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%.
To find the probability that the proportion of successes in a sample of 500 is less than 22%, we can use the normal distribution approximation to the binomial distribution.
First, we need to find the mean and standard deviation of the binomial distribution. The mean of a binomial distribution is given by np, where n is the number of trials and p is the probability of success. In this case, n = 500 and p = 0.25, so the mean is 500 * 0.25 = 125.
The standard deviation of a binomial distribution is given by sqrt(np(1-p)). In this case, the standard deviation is sqrt(500 * 0.25 * 0.75) = sqrt(93.75) ≈ 9.68.
Next, we need to convert the proportion of successes (22%) to a z-score using the formula z = (x - mean) / standard deviation. In this case, x = 500 * 0.22 = 110. So the z-score is (110 - 125) / 9.68 ≈ -1.55.
We can then use a standard normal table or a calculator to find the probability that a z-score is less than -1.55. The probability that a z-score is less than -1.55 is approximately 0.0606.
Therefore, the probability that the proportion of successes in a sample of 500 is less than 22% is approximately 0.0606, or 6.06%.