Match the LIST-I with LIST-II LIST-I (n = trials, p = probability of success)…
2025
Match the LIST-I with LIST-II
LIST-I (n = trials, p = probability of success) | LIST-II (Mean of Binomial Distribution) |
|---|---|
A. n = 16, p = 0.8 | I. 9.1 |
B. n = 21, p = 0.6 | II. 9.2 |
C. n = 13, p = 0.7 | III. 12.8 |
D. n = 23, p = 0.4 | IV. 12.6 |
- A.
A-II, B-IV, C-I, D-III
- B.
A-III, B-IV, C-I, D-II
- C.
A-I, B-III, C-II, D-IV
- D.
A-III, B-I, C-IV, D-II
Show answer & explanation
Correct answer: B
Concept: For a Binomial distribution B(n, p), the mean (expected value) equals n × p. This follows directly from linearity of expectation — the distribution is the sum of n independent Bernoulli(p) trials, and each trial contributes an expected value of p, so the sum of n such trials has expected value n × p.
Application: compute the mean for every item in LIST-I using Mean = n × p, and match each result to the corresponding value in LIST-II.
A: n = 16, p = 0.8 → Mean = 16 × 0.8 = 12.8, which is LIST-II item III.
B: n = 21, p = 0.6 → Mean = 21 × 0.6 = 12.6, which is LIST-II item IV.
C: n = 13, p = 0.7 → Mean = 13 × 0.7 = 9.1, which is LIST-II item I.
D: n = 23, p = 0.4 → Mean = 23 × 0.4 = 9.2, which is LIST-II item II.
Cross-check: the four computed means (12.8, 12.6, 9.2, 9.1) are pairwise distinct and together account for all four LIST-II entries exactly once, confirming there is no ambiguity in the matching. Recomputing item D independently (23 × 0.4 = 9.2) also confirms its pairing with II.
Result: the correct matching is A-III, B-IV, C-I, D-II.