Answer: 0.8490
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.
and
![\sigma=0.9](https://img.qammunity.org/2020/formulas/mathematics/college/fgis5weby611amgurtj46bhxcg6j9xqg2c.png)
Let x denotes the lengths of aluminum-coated steel sheets.
Required Formula :
![z=(x-\mu)/((\sigma)/(√(n)))](https://img.qammunity.org/2020/formulas/mathematics/college/kv4zbzwta1cei225xptycu57ns4dmxgoss.png)
For n= 36 , the probability that the average length of a sheet is between 29.82 and 30.27 inches long will be :-
![P(29.82<x<30.27)=P((29.82-30)/((0.9)/(√(36)))<(x-\mu)/((\sigma)/(√(n)))<(30.27-30)/((0.9)/(√(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](https://img.qammunity.org/2020/formulas/mathematics/college/wpxliljl2beb5r089oxpgoscblrvhgw36q.png)
∴ Required probability = 0.8490