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 .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?

Answer 1

A) 0.9803; 0.4803
B) 32

We calculate the z-score for this problem by using the formula:

$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$

Using our formula, we have:

$z=\frac{3.00-2.65}{\frac{0.85}{\sqrt{25}}}=2.06$

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.9803.

To find the probability it is between the mean and this, we subtract the probability associated with the mean (0.5) from this:
0.9803 - 0.5 = 0.4803.

To find B, we first find the z-score for this.  Using a z-table (http://www.z-table.com) we see that the closest z-score would be 2.33.  We then set up our equation as

$2.33=\frac{3.00-2.65}{\frac{0.85}{\sqrt{n}}}=\frac{0.35}{\frac{0.85}{\sqrt{n}}} \\ \\2.33=0.35\div \frac{0.85}{\sqrt{n}}=0.35\times \frac{\sqrt{n}}{0.85}$

Multiplying both sides by 0.85 we have
2.33(0.85) = 0.35√n
1.9805 = 0.35√n

Divide both sides by 0.35:
1.9805/0.35 = √n

Square both sides:
(1.9805/0.35)² = n
32 ≈ n