Question

A population of values has a normal distribution with j = 11.8 and o = 70.8. A random sample of sizen = 170 is drawn.a. Find the probability that a single randomly selected value is between - 1.8 and 28.6. Round your answer tofour decimal places,PC - 1.8 < X < 28.6) =b. Find the probability that a sample of size n = 170 is randomly selected with a mean between -1.8 and28.6. Round your answer to four decimal places.PC - 1.8 < M < 28.6)

Answer 1

Solution

(a) mean = 11.8

standard deviation = 70.8

P(-1.8 < x < 28.6)

$\begin{gathered} z=\frac{x-\mu}{\sigma}=\frac{-1.8-11.8}{70.8} \\ z=-\frac{13.6}{70.8} \\ z=-0.192 \end{gathered}$$\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{28.6-11.8}{70.8}=\frac{16.8}{70.8}=0.237 \end{gathered}$

therefore the z score for p(-1.8 [tex]\begin{gathered} Pr(-0.192Therefore probability that a single randomly selected value between - 1.8 and 28.6 = 0.1699 (4d.p)