The lifetime of a certain type of battery is normally distributed with mean value 10 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?

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent certain lifetime of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(10,1)[/tex]

Where [tex]\mu=10[/tex] and [tex]\sigma=1[/tex]

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X>a)=0.05[/tex] (a)

[tex]P(X<a)=0.95[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.95[/tex]

[tex]P(z<\frac{a-\mu}{\sigma})=0.95[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=1.64<\frac{a-10}{1}[/tex]

And if we solve for a we got;

[tex]10 +1.64*1=11.64[/tex]

So the value of height that separates the bottom 95% of data from the top 5% is 11.64