A variable is normally distributed with a mean of 16 and a s

A variable is normally distributed with a mean of 16 and a standard deviation of 3.

A. Determing the quartiles of the variable. (Round to two decimal places)

B. Obtain and interpret the 80th percentile. (Round to two decimal places)

C. Find the value that 65% of all values of the variable exceed. (Round to two decimal places)

D. Find the two values that divide the are under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025 (Round to two decimal places)

A. Determing the quartiles of the variable. (Round to two decimal places)

quartile1: P(X<x)=0.25

--> P((X-mean)/s <(x-16)/3) =0.25

--> P(Z<(x-16)/3) =0.25

--> (x-16)/3= -0.67 (from standard normal table)

So x= 16-0.67*3 =13.99

quartile3: P(X<x)=0.75

--> P((X-mean)/s <(x-16)/3) =0.75

--> P(Z<(x-16)/3) =0.75

--> (x-16)/3= 0.67 (from standard normal table)

So x= 16+0.67*3 =18.01

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B. Obtain and interpret the 80th percentile. (Round to two decimal places)

P(X<x)=0.8

--> P((X-mean)/s <(x-16)/3) =0.8

--> P(Z<(x-16)/3) =0.8

--> (x-16)/3= 0.84 (from standard normal table)

So x= 16+0.84*3 =18.52

the 80th percentile of the data is 18.52

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C. Find the value that 65% of all values of the variable exceed. (Round to two decimal places)

P(X>x)=0.65

--> P(Z<(x-16)/3) =1-0.65=0.35

--> (x-16)/3= -0.39 (from standard normal table)

So x= 16-0.39*3=14.83

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D. Find the two values that divide the are under the corresponding normal curve into a middle area of 0.95 and two outside areas of 0.025 (Round to two decimal places)

P(X<x) =0.025

--> P(Z<(x-16)/3) =0.025

--> (x-16)/3 =-1.96 (from standard normal table)

So x= 16-1.96*3 =10.12

P(X<x) =0.975

--> P(Z<(x-16)/3) =0.975

--> (x-16)/3 =1.96 (from standard normal table)

So x= 16+1.96*3 =21.88