Answer: 0.8490
Step-by-step explanation:
Given : The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.0 inches and a standard deviation of 0.9 inches.
i.e. [tex]\mu=30[/tex] and [tex]\sigma=0.9[/tex]
Let x denotes the lengths of aluminum-coated steel sheets.
Required Formula : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For n= 36 , the probability that the average length of a sheet is between 29.82 and 30.27 inches long will be :-
[tex]P(29.82<x<30.27)=P(\dfrac{29.82-30}{\dfrac{0.9}{\sqrt{36}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{30.27-30}{\dfrac{0.9}{\sqrt{36}}})\\\\=P(-1.2<z<1.8)\\\\=P(z<1.8)-P(z<-1.2)\\\\=P(z<1.8)-(1-P(z<1.2))\ \ {[\because\ P(Z<-z)=1-P(Z<z)]}\\\\= 0.9641-(1-0.8849)[\text{By using z-table.}]\\\\=0.8490[/tex]
∴ Required probability = 0.8490
