A random sample of 31 sales charge showed a sample standard deviation of $50.A 90% confidence interval estimate of the population standard deviation is

A) 1715.101 to 4055.589.
B) 1596.458 to 4466.679.
C) 39.956 to 66.833.
D) 41.393 to 63.684.

We use the chi-square distribution with (n-1) degrees of freedom to construct a confidence interval for population standard deviation as follows:
((n-1)*s^2)/X^2 ≤ σ^2 ≤ ((n-1)*s^2)/X^2
where X^2 is the chi-square value for the given confidence level and (n-1) degrees of freedom.
For this problem, n=31 and the sample standard deviation s=$50. Using a chi-square distribution calculator with 30 degrees of freedom and a 90% confidence level, we get X^2=42.557. Substituting the values, we get the confidence interval as:
(30*50^2)/63.69 ≤ σ^2 ≤ (30*50^2)/26.51
or 41.393 ≤ σ ≤ 63.684 (rounded to three decimal places). Therefore, the answer is D.