so the question is:

the weights of fish in a certain lake are normally distributed with a mean of 10, a standard deviation of 6. if 4 fish are selected what is the probability that the mean weight will be between 7.6 and 13.6?

so usually to find prob, we use the normcdf function on the calculator, so it would be normcdf(7.6,13.6,10,6) but that doesn't work because you are selected 4 fish. that percent would be .38% for all the fish.

the correct answer is 67.30 percent but how do you get that?

38% * not .38%

nevermind, i found a similar solution. :)

To find the probability that the mean weight of a sample of 4 fish will be between 7.6 and 13.6, you need to use the concept of the sampling distribution.

The standard deviation of the sampling distribution can be calculated using the formula:
standard deviation of the sampling distribution = standard deviation of the population / square root of the sample size

In this case, the standard deviation of the population is 6, and the sample size is 4. Therefore, the standard deviation of the sampling distribution is 6 / √4 = 6 / 2 = 3.

Once you have the standard deviation of the sampling distribution, you can calculate the z-scores for the lower and upper weights of 7.6 and 13.6. The z-score is calculated using the formula: (x - mean) / standard deviation.

For the lower weight of 7.6:
z-score = (7.6 - 10) / 3 = -0.8

For the upper weight of 13.6:
z-score = (13.6 - 10) / 3 = 1.2

Now you can use a standard normal distribution table or a calculator to find the percentage of observations within these z-scores. From the standard normal distribution table, the percentage for a z-score of -0.8 is 0.2119, and the percentage for a z-score of 1.2 is 0.8849.

To find the probability between the two z-scores, subtract the lower percentage from the upper percentage:
0.8849 - 0.2119 = 0.673

Therefore, the probability that the mean weight of the sample of 4 fish will be between 7.6 and 13.6 is 0.673, or 67.3%.