A manufacturing company produces steelbolts to be used on a certain truck. The lengths of the bolts have a mean length of 6 inches and a standard deviation of 0.1 inch. Samples of size 50 are examined at random, and if the average length is outside the interval 5.98 to 6.02 inches, the entire production for the day is examined. Find the probability that a day’s production will have to be examined.

Question
Asked Feb 27, 2019
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A manufacturing company produces steel
bolts to be used on a certain truck. The lengths of the bolts have a mean length of 6 inches and a standard deviation of 0.1 inch. Samples of size 50 are examined at random, and if the average length is outside the interval 5.98 to 6.02 inches, the entire production for the day is examined. Find the probability that a day’s production will have to be examined.

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Expert Answer

Step 1

Introduction:

A random sample of size n = 50 steel bolts from a manufacturing company is selected.

Let X denotes the length of a randomly selected steel bolt.

Denote the population mean, that is, the mean lengths of all steel bolts in manufacturing company as μ and the standard deviation in lengths of all steel bolts in manufacturing company as σ.

Here, μ =6 inches and σ = 0.1 inch.

Denote the sample mean lengths of 50 randomly selected steel bolts as X-bar.

It is given that, the entire production of the day has to be examined if the average length of steel bolts is outside the interval 5.98 to 6.02 inches.

The probability that the entire day’s production has to be examined is given below:

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Step 2

Sampling distribution of sample mean:

The sample mean of a random variable X has a sampling distribution that has population mean or expected value same as that of X and population standard deviation, called the standard error, which is the population standard deviation of X divided by the square root of the sample size n. The parameters are as follows:

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Step 3

Central limit theorem:

If X1, X2, …, Xn are independent and identically distributed (iid) random variables with finite mean, E(Xi) = μ and finite variance V(Xi) = σ2, then Z will converge to ...

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