A college has 250 full-time employees that are currently covered under the schoo

Question

A college has 250 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
with a standard deviation of $525.
The college is performing an audit of its health care plan and has randomly selected 35
employees to analyze theirout-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?
c. What is the probability that the sample mean will be more than $1,795?
d. What is the probability that the sample mean will be between $1,835
and $1,900?

Explanation / Answer

(a) The Standard error of the mean = /n = 525/35 = 88.7412

µ = 1820.

To find the probability, we need to find the Z scores first.

Z = (X - µ)/ SE

(b) For P( X < 1755)

Z = (1755 – 1820)/88.7412 = -0.73

The required probability from the normal distribution tables is = 0.2319

(c) For P (X > 1795) = 1 - P (X < 1795), as the normal tables give us the left tailed probability only.

For P( X < 1795)

Z = (1795 – 1820)/88.7412 = -0.26

The probability for P(X < 1795) from the normal distribution tables is = 0.3891

Therefore the required probability = 1 – 0.3891 = 0.6109

(d) For P (1835 < X < 1900) = P(X < 1900) – P(X < 1835)

For P( X < 1900)

Z = (1900 – 1820)/88.7412 = 0.9

The probability for P(X < 1900) from the normal distribution tables is = 0.8163

For P( X < 1835)

Z = (1835 – 1820)/88.7412 = 0.17

The probability for P(X < 1835) from the normal distribution tables is = 0.5671

Therefore the required probability is 0.8163 – 0.5671 = 0.2492

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