Question

A recent study of the lifetimes of cell phones found the average is 23.7 months. The standard deviation is 3 months. If a company provides its 13 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 24.7 months. Assume cell phone life is a normally distributed variable.

Answer 1

0.6923

1) Gathering the data we have

Average: 23.7

Standard Deviation: 3

P < 24.7

2) We need to find out the z-score for P < 24.7. So we can write out the following. Plugging the X value for 24.7, the mean is 23.7, and the Standard Deviation: 3

`Z=(x-\mu)/(\sigma)\Rightarrow Z=(24.7-23.7)/(3)\Rightarrow Z=0.33`

Now, with the Z-score table, we can locate the corresponding value:

`\begin{gathered} P(X<24.7)=P(Z<0.33)=0.6923 \\ \end{gathered}`

3) Hence, the probability that the lifetime of these phones will be less than 24.7 months is: 0.6923