The probability of success on any trail of a binomial experiment is 25%. Find the probability that the proportion of successes in a sample of 500 is less than 22%.
0.0755
0.1224
0.1863
0.0606
0.5612
To find the probability that the proportion of successes in a sample of 500 is less than 22%, we can use the normal approximation to the binomial distribution.
First we need to find the mean and standard deviation of the sample proportion. The mean of the sample proportion is equal to the probability of success, which is 0.25. The standard deviation of the sample proportion is equal to the square root of pq/n, where p = 0.25, q = 0.75, and n = 500.
Standard deviation = sqrt(0.25*0.75/500) = sqrt(0.000375) = 0.019364
Next, we need to standardize the value of 0.22 using the z-score formula:
z = (X - mean) / standard deviation
z = (0.22 - 0.25) / 0.019364
z = -1.5455
Using a standard normal distribution table, we can find the probability that the z-score is less than -1.5455, which is approximately 0.0616.
Therefore, the probability that the proportion of successes in a sample of 500 is less than 22% is 0.0616, which is closest to 0.0606.
So, the correct answer is 0.0606.