Of all 3- to 5-year old children, 56% are enrolled in school (p=0.56). We are looking at a sample of 500 (n=500) such children, and this is thought to follow a binomial distribution. Use the normal approximation to the binomial to find the probability that at least 300 of those children will be enrolled in school P(X ≥ 300)

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To find the probability that at least 300 children will be enrolled in school (P(X ≥ 300)) using the normal approximation to the binomial distribution, we need to calculate the z-score and use the standard normal distribution table.

Step 1: Calculate the mean (μ) and standard deviation (σ) of the binomial distribution.
μ = n * p = 500 * 0.56 = 280
σ = sqrt(n * p * q) = sqrt(500 * 0.56 * (1-0.56)) = sqrt(123.2) ≈ 11.1

Step 2: Calculate the z-score.
z = (x - μ) / σ
For P(X ≥ 300), we plug in x = 300.
z = (300 - 280) / 11.1 ≈ 1.80

Step 3: Look up the cumulative probability in the standard normal distribution table.
The cumulative probability for z = 1.80 is approximately 0.9641.

Step 4: Calculate the probability of at least 300 children being enrolled in school.
P(X ≥ 300) ≈ 1 - 0.9641 ≈ 0.0359

Therefore, the probability that at least 300 of these 500 children will be enrolled in school is approximately 0.0359 or 3.59%.