2 Suppose the sediment density gcm of a randomly selected sp

2. 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 .85 (suggested in Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants, Water Research. 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.65 and 3.00? b. How large a sample size would be required to ensure that the first probability in part (a) is at least .99?

Solution

(a) P(xbar<3) = P((xbar-mean)/(s/vn) <(3-2.65)/(0.85/sqrt(25)))

=P(Z<2.06) = 0.9803 (from standard normal tabe)

P(2.65<xbar<3) = P((2.65-2.65)/(0.85/sqrt(25))<Z<(3-2.65)/(0.85/sqrt(25)))

=P(0<Z<2.06) =0.4803 (from standard normal tabe)

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(b) P(xbar<3) = P((xbar-mean)/(s/vn) <(3-2.65)/(0.85/sqrt(n))) =0.99

--> P(Z<(3-2.65)/(0.85/sqrt(n))) = 0.99

--> (3-2.65)/(0.85/sqrt(n))= 2.33 (from standard normal table)

--> 0.35 = 2.33*(0.85/sqrt(n))

--> 0.35 = 1.9805/sqrt(n)

--> sqrt(n)= 1.9805/ 0.35 = 5.658571

So n= 5.658571^2 =32.01943

Take n=33