Download 52. Suppose the sediment density (g/cm^3) of a randomly selected

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
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