Question

A certain brand of flood lamps has a lifetime that is normally distributed with a mean of 4,000 hours and

a standard deviation of 310 hours.

1. What proportion of these lamps will last between 3,690 and 4,465 hours?               

b. What lifetime should the manufacturer advertise for these lamps in order that only 4% of the lamps will burn out before the advertised lifetime?

Answer 1

Solution :

Given that ,

mean = = 4000

standard deviation = = 310

a) P(3690 < x < 4465) = P[(3690 - 4000)/ 310) < (x - ) /  < (4465 - 4000) / 310) ]

= P(-1 < z < 1.5)

= P(z < 1.5) - P(z < -1)

= 0.9332 - 0.1587

0.7745

Proportion = 0.7745

b) Using standard normal table ,

P(Z < z) = 4%

P(Z < z) = 0.04

P(Z < -1.75) = 0.04

z = -1.75

Using z-score formula,

x = z * +

x = -1.75 * 310 + 4000 = 3457.50

4% of the lamps will burn out before the advertised lifetime is 3457.50