A large tank of fish from a hatchery is being delivered to a lake. The hatchery claims that the mean length of fish in the tank is 15 inches, and the standard deviation is 4 inches. A random sample of 31 fish is taken from the tank. Let x be the mean sample length of these fish. What is the probability that x is within 0.5 inch of the claimed population mean? (Round your answer to four decimal places.)

Step-by-step explanation:

Let us assume that the length of fishes in the tank is normally distributed. Thus, x is the random variable representing the length of fishes in the tank. Since the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/(σ/√n)

Where

x = sample mean

µ = population mean

σ = standard deviation

number of samples

From the information given,

µ = 15

σ = 4

n = 31

We want to find the probability that x is between (15 - 0.5) inches and (15 + 0.5) inches. The probability is expressed as

P(14.5 ≤ x ≤ 15.5)

For x = 14.5

z = (14.5 - 15)/(4/√31) = - 0.7

Looking at the normal distribution table, the probability corresponding to the z score is 0.2420

For x = 15.5

z = (15.5 - 15)/(4/√31) = 0.7

Looking at the normal distribution table, the probability corresponding to the z score is 0.7580

Therefore,

P(14.5 ≤ x ≤ 15.5) = 0.7580 - 0.2420 = 0.5160

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