Standard Normal Distribution - the mean profit

Question

The life of a certain type of car battery is known to be normally distributed with mean $30$ months and standard deviation $6$ months. The batteries cost the manufacturer $\$40$ each to make, and he sells them for $\$60$ each.

If he refunds $\$30$ for any battery which lasts less than $24$ months, what is the mean profit he will make per battery?

Answer 1

First you need to calculate the Z-Score:

$$Z_i=\frac{x_i-\mu}{\sigma}$$

where: $$\mu = \text{mean} $$ $$\sigma = \text{standard deviation}$$

since in your case $$x_i=24$$

we can calculate that

$$Z= -1$$

if you check the standard distribution tables (https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf) for the value of Z=-1.00, the area to the left is .15866 which over one and rounding is approximately 0.159.

When everything goes OK, the profit is 20 --> (60-40). When something goes KO, the profit is -10 --> (60-40-30)

The probability of getting a -10 is the area to the left calculated before, that means 0.159. The probability of getting a +20 if the difference (1-0.159=0.841)

Now you have everything you need to calculate the mean profit.

Hope I didn't make any mistake and it helps you, I am a bit rusty with maths atm to be honest.

Answer 2

$\textrm{Hints}$:

$1$. Profit for a battery which lasts more than 24 months is $\$60-\$40=\$20$

$2$. (Negative) profit for a battery which lasts less than 24 months is $\$60-\$40-\$30=-\$10$

$3$. The proportion of battteries which last less than 24 months is $P(X<24)=\Phi\left(\frac{24-30}{6} \right)$, where $\Phi(z)$ is the cdf of the standard normal distribution

$4$. The proportion of battteries which last more than 24 months is $P(X>24)=1-P(X<24)$

$5$. The expected profit ($P$) is $\mathbb E(P)=\mathbb E(P|X<24)\cdot P(X<24)+\mathbb E(P|X>24)\cdot P(X>24) \ $ $( \textrm{Law of total expectation}$)