Question

A college has 260 ​full-time employees that are currently covered under the​ school's health care plan. The average​ out-of-pocket cost for the employees on the plan is 1,820.20

with a standard deviation of ​$510. The college is performing an audit of its health care plan and has randomly selected 30 employees to analyze their​ out-of-pocket costs.

a. Calculate the standard error of the mean.

b. What is the probability that the sample mean will be less than 1,755.00

c. What is the probability that the sample mean will be more than​$1,800.00?

d. What is the probability that the sample mean will be between $1840 and $1890​?

Answer 1

Let X be the out 0f pocket cost for nan employee

X~ Normal ( 1820.20, 510)

A random sample of 30 employees is selected.

Let be the sample mean of the random sample of 30 employees out of pocket cost

~ Normal ( 1820.20,   )

a) P( < 1755.00) = P( < )

= P( z < -0.70)

= 1- P( z < 0.70)

= 1- 0.75804

=0.24196

b) P( > 1800.00) = P( > )

= P( z >   -0.21)

= P( z < 0.21)

= 0.58317

c) P( 1840.00 < < 1890.00) = P( < < < )

= P( 0.21 < z < 0.75)

= P( z < 0.75) - P(z < 0.21)

= 0.95994 - 0.58317

= 0.37677