Question

cell phone lifetimes 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 36 employees with a cell phone, find the probability that the mean lifetime of these phones will be less than 23.7 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 z - value calculations to two decimal places.

$p(\overline{x} < 23.7) = \square$

Explanation:

Step1: Identify parameters

Population mean $\mu = 24.3$, standard deviation $\sigma = 2.6$, sample size $n = 36$, sample mean $\bar{x} = 23.7$.

Step2: Calculate standard error

Standard error $SE = \frac{\sigma}{\sqrt{n}} = \frac{2.6}{\sqrt{36}} = \frac{2.6}{6} \approx 0.4333$.

Step3: Compute z-score

$z = \frac{\bar{x} - \mu}{SE} = \frac{23.7 - 24.3}{0.4333} \approx -1.38$ (rounded to two decimals).

Step4: Find probability from z-table

$P(Z < -1.38) \approx 0.0838$.

Answer:

0.0838