In a binomial distribution with n = 5 and p = 0.4, which value is the most likely for X (the mode)?

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Multiple Choice

In a binomial distribution with n = 5 and p = 0.4, which value is the most likely for X (the mode)?

Explanation:
In a binomial setting, the most probable number of successes—the mode—occurs near the mean, and there is a handy formula: the mode equals floor((n+1)p) when (n+1)p is not an integer (if it is an integer, there are two modes). Here, (n+1)p = 6 × 0.4 = 2.4, so the mode is floor(2.4) = 2. Since 2.4 isn’t an integer, there’s a single mode at 2. You can see why by comparing nearby probabilities: P(X=2) = C(5,2)(0.4)^2(0.6)^3 ≈ 0.3456, while P(X=1) ≈ 0.2592 and P(X=3) ≈ 0.2304. Thus 2 is the most likely value.

In a binomial setting, the most probable number of successes—the mode—occurs near the mean, and there is a handy formula: the mode equals floor((n+1)p) when (n+1)p is not an integer (if it is an integer, there are two modes). Here, (n+1)p = 6 × 0.4 = 2.4, so the mode is floor(2.4) = 2. Since 2.4 isn’t an integer, there’s a single mode at 2. You can see why by comparing nearby probabilities: P(X=2) = C(5,2)(0.4)^2(0.6)^3 ≈ 0.3456, while P(X=1) ≈ 0.2592 and P(X=3) ≈ 0.2304. Thus 2 is the most likely value.

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