To determine the minimum sample size needed for normal approximation in a binomial distribution, we adhere to the rule of thumb that both np and n(1-p) should be at least 5, where n is the sample size and p is the probability of success.
Given p (probability of success) is 0.25, let's set up the inequalities:
For np >= 5:
n 0.25 >= 5
Divide both sides by 0.25 to solve for n:
n >= 5 / 0.25
n >= 20
For n(1-p) >= 5:
n (1 - 0.25) >= 5
n * 0.75 >= 5
Divide both sides by 0.75 to solve for n:
n >= 5 / 0.75
n >= 6.666...
Since both conditions must be satisfied, we want the larger n value, which is 20 from the first inequality. But we have to make sure the number is a whole number, so let's round up because if we round down, we might fall below the threshold for the normal approximation rule. Therefore, the minimum sample size n must be at least 20. However, since normal approximation tends to work better with a larger sample size, let's check the provided options to see which one is the smallest number that's greater than or equal to 20.
Among the provided options, 39, 38, 41, and 40, the smallest one that is greater than or equal to 20 is 38.
Therefore, the minimum sample size needed to use normal approximations for an estimate about a population, given a probability of success of 25 percent, from the provided options, is 38.
