

Accepted Answer

P(King) = 4/52, P(Heart) = 13/52. King and Heart are NOT mutually exclusive — the king of hearts belongs to both — so P(King∩Heart) = 1/52.

P(King or Heart) = P(King) + P(Heart) − P(King∩Heart) = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

Answer: (a). Trap (b) 17/52 = 4/52+13/52 naively adds the two probabilities as if King and Heart were mutually exclusive, without subtracting the king-of-hearts overlap — the single most common addition-theorem error. Trap (c) 1/52 multiplies P(King)×P(Heart) as if using the independence rule instead of the addition theorem, which is the wrong operation for an 'or' question. Trap (d) 1/4 = 13/52 counts only the hearts, ignoring the three kings that are not hearts.

Question 9

A fair die is rolled once. Given that the number obtained is even, what is the probability that it is greater than 4?
(a) 1/3  (b) 1/2  (c) 1/6  (d) 2/3

Accepted Answer

B = even numbers = {2,4,6}, P(B) = 3/6 = 1/2. A = numbers greater than 4 = {5,6}. A∩B (even AND greater than 4) = {6}, so P(A∩B) = 1/6.

P(A|B) = P(A∩B)/P(B) = (1/6)/(1/2) = 1/3.

Answer: (a). Trap (b) 1/2 reports P(B) itself, confusing the unconditional probability of the given event with the conditional probability being asked for. Trap (c) 1/6 reports P(A∩B) alone, skipping the division by P(B) that conditioning requires. Trap (d) 2/3 mistakenly includes 4 in A∩B (treating 4 as 'greater than 4'), giving A∩B={4,6}, P=2/6, and (2/6)/(1/2) = 2/3 — a boundary error on the strict inequality.

Question 10

A bag contains 5 red and 7 black balls. Two balls are drawn one after another WITH replacement. What is the probability that both balls drawn are red?
(a) 25/144  (b) 5/6  (c) 5/12  (d) 5/33

Accepted Answer

With replacement, the two draws are independent, and P(red) = 5/12 on each draw.

P(both red) = P(red) × P(red) = (5/12) × (5/12) = 25/144.

Answer: (a). Trap (b) 5/6 = 10/12 wrongly ADDS the two probabilities (5/12+5/12) instead of multiplying them, confusing the multiplication theorem for independent events with the addition theorem. Trap (c) 5/12 reports the probability for only a single draw, forgetting the second draw entirely. Trap (d) 5/33 = (5/12)×(4/11) wrongly applies the WITHOUT-replacement (dependent-events) formula to a question that explicitly states the balls are drawn WITH replacement.

Question 11

A factory has two machines, A and B. Machine A produces 60% of the total output and Machine B produces 40%. 5% of the items from A are defective, and 3% of the items from B are defective. An item is picked at random and found defective. What is the probability that it was produced by Machine A?
(a) 5/7  (b) 3/100  (c) 2/7  (d) 5/8

Accepted Answer

P(A) = 0.6, P(B) = 0.4, P(D|A) = 0.05, P(D|B) = 0.03.

P(D) = P(A)P(D|A) + P(B)P(D|B) = 0.6(0.05) + 0.4(0.03) = 0.03 + 0.012 = 0.042.

P(A|D) = P(A)P(D|A)/P(D) = 0.03/0.042 = 30/42 = 5/7.

Answer: (a). Trap (b) 3/100 = 0.03 is just the numerator P(A)P(D|A), with the division by the total-probability denominator P(D) forgotten entirely. Trap (c) 2/7 = 0.012/0.042 is P(B|D), the probability the item came from the OTHER machine — answering for the wrong cause (note these two answers correctly sum to 1). Trap (d) 5/8 comes from wrongly assuming both machines produce an equal 50% share of output instead of using the actual given 60:40 split: (0.5×0.05)/(0.5×0.05+0.5×0.03) = 0.025/0.04 = 5/8.

Question 12

A fair coin is tossed 8 times. If X denotes the number of heads obtained, the mean and variance of X are respectively:
(a) 4 and 2  (b) 4 and 4  (c) 8 and 2  (d) 2 and 4

Accepted Answer

n = 8, p = q = 1/2 (fair coin).

Mean = np = 8 × (1/2) = 4. Variance = npq = 8 × (1/2) × (1/2) = 2.

Answer: (a) 4 and 2. Trap (b) 4 and 4 uses variance = np (forgetting the q factor entirely — using the mean formula for variance too), instead of npq. Trap (c) 8 and 2 uses n itself as the mean (forgetting to multiply by p) while getting the variance right. Trap (d) 2 and 4 swaps the mean and variance values around — the mean (np=4) must be larger than the variance (npq=2) here since q=1/2<1, so a swap is easy to catch by comparing the two numbers.

Question 13

A discrete random variable X has the probability distribution: X = 0, 1, 2, 3 with P(X) = k, 2k, 3k, 4k respectively. The value of k and E(X) are:
(a) k = 0.1, E(X) = 2  (b) k = 0.1, E(X) = 5  (c) k = 1/6, E(X) = 10/3  (d) k = 0.1, E(X) = 1.5

Accepted Answer

Since probabilities must sum to 1: k+2k+3k+4k = 10k = 1, so k = 0.1.

E(X) = Σxipi = 0(k)+1(2k)+2(3k)+3(4k) = 2k+6k+12k = 20k = 20(0.1) = 2.

Answer: (a) k=0.1, E(X)=2. Trap (b) uses the correct k but computes E(X²) = 0(k)+1(2k)+4(3k)+9(4k) = 2k+12k+36k = 50k = 5 instead of E(X), confusing the first and second moments. Trap (c) finds k by mistakenly omitting the X=3 row when normalising (k+2k+3k=1 ⇒ k=1/6), then computes E(X) with all four terms using this wrong k: 20(1/6) = 10/3. Trap (d) uses the correct k=0.1 but computes E(X) as a plain unweighted average of the X-values (0+1+2+3)/4 = 1.5, ignoring the probability weights entirely.

Practice — Statistics & Probability

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