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The lifetime of a certain type of battery is normally distributed with mean value 13 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 :

Answer:

13.55 hours

Step-by-step explanation:

given that the  lifetime of a certain type of battery is normally distributed with mean value 13 hours and standard deviation 1 hour.

Sample size=n =9

Sample mean = [tex]\bar X is N(13, 1/3)[/tex]

Let c be the  lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages

Then we have [tex]P(X>c) =0.05[/tex]

Convert into Z score to get

[tex]P(\frac{x-13}{1/3} >z) = 0.05\\[/tex]

z =1.645 from std normal distribution table

Hence x = [tex]13+\frac{1.645}{3} \\=13+0.548\\=13.548[/tex]

Answer is 13.55 hours

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