Question

1. A company offers its employees dental care insurance. A recent study shows that the annual cost per employee had a normal distribution, with an average of $ 1280 and a standard deviation of $ 420 annually.

1. What is the probability that a randomly chosen employee spends between $ 1,500 and $ 2,000 per year on dental expenses?
2. What is the minimum cost to be in the top 10% for annual dental care?
3. 8,000 employees are chosen at random. Approximately how much do you spend more than 1800 USD?
4. If 7 employees are chosen at random, what is the probability that between 3 (not inclusive) and 6 (inclusive) of them will spend between 1800 and 2000 USD?

Answer 1

the annual cost per employee had a normal distribution

a) the probability that a randomly chosen employee spends between $ 1,500 and $ 2,000 per year on dental expenses =

b) Let the minimum cost be A

critical Z value corresponding to probability 0.90 is 1.2816

c) proportion of the employee that spend more than $1800 is

of 8000 employee chosen at random, you can expect 0.0765*8000=612employee to spend more than $1800

d) Probability that employee spend between 1800 and 2000 USD is

If 7 employees are chosen at random, the probability that between 3 (not inclusive) and 6 (inclusive) of them will spend between 1800 and 2000 USD is binomial.

With probability of success p = 0.0481

number of trials n= 7

the probability that between 3 (not inclusive) and 6 (inclusive) of them will spend between 1800 and 2000 USD =0.0002