Part 3 of 3 Points: 0.67 of 1 Save In a recent year, the mean number of strokes per hole for a famous golfer was approximately 3.9. (a) Find the variance and standard deviation using the fact that the variance of a Poisson distribution is $\sigma^{2}=\mu$. (b) How likely is this golfer to play an 18 -hole round and have more than 72 strokes? (a) The variance is 3.9 . (Round to the nearest tenth as needed.) The standard deviation is 2 strokes. (Round to the nearest tenth as needed.) (b) The probability of more than 72 strokes in an 18 -hole round is $\square$. (Round to the nearest thousandth as needed.)

Step 1 ：The variance of a Poisson distribution is equal to its mean, so the variance is \( \boxed{3.9} \).

Step 2 ：The standard deviation is the square root of the variance, so the standard deviation is \( \boxed{2.0} \).

Step 3 ：The mean number of strokes for an 18-hole round is 18 * 3.9 = 70.2.

Step 4 ：The probability of having more than 72 strokes is the same as 1 minus the probability of having 72 or fewer strokes.

Step 5 ：We can use the cumulative distribution function (CDF) of the Poisson distribution to find the probability of having a certain number of strokes or fewer.

Step 6 ：The probability of the golfer having more than 72 strokes in an 18-hole round, rounded to the nearest thousandth, is \( \boxed{0.385} \).