# 2 companies, A and B, drill wells in a rural area. Company A charges a flat fee of RM3500 to drill a well regardless of its depth. Company B charges RM1000 plus RM12 per foot to drill a well. ?

## The depths of wells in this area have a normal distribution with a mean of 250 feet and a standard deviation of 40 feet. i) What is the probability that Company B would charge more than Company A to drill a well? ii) Find the mean amount charged by Company B to drill a well.

Nov 10, 2017

i) "P"(X > 208.33)=0.8508, or 85.08%.
ii) Company B charges an average amount of RM4000.

#### Explanation:

Part i) First, find out the well depth needed for Company B to charge more than Company A.

This is determined by finding the depth where their costs are equal:

$\text{A's cost " = " B's cost}$
$\text{ "3500 = 1000 + 12x" }$ (where $x$ is number of feet)
$\text{ } 2500 = 12 x$

$x = \frac{2500}{12} = \text{RM} 208.33$

Since the price for Company A remains fixed, then for any depth exceeding 208.33 feet, Company B charges more than Company A.

Then, find the probability that a well depth exceeds this threshold.

What is the probability that a well depth is greater than 208.33 feet?

Let $X$ be the depth of a random well. Then $X \text{ ~ Normal} \left(\mu = 250 , {\sigma}^{2} = {40}^{2}\right) .$

We seek $\text{P} \left(X > 208.33\right)$. This can be translated into a probability for the standard normal variable $Z$ using the formula $z = \frac{x - \mu}{\sigma} .$

z=(x-mu)/sigma = (208.33-250)/40=–1.04175

Thus,

"P"(X>208.33)="P"(Z>–1.04175)
color(white)("P"(X>208.33))=1-"P"(Z<=–1.04175)

By table lookup, "P"(Z <= –1.04) = 0.1492, so we get

$\text{P} \left(X > 208.33\right) = 1 - 0.1492$
$\textcolor{w h i t e}{\text{P} \left(X > 208.33\right)} = 0.8508$.

Part ii) If well depth has an average value of 250 feet, then the mean amount Company B will charge to drill a well is

$1000 + 12 x = 1000 + 12 \left(250\right)$
$\textcolor{w h i t e}{1000 + 12 x} = 1000 + 3000$
$\textcolor{w h i t e}{1000 + 12 x} = \text{RM} 4000.$