The average annual out-of-pocket costs for an implant for Americans last year is estimated to be $2,488. Assume that this cost f
ollows 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
1 answer:
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%
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