Question

The life of light bulbs is distributed normally. The variance of the lifetime is 400400 and the mean lifetime of a bulb is 600600 hours. Find the probability of a bulb lasting for at most 633633 hours. Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 600

2= 400

standard deviation = = 20

P(X< 633) = P[(X- ) / < (633-600) /20 ]

= P(z <1.65 )

Using z table

= 0.9505