Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.68 and standard deviation 0.81. (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 2.68 and 3.00? (round your answers to four decimal places.) at most 3.00 between 2.68 and 3.00 (b) how large a sample size would be required to ensure that the probability in part (a) is at least 0.99? (round your answer up to the nearest whole number.) specimens

a) 0.9761; 0.4761; b) 35

Step-by-step explanation:

We use the z score formula

Our mean, μ, is 2.68 and our standard deviation, σ, is 0.81.  The sample size, n, is 25.

We want to find P(X ≤ 3.00):

z = (3.00-2.68)/(0.81÷√25) = 0.32/(0.81÷5) = 0.32/0.162 = 1.98

Using a z table, we see that the area under the curve to the left of this is 0.9761.

We now want to find P(2.68 ≤ X ≤ 3.00):

z = (2.68-2.68)/0.162 = 0/0.162 = 0

z = (3.00-2.68)/0.162 = 0.32/0.162 = 1.98

The area under the curve to the right of z = 0.00 is 0.5.  The area under the curve to the right of z = 1.98 is 0.9761.  This makes the area between them

0.9761-0.5000 = 0.4761.

For part b,

We use the same information, except we are now finding the sample size.  We want the probability to be 0.99.  Looking in the z table, we see that the closest number to this is 0.9901, which corresponds to a z score of 2.33:

2.33 = (3.00-2.68)/(0.81÷√n)

2.33 = 0.32/(0.81÷√n)

2.33 = 0.32×(√n/0.81)

2.33 = 0.32√n/0.81

Multiply both sides by 0.81:

0.81(2.33) = (0.32√n/0.81)(0.81)

1.8873 = 0.32√n

Divide both sides by 0.32:

1.8873/0.32 = 0.32√n/0.32

5.8978 = √x

Square both sides:

(5.8978)² = (√x)²

34.78 = x

This rounds to 35.

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70 years.

step-by-step explanation: