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)

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

the answer is 375 miles

step-by-step explanation: