Question

The lifetime of a certain type of battery is normally distributed with mean value 12 hours and standard deviation 1 hour. there are nine batteries in a package. what lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages? (round your answer to two decimal places.)

Answer 1

The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by


`P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \ (x-\mu)/((\sigma)/(√(n))) \right)`

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


`P(X\ \textgreater \ x)=0.05 \\ \\ \Rightarrow1-P\left(z\ \textless \ (x-12)/((1)/(√(9))) \right)=0.05 \\ \\ \Rightarrow P\left(z\ \textless \ (x-12)/((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= (1.645)/(3) =0.5483 \\ \\ \Rightarrow x=0.5483+12\approx12.55`