Question

Suppose LCD screens have lifetimes that are normally distributed with a mean lifetime of 18000 hours with a variance of 1000000 hours. (a) Find the probability a screen lifetime is under 15000 hours. (b) Find the probability a lifetime is between 15000 and 20000 hours. (c) The top 10% of lifetimes will be at least how long?

Please show all work for review

Answer 1

Solution :

Given that ,

mean = = 18000

variance = 2 = 1000000

standard deviation = = 2 = 1000000 = 1000

a) P(x < 15000)

= P[(x - ) / < (15000 - 18000) / 1000 ]

= P(z < -3.00)

Using z table,

= 0.0013

b) P(15000 < x < 20000) = P[(15000 - 18000)/ 1000) < (x - ) /  < (20000 - 18000) / 1000 ) ]

= P(-3.00 < z < 2.00)

= P(z < 2.00 ) - P(z < -3.00 )

Using z table,

= 0.9772 - 0.0013

= 0.9759

c) Using standard normal table,

P(Z > z) = 10%

= 1 - P(Z < z) = 0.10  

= P(Z < z) = 1 - 0.10

= P(Z < z ) = 0.90

= P(Z < 1.28 ) = 0.90  

z = 1.28

Using z-score formula,

x = z * +

x = 1.28 * 1000 + 18000

x = 19280 hours