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

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

Step-by-step 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%

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