The lifetime of a certain type of battery is normal distributed with mean value

Question

The lifetime of a certain type of battery is normal distributed with mean value of 12 hours and standard deviation 1.5 hours. There are 6 batteries in a package. What total lifetime value of all batteries in a package exceeds the value for 10% of all values The lifetime of a certain type of battery is normal distributed with mean value of 12 hours and standard deviation 1.5 hours. There are 6 batteries in a package. What total lifetime value of all batteries in a package exceeds the value for 10% of all values

Explanation / Answer

Given a normal distribution

Mean =12

Standard Deviation= 1.5

n=6

We are asked to calculate P(X> 10%(6*12))

P(X>7.2)

We have to convert it into Z values

We know Z=(X-mu)/sigma

Here X=7.2

Mu=12

Sigma=1.5

Z=(7.2-12)/1.5=-3.2

P(Z>-3.2) has to be found

P(Z>-3.2)=1-P(Z<-3.2)=1-0.0007=0.9993 ( We take the value from the Z table)

or 99.93%

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