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.
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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.