Question

the weights of the fish in a certain lake are normally distributed with a mean of 20 lb and astandard deviation of 9. if 9 fish are randomly selected, what is the probability that the meanweight will be between 18 and 23 lb?

Answer 1

The probability that the mean weight of 9 randomly selected fish will be between 18 lb and 23 lb is found by converting the weights into z-scores and looking up the corresponding probabilities in the standard normal distribution.

Step-by-step explanation:

The question asks for the probability that the mean weight of 9 randomly selected fish from a lake with normally distributed weights will be between 18 lb and 23 lb, given that the mean weight is 20 lb and the standard deviation is 9 lb. To answer this, we use the concept of the sampling distribution of the sample mean. Since the weights are normally distributed, the sampling distribution of the sample mean for n = 9 fish will also be normal with mean (μ) equal to the population mean and standard deviation (σ/√n) equal to the population standard deviation divided by the square root of the sample size. In this case, we have μ = 20 lb and σ/√n = 9 lb/√9 = 3 lb.

To find the probability, we convert the weights to z-scores using the formula: z = (X - μ) / (σ/√n). For X = 18 lb, z = (18 lb - 20 lb) / 3 lb = -0.67. For X = 23 lb, z = (23 lb - 20 lb) / 3 lb = 1.00. We then look up these z-scores in a standard normal distribution table or use a calculator to find the probabilities that correspond to these z-scores. The final probability is the difference between these two probabilities.