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.) hours

Answer 1

Answer:

33.68 hours

Step-by-step explanation:

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour then

The lifetime of a battery package consisted of 4 batteries is normally distributed with mean value 60(15×4) hours and standard deviation 16 (1 × $4^{2}$) hour

If the total lifetime of all batteries in a package exceeds lifetime value for only 5% of all packages then P(z<Z)=0.05 where

• Z is the corresponding z-score for the lifetime value that exceeds 5% of the all packages' lifetime value.

From z-table we can conclude that Z=-1.645

Z is also calculated as:

Z=-1.645=$\frac{X-M}{s}$ where

• X is the cut-off time for exceeding 5%

• M is the mean lifetime of packages (60)

• s is the standard deviation of package lifetime values (16)

Solving the equation for X we get X=33.68