Thread: Standard Deviation?

the life expectancy of a particular brand of hair dryer is normally distributed with a mean of four years and a standard deviation of eight months. this Equates to (60-48)/8 = 1.5 and .9332 then1- .9332=.0668

The company has a three year warranty period on its hair dryers. what percentage of its hair dryers will be in operating condition after the warranty period? (36-48)/8 = - 1.5 and .0068 then1- ..0068=.9332

what is the minimum and the maximum life expectancy of the 90% of the hair dryers? on this problem I am lost.

Originally Posted by hrunup

the life expectancy of a particular brand of hair dryer is normally distributed with a mean of four years and a standard deviation of eight months. this Equates to (60-48)/8 = 1.5 and .9332 then1- .9332=.0668

The company has a three year warranty period on its hair dryers. what percentage of its hair dryers will be in operating condition after the warranty period? (36-48)/8 = - 1.5 and .0068 then1- ..0068=.9332

what is the minimum and the maximum life expectancy of the 90% of the hair dryers? on this problem I am lost.

Hi,
Make X be the first variable (the life expectancy that is normally distributed).
The thing is to make X be Z, where Z is a standard normally distributed variable. And we know that Z=(X-48)/8.
Then use the z-table like you did in the first question (don't laugh, I didn't understand what you wrote in a first time lol)

In the second question, you're asked to find z such that P(Z>z)=90%, then use the definition of the standard deviation.