Question

A recent study of the lifetimes of cell phones found the average is 24.3 months. The standard deviation is 2.6 months. If a company provides its 33 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.8 months. Assume cell phone life is a normally distributed variable, the sample is taken from a large population and the correction factor can be ignored. Round the final answer to at least four decimal places and intermediate -value calculations to two decimal places.

Answer 1

Answer:

$P(\bar X$

And we can find this probability using the normal standard distribution or excel and we got:

$P(z$

Step-by-step explanation:

Let X the random variable that represent the lifetimes of cell phones, and for this case we know the distribution for X is given by:

$X \sim N(24.3,2.6)$  

Where $\mu=24.3$ and $\sigma=2.6$

We select a sample size of n = 33 emplloyees and we are interested on this probability

$P(\bar X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

The standard deviation would be:

$\sigma_{\bar X} = \frac{2.6}{\sqrt{33}}= 0.4526$

If we apply this formula to our probability we got this:

$P(\bar X$

And we can find this probability using the normal standard distribution or excel and we got:

$P(z$