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Homework 14 52. a. Brad King Suppose the sediment density (g/cm^3) 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, pp. 1169-1174) 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? Begin by stating what the random variable is exactly. Xi = sediment density (g/cm^3) of a specimen from a certain region. Where, Xi ~ N(µ = 2.65,σ = .85) If the Xi 's are independent random variable and every Xi has the same distribution then the random variables, X1, X2, X3, … , Xn form a random sample of size n. It is also stated, on page 231, in Chapter 5 that if a set of random sample normally distributed, then for any number of n samples, the average of the samples is also normally distributed. So, to find the probability of the average sediment density of sample size of 25 specimens, we need to calculate average expected value and the average standard deviation. Refer to Section 5.4 on page 230. It has been proposed that, Therefore, σ2 E ( X ) = µ X = µ And V ( X ) = n E ( X ) = 2.65 g/cm^3 and σ X = X ~ N ( µ X = 2.65,σ X = .170) σ .85 = = .170 g/cm^3 n 25 Homework 14 Brad King Now, determine the requested probability of. P ( X ≤ 3.00) = P (Z ≤ 3.00 − 2.65 ) = P ( Z ≤ 2.058) = Φ (2.056) = .9803 .170 The second part of part a. asks… P (2.65 ≤ X ≤ 3.00) = P ( X ≤ 3.00) − P ( X ≤ 2.65) Since we already know the probability of the average sediment density at most 3.00, solve for the second half of the probability equation and standardize it… P ( X ≤ 2.65) = P ( Z ≤ therefore, 2.65 − 2.65 ) = P ( Z ≤ 0) = Φ (0) = .50 .170 P ( X ≤ 3.00) − P ( X ≤ 2.65) = .98 − .50 = .48 Homework 14 b. Brad King How large a sample size would be required to ensure that the first probability in part (a) is at least .99? Since we already are given σ, and have calculated the average of the standard deviation, all we have to do is stick the average of the population standard deviation into the standardized equation for the probability and solve for n. First, discover the Z value for the probability of 99%. Φ (2.33) = .99 µ X = µ = 2.65 σX = σ n P ( X ≤ 3.00) = P (Z ≤ Z = 2.33 = 6.657 = n 3.00 − 2.65 ) = P ( Z ≤ 2.33) = Φ (2.33) = .99 .85 n 3.00 − 2.65 .35 = .85 .85 n n n = 32.02 .85 To assure the average sample density is less than 3.00 g/cm^3 99% of the time, 33 specimens would insure this probability. 2.5 2 1.5 f(x) Sample Sample Mean 1 0.5 0 0 1 2 3 4 5 sediment density grams/cubic meter Figure 1 Shows the pdf’s of X1 and X2 6 7