Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.66 and standard deviation 0.87. (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 2.66 and 3.00? (round your answers to four decimal places.) at most 3.00 between 2.66 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.)

a) 0.9744; 0.4744; b) 36

Step-by-step explanation:

The formula for a z score of a mean of a sample is

For part a,

Our mean, μ, is 2.66 and our standard deviation, σ, is 0.87.  Our sample size, n, is 25.  We want P(X ≤ 3.00):

z = (3.00-2.66)/(0.87÷√25)

z = 0.34/(0.87÷5)

z = 0.34/0.174 = 1.95

Using a z chart, we see that the area to the left of this value is 0.9744.

For the second question,

We want P(2.66 ≤ X ≤ 3.00):

z = (2.66-2.66)/(0.87÷√25) = 0/0.174 = 0

z = (3.00-2.66)/(0.87÷√25) = 0.34/0.174 = 1.95

The area under the curve to the left of z = 0 is 0.5000; the area under the curve to the left of z = 1.95 is 0.9744.  This means the area between them is

0.9744-0.5000 = 0.4744.

For part b,

We want our probability to be at least 0.99.  Since we have the area under the curve to the left in a z table, we will subtract from 1:

1-0.99 = 0.01

In the z table, we see the cell with the value closest to 0.01 is 0.0099; this corresponds with a z score of -2.33:

Multiply both sides by 0.87:

0.87(-2.33) = ((0.34√n)/0.87)

-2.0271 = 0.34√n

Divide both sides by 0.34:

-2.0271/0.34 = 0.34√n/0.34

-5.9621 = √n

Square both sides:

(-5.9621)² = (√n)²

35.55 = n

This rounds to 36.

associative property hopes this ! : )

step-by-step explanation:

answer: get a conversion sheet out i dont know the exact measurments

step-by-step explanation: divide the number of feet with the nu,ber of milrs and there your answer