Question

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.61 and standard deviation 0.82. (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 2.61 and 3.00

Answer 1

Answer: The probability that the sample average sediment density is at most 3.00 = 0.9913

The probability that the sample average sediment density is between 2.61 and 3.00 = 0.4913

Explanation:

Given : Mean : $\mu=2.61$

Standard deviation : $\sigma =0.82$

Sample size : $n=25$

The value of z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

a) For x= 3.00

$z=\dfrac{3.00-2.61}{\dfrac{0.82}{\sqrt{25}}}=2.38$

The p-value : $P(z\leq2.38)=0.9913$

b) For x= 2.61

$z=\dfrac{2.61-2.61}{\dfrac{0.82}{\sqrt{25}}}=0$

The p-value : $P(0<z\leq2.38)=P(2.38)-P(0)=0.9913-0.5=0.4913$