I need help with this problem pls help!
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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 × ) 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= 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