A variable x is normally distributed with a mean of 500 and

A variable x is normally distributed with a mean of 500 and a standard deviation of 20. Find
a. the 70th percentile.
b. the 30th percentile.
c. the x value which exceeds 90% of all x values.

Solution

a. the 70th percentile.

P(X<x)=0.7

--> P((X-mean)/s <(x-500)/20) = 0.7

--> P(Z<(x-500)/20) =0.7

--> (x-500)/20= 0.52 (from standard normal table)

So x= 500+0.52*20 =510.4

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b. the 30th percentile.

P(X<x)=0.3

--> P(Z<(x-500)/20) = 0.3

--> (x-500)/20 = -0.52 (from standard normal table)

So x= 500+0.52*20=510.4

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c. the x value which exceeds 90% of all x values.

P(X<x) =0.9

--> P(Z<(x-500)/20) = 0.9

--> (x-500)/20= 1.28(from standard normal table)

So x= 500 +1.28*20 =525.6