Answer to Question #343259 in Statistics and Probability for mandy

Question

For a certain type of fluorescent light in a large building, the cost per bulb of replacing bulbs all at once is much less than if they are replaced individually as they burn out. It is known that the lifetime of these bulbs is normally distributed, and that 60% last longer than 2500 hours, while 30% last longer than 3000 hours.

a) What are the approximate mean and standard deviation of the lifetimes of the bulbs?

 

If the light bulbs are completely replaced when more than 20% have burned out, what is the time between complete replacements

Answer 1

a)

$$P(Z>\dfrac{2500-\mu}{\sigma})=0.6$$

$$\dfrac{2500-\mu}{\sigma}=-0.2533$$

$$\mu=2500+0.2533\sigma$$

$$P(Z>\dfrac{3000-\mu}{\sigma})=0.3$$

$$\dfrac{3000-\mu}{\sigma}=0.5244$$

$$\mu=3000-0.5244\sigma$$

$$2500+0.2533\sigma=3000-0.5244\sigma$$

$$\sigma=642.921435$$

$$\mu=3000-0.5244(642.921435)=2662.852$$

$$\mu\approx2663.85, \sigma\approx642.92$$

b)

$$P(X>x)=1-P(Z\le\dfrac{x-2662.852}{642.921435})=0.2$$

$$\dfrac{x-2662.852}{642.921435}=0.8416$$

$$x=2662.852+642.921435(0.8416)$$

$$x=3204$$