A college has 230 full-time employees that are currently covered under the schoo

Question

A college has 230 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 $1,850 with a standard deviation of $505. The college is performing an audit of its health care plan and has randomly selected 35 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 $1,785?

c. What is the probability that the sample mean will be more than $1,810?

d. What is the probability that the sample mean will be between $1,865 and $1,915?

Explanation / Answer

Popuation Size N = 230

M = 1850

s = 505

Sample size n = 35

.

Part a)

SE = standard deviation / sqrt(sample size)

SE = 505 / 35

SE = 14.42

.

Part b)

P(x < 1785) = P(z < (1785-1850 / 14.42)

[formula of Z is apllied here, z = (x - M) / SE]

P(X < 1890) = P(z < -4.50)

.

Now we refer to the Z table to find the probability value corresponding to z = -2.88

P = 0.000003.

Part c)

P(x > 1810) = 1 - P(X < 1810)

[this is done because the Z table provides us the left tailed area ]

.

P(x < 1810) = P(z < 1810-1850 /14.42) = P(z < -2.77)

P(x < 1920) = 0.002803

.

Part d)

P(1865 < x < 1915) = P(x < 1915) - P(x < 1865)

P(x < 1915) = P(z < 1915-1850/14.42) = P(z < 4.50) = 0.9997

P(x < 1865) = P(Z < 1865-1850/14.42) = P(Z < 1.40) = 0.9192

P(1865 < x < 1915) = 0.9997 - 0.9192 = 0.0805

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