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The amount of time that you have to wait before seeing the doctor in the doctors office is normally distributed with a mean
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Answer #1

The amount of waiting before seeing a doctor is normally distributed with mean = 15.2 minutes and standard deviation = 15.2 minutes

Since we have to find probabilty of average wait time greater than 20 minutes, we will find z value for this sample mean and then corresponidng p value, since this p value will show probability of seeing doctor in less than or equal to 20 minutes, so we will substract this value from 1 and then our required value will appear.

So sample size = n = 64

Sample mean = X = 20 minutes

Population mean = μ = 15.2 minutes

Population standard deviation = σ = 15.2 minutes

z= \frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}

z= \frac{20-15.2}{\frac{15.2}{\sqrt{64}}}

z= 2.526

p value for z = 2.526 is 0.9942, which means probabilty for average waiting time less than or equal to 20 minute is 0.9942

So for probabilty of average waiting time to be greater than 20 minute is

=1-0.9942 = 0.0058

I have also shown in z table , corresponding p value,

Number in the table represents PIZsz) 02 Z 0.01 0.04 0.05 0.06 0.07 0.08 0.09 0.00 .5000 .5040 .5160 .5199 5319 .5359 0.0 0.1

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