Respuesta :
Answer:
Probability that the sample average is at most 3.00 = 0.98030
Probability that the sample average is between 2.65 and 3.00 = 0.4803
Step-by-step explanation:
We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.
Also, a random sample of 25 specimens is selected.
Let X bar = Sample average sediment density
The z score probability distribution for sample average is given by;
Z = [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 2.65
[tex]\sigma[/tex] = standard deviation = 0.85
n = sample size = 25
(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)
P(X bar <= 3) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }[/tex] ) = P(Z <= 2.06) = 0.98030
(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)
P(X bar < 3) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{3-2.65}{\frac{0.85}{\sqrt{25} } }[/tex] ) = P(Z < 2.06) = 0.98030
P(X bar <= 2.65) = P( [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] <= [tex]\frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } }[/tex] ) = P(Z <= 0) = 0.5
Therefore, P(2.65 < X bar < 3) = 0.98030 - 0.5 = 0.4803 .
