Question

Of all 3- to 5-year old children, 56% are enrolled in school. If a sample of 500 such children is randomly selected, find the probability that at least 253 will be enrolled in school. Round the final answer to 4 decimal places and round intermediate-value calculations to 2 decimal places.

What is the probability that, in a randomly selected sample of 500 children aged 3 to 5, at least 253 will be enrolled in school? Use the normal approximation to the binomial distribution with parameters n = 500 and p = 0.56. Please provide your answer rounded to 4 decimal places.

Answer 1

To find the probability using the normal approximation to the binomial distribution, calculate the mean and standard deviation, then use the z-score formula to find the probability.

Step-by-step explanation:

To find the probability that at least 253 children will be enrolled in school out of a randomly selected sample of 500 children aged 3 to 5, we can use the normal approximation to the binomial distribution. Since np = 500 * 0.56 = 280 > 5 and nq = 500 * 0.44 = 220 > 5, we can assume that the distribution is approximately normal. The mean (μ) is given by μ = np = 500 * 0.56 = 280, and the standard deviation (σ) is given by σ = sqrt(npq) = sqrt(500 * 0.56 * 0.44) ≈ 11.05.

Now we can use a z-score to find the probability. The formula for the z-score is z = (x - μ) / σ, where x is the number of children enrolled in school that we want to find the probability for. In this case, we want to find the probability for at least 253 children, so x = 253.

Plugging the values into the formula, we get z = (253 - 280) / 11.05 ≈ -2.44. Now we can use a standard normal distribution table or a calculator to find the probability corresponding to this z-score. The probability for a z-score of -2.44 or less is approximately 0.0071.