A variable is normally distributed with mean 16 and standard deviation 2. Find t

Question

A variable is normally distributed with mean 16 and standard deviation 2. Find the percentage of all possible values of the variable that lie between 11 and 20. Find the percentage of all possible values of the variable that exceed 14 Find the percentage of all possible values of the variable that are less than 15. The percentage of all possible values of the variable that lie between 11 and 20 is 1.02 % (Round to two decimal places as needed.) The percentage of all possible values of the variable that exceed 14 is % (Round to two decimal places as needed.) The percentage of all possible values of the variable that are less than 15 is % (Round to two decimal places as needed.)

Explanation / Answer

Answer:

a).

z value for 11, z =(11-16)/2 =-2.5

z value for 20, z =(20-16)/2 =2

P( 11<x<20) = P( -2.5<z<2)=P( z <2)-P( z <-2.5)

=0.9772-0.0062

=0.9710

The required percentage = 97.10%

b).

z value for 14, z =(14-16)/2 =-1

P( x >14) = P( z >-1) =0.8413

The required percentage = 84.13%

c).

z value for 15, z =(15-16)/2 =-0.5

P( x <15) = P( z <-0.5) =0.3085

The required percentage = 30.85%

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