Question

A TV manufacturer knows that their TV's have a normally distributed lifespan with a mean of 3.6 years and standard deviation of 1.2 years if a TV is picked at random 8% percent of the time its life will be less than how many years give your answer to one decimal place

Answer 1

Solution :

Given that,  

mean = = 3.6

standard deviation = = 1.2

Using standard normal table ,

P(Z < z) = 8%

P(Z < z) = 0.08

P(Z < -1.41) = 0.08

z = -1.41

Using z-score formula,

x = z * +

x = -1.41 * 1.2 + 3.6 = 1.9

random 8% percent of the time its life will be less than 1.9 years .