Question

A variable is normally distributed with mean 15 and standard deviation 4.

a. Find the percentage of all possible values of the variable that lie between 8 and 19.

b. Find the percentage of all possible values of the variable that are at least 12.

c. Find the percentage of all possible values of the variable that are at most 13.

Answer 1

Solution :

Given that ,

mean = = 15

standard deviation = = 4

a.

P(8 < x < 19) = P[(8 - 15)/ 4) < (x - ) /  < (19 - 15) / 4) ]

= P(-1.75 < z < 1)

= P(z < 1) - P(z < -1.75)

= 0.8413 - 0.0401

= 0.8012

percentage = 80.12%

b.

P(x 12) = 1 - P(x   12)

= 1 - P[(x - ) / (12 - 15) / 4]

= 1 -  P(z -0.75)   

= 1 - 0.2266

= 0.7734

percentage = 77.34%

c.

P(x 13)

= P[(x - ) / (13 - 15) / 4]

= P(z -0.5)

= 0.3085

percentage = 30.85%