30% of the fifth grade students in a large school district read below grade level. The distribution of sample proportions of samples of 100 students from this population is normal with a mean of 0.30 and a standard deviation of 0.045. Suppose that you select a sample of 100 fifth grade students from this district and find that the proportion that reads below grade level in the sample is 0.36. What is the probability that a second sample would be selected with a proportion less than 0.36?

A. 0.8932
B. 0.8920
C. 0.9032
D. 0.9048

answer :D

To solve this problem, we need to use the concept of the sampling distribution of sample proportions. The sampling distribution is approximately normal when the sample size is large (greater than or equal to 30) and the population follows a binomial distribution.

Given that the distribution of sample proportions is normal with a mean of 0.30 and a standard deviation of 0.045, we can calculate the z-score for the given sample proportion.

The formula for calculating the z-score is:

z = (x - μ) / σ

Where:
x = Sample proportion
μ = Mean of the distribution (0.30)
σ = Standard deviation of the distribution (0.045)

In this case, x = 0.36, μ = 0.30, and σ = 0.045.

Plugging in the values, we get:
z = (0.36 - 0.30) / 0.045 = 1.3333

Next, we need to find the probability of getting a sample proportion less than 0.36, which is equivalent to finding the area under the normal curve to the left of the z-score (-1.3333).

Using a standard normal table or a calculator, we can find that the probability corresponding to a z-score of -1.3333 is approximately 0.0916.

However, the question asks for the probability that a second sample would be selected with a proportion less than 0.36. Since the sampling distribution is approximately normal, the probability of selecting a second sample with a proportion less than 0.36 would be the same as the probability of selecting the first sample with a proportion less than 0.36.

Therefore, the answer is:
A. 0.8916 (rounded to four decimal places)