The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by
[tex]P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \ \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)[/tex]
Given that the lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour and that the probability that nine
batteries in a package exceeds a certain value, x, is 5% or 0.05. Then the value x is given by
[tex]P(X\ \textgreater \ x)=0.05 \\ \\ \Rightarrow1-P\left(z\ \textless \ \frac{x-12}{\frac{1}{\sqrt{9}}} \right)=0.05 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-12}{\frac{1}{3}} \right)=1-0.05=0.95 \\ \\ \Rightarrow P(z\ \textless \ 3(x-12))=P(z\ \textless \ 1.645) \\ \\ \Rightarrow3(x-12)=1.645 \\ \\ \Rightarrow x-12= \frac{1.645}{3} =0.5483 \\ \\ \Rightarrow x=0.5483+12\approx12.55[/tex]
