# The weights of the fish in a certain lake are normally distributed with a mean of 11 lb and a standard deviation of 6. if 4 fish are randomly selected, what is the probability that the mean weight will be between 8.6 and 14.6 lb? your answer should be a decimal rounded to the fourth decimal place.

Since the sample standard deviation is now known, we use the z-score test. The formula is given as: z= (X – μ) / (s / sqrt(n)) Where, X = sample mean = 8.6 lb to 14.6 lb μ = population mean = 11 lb s = population standard deviation = 6 n = sample size = 4 1st: Calculating for z when x = 8.6 lb z = (8.6 – 11) / (6 / sqrt4) z = - 0.8 Using the standard distribution table for z: Probability (x = 8.6 lb) = 0.2119 2nd: Calculating for z when x = 14.6 lb z = (14.6 – 11) / (6 / sqrt4) z = 1.2 Using the standard distribution table for z: Probability (x = 14.6 lb) = 0.8849 Therefore the probability that the mean weight will be between 8.6 and 14.6 lb: Probability (8.6 ≤ x ≤ 14.6 ) = 0.8849 - 0.2119 Probability (8.6 ≤ x ≤ 14.6 ) = 0.673 (ANSWER)