Question

General Electric manufactures a decorative Crystal Clear 60-watt light bulb. Suppose that the lifetimes of the light bulbs are approximately normally distributed, with a mean of 1550 hours and a standard deviation of 57 hours. What would be the cutoff value for a light bulb that lasts in the top 10% of all bulbs

Answer 1

The lifetime of bulb should be 1623.074 hours or more to lie in the top 10%.

Explanation:

We are given the following information in the question:

Mean, μ = 1550 hours

Standard Deviation, σ = 57 hours

We are given that the distribution of lifetimes of the light bulbs is a bell shaped distribution that is a normal distribution.

Formula:

`z_(score) = \displaystyle(x-\mu)/(\sigma)`

We have to find the value of x such that the probability is 0.10

`P( X > x) = P( z > \displaystyle(x - 1550)/(57))=0.10`

`= 1 -P( z < \displaystyle(x - 1550)/(57))=0.10`

`P( z < \displaystyle(x - 1550)/(57))=0.9`

Calculation the value from standard normal z table, we have,

`\displaystyle(x - 1550)/(57) = 1.282\\\\x = 1623.074`

Thus, the lifetime of bulb should be 1623.074 hours or more to lie in the top 10%.