Question

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.)

Answer 1

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 = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5$

We have to find the value of x such that the probability is 0.05

P(X > x)  = 0.05

$P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05$  

$= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05$  

$=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83$  

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.