Answered: The patient director of Health plus…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

The patient director of Health plus clinic has just conducted a study of customer waiting time before seeing a doctor. For the study, he randomly observed 1045 patients throughout one-week. Let assume that the waiting time has an approximately normal distribution with a mean time of 22.14 and a standard deviation of 6.09. Based on this assumption, the patient director is considering providing a gift certificate to any customer required to wait more than 30 minutes.

a. What is the probability of having to pay off on this offer?

Suppose the patient director believes this procedure is too generous. He would like to set the time limit so that the likelihood of giving out the gift is at most only 5%.

b. What will be the required waiting time for any customer to receive a gift certificate?

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1 A.

Let X be the Costumer waiting time before seeing a doctor.

We assume that that waiting time has an normal distribution with a mean time of 22.14 and Standard Deviation of 6.09

$\Rightarrow μ = 22.14$ and $σ = 6.09$

According the question ,

Director has considered providing a gift certificate to all costumers who have waited for 30 minutes.

We have to calculate Probablity to pay off i.e P(X>30)

$P (X > 30) = P [\frac{X - μ}{σ} > \frac{(30 - 22.14)}{(6.09)}] = P (z > 1.29) = 1 - P (z \leq 1.29) = 1 - 0.9015 = 0.0985$

(Using Z-Table)

Hence P(X>30) = 0.0985

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