Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standar

Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? eetween 2.65 and 3.00?

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Maria 1 month 2021-09-15T20:10:05+00:00 1 Answer 0

Answers ( )

  1. Charlotte
    0
    2021-09-15T20:11:54+00:00

    Answer:

    Probability that the sample average is at most 3.00 = 0.98030

    Probability that the sample average is between 2.65 and 3.00 = 0.4803

    Step-by-step explanation:

    We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

    Also, a random sample of 25 specimens is selected.

    Let X bar = Sample average sediment density

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

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

    where, \mu = population mean = 2.65

               \sigma  = standard deviation = 0.85

                n = sample size = 25

    (a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

        P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

    (b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) – P(X bar <= 2.65)

    P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

     P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

    Therefore, P(2.65 < X bar < 3)  = 0.98030 – 0.5 = 0.4803 .

                                                                                 

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