Compute $P(X)$ using the binomial probability formula. Then determine whether the normal distribution can be used to estimate this probability. If so, approximate $P(X)$ using the normal distribution and compare the result with the exact probability.
\[
n=54, p=0.6 \text {, and } X=36
\]
For $n=54, p=0.6$, and $X=36$, find $P(X)$.
$P(X)=0.0687$ (Round to four decimal places as needed.)
Can the normal distribution be used to approximate this probability?
A. No, the normal distribution cannot be used because $n p(1-p) \geq 10$.
B. No, the normal distribution cannot be used because $n p(1-p)< 10$.
C. Yes, the normal distribution can be used because $n p(1-p) \leq 10$.
D. Yes, the normal distribution can be used because $n p(1-p) \geq 10$.

Step 1 ：Given the values: \(n = 54\), \(p = 0.6\), and \(X = 36\).

Step 2 ：Calculate the binomial probability \(P(X)\) using the formula: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\).

Step 3 ：Substitute the given values into the formula to get: \(P(X=36) = C(54, 36) * (0.6^{36}) * ((1-0.6)^{54-36})\).

Step 4 ：Compute the above expression to get the probability \(P(X=36)\) which is approximately \(\boxed{0.0687}\).

Step 5 ：Next, check if the normal distribution can be used to approximate this probability. The rule of thumb is that the normal distribution can be used if \(np(1-p) \geq 10\).

Step 6 ：Calculate the value of \(np(1-p)\) by substituting the given values: \(54 * 0.6 * (1-0.6)\).

Step 7 ：The calculated value is approximately 12.96, which is greater than 10.

Step 8 ：Therefore, the normal distribution can be used to approximate this probability.

Step 9 ：Final Answer: The probability \(P(X=36)\) is approximately \(\boxed{0.0687}\). The normal distribution can be used to approximate this probability because \(np(1-p)\) is \(\boxed{12.96}\), which is greater than 10. Therefore, the answer is \(\boxed{\text{D. Yes, the normal distribution can be used because } np(1-p) \geq 10}\).