Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with a mean of 2.65 and a standard deviation of .85.

a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is more than 2.35?,

b. at most 3.0 and

c. between 2.35 and 3.00

d. How large a sample would be required to insure that the probability in part a is at least 0.99

Answer 1

ANSWER:

Given that:

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with a mean of 2.65 and a standard deviation of .85.

here std error of mean =std deviaiton/(n)1/2 =0.17

a)

If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is more than 2.35

probability that the sample average sediment density is more than 2.35 =P(X>2.35)=P(Z>(2.35-2.65)/0.17)

=P(Z>-1.7647)=0.9612

b)

at most 3.0

P(X<3.0)=P(Z<(3.0-2.65)/0.17)=P(Z<2.0588)=0.9802

c)

between 2.35 and 3.00

P(2.35<X<3.0)=P(-1.7647<Z<2.0588)=0.9802-0.0388=0.9414

d)

How large a sample would be required to insure that the probability in part a is at least 0.99

std deviaiton =0.85

for 99 percentile z =2.326

therefore std error =(X-mean)/z =(2.35-2.65)/2.326=-0.129

therefore sample size =(0.85/0.129)2 =~44