Question

A variable is normally distributed with mean 16 and and standard deviation 2. a. Find the percentage of all possible values of the variable that lie between 11 and 17. b. exceed 15. c less than 12.

Answer 1

Given,

= 16 , = 2

We convert this to standard normal as

P( X < x) = P( Z < x - / )

a)

P(11 < X < 17) = P (X < 17) - P( X < 11)

= P( Z < 17 - 16 / 2) - P( Z < 11 - 16 / 2)

= P( Z < 0.5) - P( Z < -2.5)

= P( Z < 0.5) - ( 1 - P (Z < 2.5 ) )

= 0.6915 - ( 1 - 0.9938 ) (Probability calculated from Z table)

= 0.6853

= 68.53%

b)

P( X > 15) = P( Z > 15 - 16 / 2)

= P( Z > -0.5)

= P( Z < 0.5)

= 0.6915 (Probability calculated from Z table)

= 69.15%

c)

P( X < 12) = P( Z < 12 - 16 / 2)

= P( Z < -2)

= 1 - P( Z < 2)

= 1 - 0.9772 (Probability calculated from Z table)

= 0.0228

= 2.28%