The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds with a standard deviation of 0.8 pound. Assuming the weights of catfish are normally distributed, the probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is .

The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is 0.596

Step-by-step explanation:

Since the weights of catfish are assumed to be normally distributed,

we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = weights of catfish.

µ = mean weight

σ = standard deviation

From the information given,

µ = 3.2 pounds

σ = 0.8 pound

The probability that a randomly selected catfish will weigh between 3 and 5.4 pounds is is expressed as

P(x ≤ 3 ≤ 5.4)

For x = 3

z = (3 - 3.2)/0.8 = - 0.25

Looking at the normal distribution table, the probability corresponding to the z score is 0.401

For x = 5.4

z = (5.4 - 3.2)/0.8 = 2.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.997

Therefore,.

P(x ≤ 3 ≤ 5.4) = 0.997 - 0.401 = 0.596

perimeter of the base times height

step-by-step explanation: