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A large tank of fish from a hatchery is being delivered to a lake. The hatchery claims the mean of fish in the tank is 15 inches, and the standard deviation is 2 inches. A random sample of 35 fish is taken from the tank. Let x bar be the mean sample length of these fish. What is the probability that x bar is within 0.5 inch of the claimed population mean? (Around your answer to four decimal places.)

User Nambrot
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Final answer:

The probability that x bar is within 0.5 inch of the claimed population mean is approximately 0.8584.

Step-by-step explanation:

To find the probability that x bar is within 0.5 inch of the claimed population mean, we can use the standard normal distribution and the Central Limit Theorem.

The Central Limit Theorem states that for a large enough sample size, the distribution of sample means will be approximately normally distributed, regardless of the shape of the population distribution.

In this case, we have a sample size of 35, which is considered large enough. So, we can assume that the sample means follow a normal distribution.

First, we need to find the standard error of the mean. The standard error of the mean is calculated by dividing the standard deviation of the population by the square root of the sample size.

Standard error of the mean = standard deviation / sqrt(sample size)

= 2 / sqrt(35) ≈ 0.339

Next, we need to calculate the z-score corresponding to 0.5 inch above and below the claimed population mean.

The z-score formula is z = (x - μ) / σ, where x is the value of interest, μ is the population mean, and σ is the standard deviation.

For the upper limit, the z-score is z = (15 + 0.5 - 15) / 0.339 ≈ 1.472

For the lower limit, the z-score is z = (15 - 0.5 - 15) / 0.339 ≈ -1.472

We can use the standard normal distribution table or a calculator to find the probabilities corresponding to these z-scores. The probability that x bar is within 0.5 inch of the claimed population mean is the sum of the probabilities for the upper and lower limits.

Probability = P(z < 1.472) - P(z < -1.472)

Using a standard normal distribution table or a calculator, we find that P(z < 1.472) ≈ 0.9292 and P(z < -1.472) ≈ 0.0708.

So, the probability that x bar is within 0.5 inch of the claimed population mean is approximately:

=> 0.9292 - 0.0708 = 0.8584.

User Manuel De Leon
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