Z=standard score
x=observed value
\mu=mean of the sample
\sigma=standard deviation of the sample
[tex]\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}[/tex][tex]\begin{gathered} Pr(0\leq z)\text{ + Pr(0 }\leq z\leq2.5) \\ 0.5\text{ + 0.4938} \\ 0.9938 \end{gathered}[/tex]Hence the probability that a battery will last at most 300 days = 0.9938
