Score: $32.5 / 100 \quad 7 / 20$ answered
Question 1
A manufacturer produces a commodity where the length of the commodity has approximately normal distribution with a mean of 5.5 inches and standard deviation of 1.7 inches. If a sample of 31 items are chosen at random, what is the probability the sample's mean length is greater than 4.8 inches? Round answer to four decimal places.
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Step 1 ：The problem is asking for the probability that the sample mean length is greater than 4.8 inches. This is a problem of normal distribution. We know that the sample mean of a normal distribution also follows a normal distribution. The mean of this distribution is the same as the population mean, and the standard deviation is the population standard deviation divided by the square root of the sample size.

Step 2 ：We can use the z-score formula to find the z-score of 4.8. The z-score is calculated as \(z = \frac{x - \mu}{\sigma/\sqrt{n}}\), where \(x\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size.

Step 3 ：Substituting the given values into the formula, we get \(z = \frac{4.8 - 5.5}{1.7/\sqrt{31}} = -2.2926088552829507\).

Step 4 ：We then use the standard normal distribution table or a function to find the probability that the z-score is greater than the calculated value. The probability is 0.9891.

Step 5 ：Final Answer: The probability that the sample's mean length is greater than 4.8 inches is \(\boxed{0.9891}\).