Question

A 60-watt light bulb advertises that it will last 1500 hours. The lifetimes of these light bulbs is approximately normally distributed with a mean of 1550 hours and a standard deviation of 57 hours. What proportion of these light bulbs will last less than the advertised time

Answer 1

The proportion of these light bulbs that will last less than the advertised time is 18.94% or 0.1894

Explanation:

The first thing to do here is to calculate the z-score

Mathematically;

z-score = (x - mean)/SD

= (1500-1550)/57 = -50/57 = -0.88

So the proportion we will need to find is;

P( z < -0.88)

We shall use the standard score table for this and our answer from the table is 0.1894 which is same as 18.94%