Mean and variance for four different Binomial distributions with parameters n…
2026
Mean and variance for four different Binomial distributions with parameters n (number of trials) and p (probability of success) are given in A-D below. Compute the number of trials (n) for each Binomial distribution and arrange in ascending order.
A. Mean = 2.1, Variance = 1.47
B. Mean = 2.0, Variance = 1.20
C. Mean = 4.0, Variance = 2.00
D. Mean = 3.6, Variance = 1.44
Choose the correct answer from the options given below:
- A.
C, D, A, B
- B.
A, C, B, D
- C.
D, A, C, B
- D.
B, D, A, C
Show answer & explanation
Correct answer: D
Concept: For a Binomial distribution X ~ B(n, p), Mean = np and Variance = np(1 − p). Dividing Variance by Mean cancels the common factor np, leaving Variance ÷ Mean = 1 − p. So p = 1 − (Variance ÷ Mean), and once p is known, n = Mean ÷ p.
Application: Apply this to each case.
Case A: Mean = 2.1, Variance = 1.47, so 1 − p = 1.47 ÷ 2.1 = 0.7, giving p = 0.3 and n = 2.1 ÷ 0.3 = 7.
Case B: Mean = 2.0, Variance = 1.20, so 1 − p = 1.20 ÷ 2.0 = 0.6, giving p = 0.4 and n = 2.0 ÷ 0.4 = 5.
Case C: Mean = 4.0, Variance = 2.00, so 1 − p = 2.00 ÷ 4.0 = 0.5, giving p = 0.5 and n = 4.0 ÷ 0.5 = 8.
Case D: Mean = 3.6, Variance = 1.44, so 1 − p = 1.44 ÷ 3.6 = 0.4, giving p = 0.6 and n = 3.6 ÷ 0.6 = 6.
Cross-check: Substitute each (n, p) pair back into Variance = np(1 − p). For Case B, 5 × 0.4 × 0.6 = 1.2, which matches the given variance, confirming the values were derived correctly rather than assumed.
Arranging the four number-of-trials values in ascending order (5, 6, 7, 8) places the cases in the sequence B, D, A, C.