The director of the library assumes that 20 % of the library's collection is checked out. If the director is correct, what is the probability that the proportion of books checked out in a sample of 483 books would differ from the population proportion by less than 3% ? Round your answer to four decimal places.

Step 1 ：We are given a problem of sampling distribution of proportions. The population proportion (p) is 0.20, the sample size (n) is 483, and we are asked to find the probability that the sample proportion (p_hat) differs from the population proportion by less than 3% (0.03).

Step 2 ：The sampling distribution of proportions is approximately normal if np and n(1-p) are both greater than 5, which is true in this case.

Step 3 ：The mean of the sampling distribution is equal to the population proportion (p), and the standard deviation is calculated as \(\sqrt{p(1-p)/n}\).

Step 4 ：We calculate the z-scores for the boundaries of the interval (p - 0.03 and p + 0.03), and then use the standard normal distribution to find the probability that the sample proportion falls within this interval.

Step 5 ：Given p = 0.2, n = 483, and the difference = 0.03, we calculate the standard deviation as 0.018200630207731605.

Step 6 ：The z-scores for the boundaries of the interval are calculated as -1.648294573187693 and 1.648294573187693.

Step 7 ：Using the standard normal distribution, we find the probability that the sample proportion falls within this interval as 0.9007077621842103.

Step 8 ：Final Answer: The probability that the proportion of books checked out in a sample of 483 books would differ from the population proportion by less than 3% is \(\boxed{0.9007}\).