Question 3 of 18 Step 1 of 1 $11: 27: 40$ The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shrimp is randomly selected, what is the probability that the sample mean would be less than 40.6 dollars? Round your answer to four decimal places. Answer How to enter your answer (opens in new window) 5 Points Tables Keypad Keybourd Shortcuts Prev Next

Step 1 ：First, we need to calculate the standard error, which is the standard deviation divided by the square root of the sample size. In this case, the standard deviation is 7 and the sample size is 51. So, the standard error is \( \frac{7}{\sqrt{51}} \approx 0.9802 \).

Step 2 ：Next, we need to calculate the z-score, which is \( \frac{value - mean}{standard \, error} \). The value is 40.6, the mean is 40, and the standard error is 0.9802. So, the z-score is \( \frac{40.6 - 40}{0.9802} \approx 0.6121 \).

Step 3 ：Finally, we need to use a z-table to find the probability that the sample mean would be less than 40.6 dollars. The z-score is 0.6121, so the probability is approximately 0.7298.

Step 4 ：\(\boxed{0.7298}\) is the final answer, which is the probability that the sample mean would be less than 40.6 dollars.