Q:

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There are four 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.) hours

Accepted Solution

A:
Answer:If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.Step-by-step explanation:We are given the following information in the question:Mean, μ = 15Standard Deviation, σ = 1Sample size = 4Total lifetime of 4 batteries = 40 hoursWe are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.Formula:[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]Standard error due to sampling:[tex]\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5[/tex]We have to find the value of x such that the probability is 0.05P(X > x)  = 0.05[tex]P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05[/tex]  [tex]= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05 [/tex]  [tex]=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95 [/tex]  Calculation the value from standard normal z table, we have,  [tex]\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83[/tex]  Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.