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.
\[
n=50, p=0.7, \text { and } X=39
\]
For $n=50, p=0.7$, and $X=39$, find $P(X)$.
$P(X)=\square$ (Round to four decimal places as needed.)

Step 1 ：First, we need to compute the binomial probability $P(X)$ using the formula $P(X) = \binom{n}{X}p^X(1-p)^{n-X}$, where $n=50$, $p=0.7$, and $X=39$.

Step 2 ：Substitute the given values into the formula, we get $P(X) = \binom{50}{39}(0.7)^{39}(1-0.7)^{50-39}$.

Step 3 ：Calculate $\binom{50}{39}$ using the formula $\binom{n}{X} = \frac{n!}{X!(n-X)!}$, we get $\binom{50}{39} = \frac{50!}{39!(50-39)!} = \frac{50!}{39!11!}$.

Step 4 ：Calculate $50!$, $39!$, and $11!$ respectively, we get $50! = 30414093201713378043612608166064768844377641568960512000000000000$, $39! = 20397882081197443358640281739902897356800000000$, and $11! = 39916800$.

Step 5 ：Substitute these values into the formula, we get $\binom{50}{39} = \frac{30414093201713378043612608166064768844377641568960512000000000000}{20397882081197443358640281739902897356800000000 \times 39916800} = 122977956517698282$.

Step 6 ：Substitute $\binom{50}{39}$, $p$, and $n-X$ into the formula, we get $P(X) = 122977956517698282 \times (0.7)^{39} \times (1-0.7)^{11}$.

Step 7 ：Calculate $(0.7)^{39}$ and $(1-0.7)^{11}$ respectively, we get $(0.7)^{39} = 1.6069380442589902755419620923412e-10$ and $(1-0.7)^{11} = 5.904899999999999e-6$.

Step 8 ：Substitute these values into the formula, we get $P(X) = 122977956517698282 \times 1.6069380442589902755419620923412e-10 \times 5.904899999999999e-6 = 0.0117$.

Step 9 ：Round to four decimal places as needed, we get $P(X) = \boxed{0.0117}$.

Step 10 ：Next, we need to determine whether the normal distribution can be used to estimate this probability. The rule of thumb is that the normal distribution can be used if both $np$ and $n(1-p)$ are greater than 5.

Step 11 ：Calculate $np$ and $n(1-p)$ respectively, we get $np = 50 \times 0.7 = 35$ and $n(1-p) = 50 \times (1-0.7) = 15$.

Step 12 ：Since both $np$ and $n(1-p)$ are greater than 5, the normal distribution can be used to estimate this probability.

Step 13 ：Finally, we need to approximate $P(X)$ using the normal distribution and compare the result with the exact probability. The formula for the normal distribution is $P(X) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(X-\mu)^2}{2\sigma^2}}$, where $\mu = np$ and $\sigma = \sqrt{np(1-p)}$.

Step 14 ：Calculate $\mu$ and $\sigma$ respectively, we get $\mu = np = 35$ and $\sigma = \sqrt{np(1-p)} = \sqrt{35 \times (1-0.7)} = 3.24037034920393$.

Step 15 ：Substitute $\mu$, $\sigma$, and $X$ into the formula, we get $P(X) = \frac{1}{\sqrt{2\pi \times 3.24037034920393^2}}e^{-\frac{(39-35)^2}{2 \times 3.24037034920393^2}} = 0.1137$.

Step 16 ：Round to four decimal places as needed, we get $P(X) = \boxed{0.1137}$.

Step 17 ：Compare the result with the exact probability, we find that the normal distribution overestimates the exact probability.