For a population with µ = 40 and σ = 8, what is the z-score corresponding to X = 34?

A scuba diving company currently charges $100per dive. On average, there are 30 customers per day. The company performed a study

Semenov [28]

A. Total Revenue (R) is equal to price per dive (P) multiplied by number of customers (C). We can write $R=PC$ .

Per price increase is $20. So four price increase is $ $20*4=$80$ . Hence, price per dive is 100+80=$180.

Also per price increase, 2 customers are reduced from 30. For 4 price increases, $4*2=8$ customers are reduced. Hence, total customers is $30-8=22$ .

So Total Revenue is:

$R=180*22=3960$

B. Each price increase is 20. So x price increase is 20x. Hence, new price per dive would be equal to the sum of 100 and 20x.

Also per price increase, customers decrease by 2. So per x price increases, the customer decrease is 2x. Hence, new number of customers is the difference of 30 and 2x.

Therefor we can write the quadratic equation for total revenue as the new price times the new number of customers.

$R=(100+2x)(30-2x)=-40x^{2}+400x+3000$

C. We are looking for the point (x) at which the equation modeled in part (B) gives a maximum value of revenue (y). That x value is given as $x=-\frac{b}{2a}$ , where a is the coefficient of $x^{2}$ and b is the coefficient of x. So we have,

$x=-\frac{b}{2a}=- \frac{400}{(2)(-40)}=5$

That means, the greatest revenue is achieved after 5 price increases. Each price increase was 20, so 5 price increase would be $5*20=100$ . So the price that gives the greatest revenue is $100+100=200$ .