Q:

Furnace repair bills are normally distributed with a mean of 267 dollars and a standard deviation of 20 dollars. If 64 of these repair bills are randomly selected, find the probability that they have a mean cost between 267 dollars and 269 dollars.

Accepted Solution

A:
Answer: 0.7881446Step-by-step explanation:Given : Mean : [tex]\mu = 267\text{ dollars} [/tex]Standard deviation : [tex]\sigma =20 \text{ dollars}[/tex]Sample size : [tex]n=64[/tex]The formula to calculate the z-score :-[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]For x=  267 dollars [tex]z=\dfrac{267-267}{\dfrac{20}{\sqrt{64}}}=0[/tex]For x= 269 dollars.[tex]z=\dfrac{269-267}{\dfrac{20}{\sqrt{64}}}=0.80[/tex]The P-value : [tex]P(0<z<0.8)=P(z<0.8)-P(z<0)[/tex][tex]= 0.7881446-0.5= 0.2881446\approx 0.7881446[/tex]Hence, the probability that they have a mean cost between 267 dollars and 269 dollars.= 0.7881446