The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.

Question

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches. A sample of four metal sheets is randomly selected from a batch. What is the probability that the average length of a sheet is between 30.25 and 30.35 inches long

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Clara 2 hours 2021-09-13T15:33:42+00:00 1 Answer 0

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    2021-09-13T15:35:33+00:00

    Answer:

    Probability that the average length of a sheet is between 30.25 and 30.35 inches long is 0.0214 .

    Step-by-step explanation:

    We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard deviation of 0.2 inches.

    Also, a sample of four metal sheets is randomly selected from a batch.


    Let X bar = Average length of a sheet

    The z score probability distribution for average length is given by;

                    Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

    where, \mu = population mean = 30.05 inches

               \sigma   = standard deviation = 0.2 inches

                 n = sample of sheets = 4


    So, Probability that average length of a sheet is between 30.25 and 30.35 inches long is given by = P(30.25 inches < X bar < 30.35 inches)

    P(30.25 inches < X bar < 30.35 inches)  = P(X bar < 30.35) – P(X bar <= 30.25)

    P(X bar < 30.35) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{30.35-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z < 3) = 0.99865

     P(X bar <= 30.25) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{30.25-30.05}{\frac{0.2}{\sqrt{4} } } ) = P(Z <= 2) = 0.97725

    Therefore, P(30.25 inches < X bar < 30.35 inches)  = 0.99865 – 0.97725

                                                                                           = 0.0214

                                           

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