The lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.

a. What is the probability that a battery will last more than 19 hours?

b. Find the 10th percentile of the lifetimes.

c. A particular battery lasts 14.5 hours. What percentile is its lifetime on?

d. What is the probability that the lifetime of a battery is between 14.5 and 17 hours?

Step 1 of 4:

Let X denotes the lifetime of a battery.

Then the mean 16 hours and standard deviation is 2 hours.

Here hours and = 2 hours.

So XN()

XN().

Our goal is:

a). We need to find the probability that a battery will last more than 19 hours.

b). We need to find the 10th percentile of the lifetimes.

c). We need to find the percentile is its lifetime on.

d). We need to find the probability that the lifetime of a battery is between 14.5 and 17 hours.

a).

Now we have to find the probability that a battery will last more than 19 hours.

Here X is 19 hours.

Then we have to calculate .

The formula of the z is

z=

We know that mean 16 and standard deviation 2.

Then,

= 1-

Using area under the normal curve table.

= 0.9332.

= 1- 0.9332

= 0.0668

Therefore the probability that a battery will last more than 19 hours is 0.0668.

Step 2 of 4:

b).

Now we have to find the 10th percentile of the lifetimes.

The z score of the 10th percentile it is corresponding to the -1.28 on the chart book.

Then the formula of the z is

z=

We know that z value is -1.28.

-1.28=

=(-1.28)(2)

X= 16-2.56

X= 13.44

Therefore the 10th percentile of the lifetimes is 13.44.