Question

he health center of a small, private university requires students to have appointments for visits. Each appointment takes an average of

5.7

minutes with a standard deviation of

3.7

minutes. The health center makes

100

appointments each day and is scheduled to be open for

10

hours. What is the probability that the total time spent for

100

students who have appointments for tomorrow will exceed

10

hours? (Hint:

10

hours

=600

minutes.)
Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Answer 1

Here, μ = 5.7*100 = 570,
σ = 3.7*sqrt(100) = 37 and x = 600.

We need to compute P(X >= 600). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (600 - 570)/37 = 0.81

Therefore,
P(X >= 600) = P(z <= (600 - 570)/37)
= P(z >= 0.81)
= 1 - 0.791 = 0.209