How would I solve for probability.

Question

The lifetime of lightbulbs that are advertised to last for 3500 hours are normally distributed with a mean of 3700 hours and a standard deviation of 150 hours. What is the probability that a bulb lasts longer than the advertised figure?

Answer 1

We accept as given that the population mean is

mu = 3700 hr

and the population standard deviation is

sigma = 150 hr

We want the fraction of the population that should last

x = 3500 hours or longer.

The normal distribution variable z is given by

z = (x - mu) / sigma

z = (3500 - 3700) / 150 = -1.33

This is 1.33 standard deviations below the mean.

From tables of the normal probability distribution, we find

that 0.091 = 9.1% fall below this z = -1.33 level,

so 1 - 0.091 = 0.909 = 90.9%

are expected to exceed this value of 3500 hours.