Question

. Consider two brands of batteries for calculators, Xmax and Y-cell, manufactured independently by two companies. Assume that the lifetimes of each battery has a Normal distribution. The Xmax batteries has a mean of 1000 hours with a standard deviation of 100 hours, and Y-cell batteries has a mean of 1200 hours with a standard deviation of 200 hours. For each of the following questions, you must state the random variable you are using and the distribution assumption you make.

(a) Assume, two batteries are randomly selected, one Xmax and one Y-cell. What can you say about the difference D in their lifetimes? It turns out that D is also Normally distributed. Find the expected value and variance of the random variable D.

(b) You notice that, on average, Y-cell batteries will last longer than Xmax batteries. But what is the exact probability that the Y-cell batteries will last longer than Xmax batteries?

Answer 1

(a)

By theorem:

If X is Normally distributed with mean $\mu _{X}$ and standard deviation $\sigma _{X}$ and If Y is Normally distributed with mean $\mu _{Y}$ and standard deviation $\sigma _{Y}$ , then the difference D = X - Y is Normally distributed with mean
Art – X
and standard deviation
Vox toy
.

Given:

$\mu _{X}$ = 1000

$\sigma _{X}$ = 100

$\mu _{Y}$ = 1200

$\sigma _{Y}$ = 200

Substituting, we get:

the difference D = X - Y is Normally distributed with mean
ux - My = 1000 - 1200 = -200
and standard deviation =
Voof = V1002 + 2002 = V10000 + 40000 = 223,6068

So,

Expected Value of D = .

ux - My = 1000 - 1200 = -200

Variance of D =

of + 0 = 1002 + 2002 = 10000 + 40000 = 50000

(b)

$\mu$ = - 200

$\sigma$ = 223.6068

To find P(D<0):

Z = (0 -(-200))/223.6068

= 0.8944

By Technology, Cumulative Area Under Standard ormal Curve = 0.8144

So,

Answer is:

0.8144