Question

The life of batteries is normally distributed with a mean of 200 days and a standard deviation of 40 days. What is the probability that a battery will last at most 300 days?

Answer 1

$z\text{ = }\frac{x\text{ - }\mu}{\sigma}$

Z=standard score

x=observed value

\mu=mean of the sample

\sigma=standard deviation of the sample

$\begin{gathered} \sigma\text{ = 40 days} \\ \mu\text{ = 200 days} \\ x\text{ = 300 days} \\ z\text{ = }\frac{x\text{ - }\mu}{\sigma} \\ z\text{ = }\frac{300\text{ - 200}}{40} \\ z\text{ = }\frac{100}{40} \\ z\text{ = 2.5} \end{gathered}$$\begin{gathered} Pr(0\leq z)\text{ + Pr(0 }\leq z\leq2.5) \\ 0.5\text{ + 0.4938} \\ 0.9938 \end{gathered}$

Hence the probability that a battery will last at most 300 days = 0.9938