Find the minimum sample size $n$ needed to estimate $\mu$ for the given values of $c, \sigma$, and $E$. \[ c=0.90, \sigma=8.7 \text {, and } E=2 \] Assume that a preliminary sample has at least 30 members. $n=$ (Round up to the nearest whole number.)

Step 1 ：Given values are: confidence level c = 0.90, standard deviation σ = 8.7, and margin of error E = 2.

Step 2 ：The z-score Zc corresponding to the confidence level c = 0.90 is 1.645 (for a two-tailed test).

Step 3 ：The formula to calculate the sample size n is given by: \(n = \left(\frac{Z_c * \sigma}{E}\right)^2\)

Step 4 ：Substitute the given values into the formula: \(n = \left(\frac{1.645 * 8.7}{2}\right)^2\)

Step 5 ：Calculate the sample size n to get n = 52.

Step 6 ：So, the minimum sample size n needed to estimate μ for the given values of c, σ, and E is \(\boxed{52}\).