Question

The average annual out-of-pocket costs for an implant for Americans last year is estimated to be $2,488. Assume that this cost follows the normal distribution with a standard deviation of $250. What is the probability that a randomly selected American who has dental insurance will spend between $2,400 and $2,900 for an implant

Answer 1

Answer:

58.73%

Step-by-step explanation:

What we must do is calculate the z value for each value and thus find what percentage each represents and the subtraction would be the percentage between those two values.

We have that z is equal to:

z = (x - m) / (sd)

x is the value to evaluate, m the mean, sd the standard deviation

So for 2400 we have:

z = (2400 - 2488) / (250)

z = -0.35

and this value represents 0.3632

for 89 we have:

z = (2900 - 2488) / (250)

z = 1.65

and this value represents 0.9505

we subtract:

0.9505 - 0.3632 = 0.5873

Therefore the probability would be 58.73%