Question

Statistics questions

In a Doctor’s office, the average wait time before a patient sees the doctor is normally distributed with mean 20minutes. Because of some actions taken at the office, the standard deviation has now reduced by 1 minute so, the standard deviation is now 2

7. Approximately 95% of the patients will now see a doctor between _______ and _________minutes

8. How unusual it is now to wait to be seen by the doctor for 30mins or more? _______% of patients wait beyond 30mins to be seen by a doctor.

9. 5% of the patients will now wait beyond______minutes

10. 10% of the (lucky!) patients will now be seen faster than how ________minutes

11. _______% of the patients wait more than 20mins to be seen by the doctor

Answer 1

(7)

= 20

= 2

= 0.05

From Table, critical values of Z = 1.96

95% Confidence Interval:

20 (1.96 X 2)

= 20 3.92

= (16.08, 23.92)

So,

Answer is:

Approximately 95% of the patients will now see a doctor between 16.08 and 23.92 minutes

(8)

X = 30

Z = (30 - 20)/2

= 5

By Technology, Cumulative Area Under Standard Normal Curve = 0.9999997

So,

P(X>30) = 1 - 0.9999997 = 0.0000003 = 0.00003 %

So,

Answer is:

0.00003 % of patients wait beyond 30mins to be seen by a doctor.

(9)

= 0.05

From Table, critical value of Z = 1.645

So,

5% of patients will wait beyond 20 + 1.645 X 2 = 20 + 3.29 = 23.29

Answer is:

5% of the patients will now wait beyond 23.29 minutes

(10)

= 0.10

From Table, critical value of Z = 1.28

So,

10% of patients will wait less than 20 - 1.28 X 2 = 20 - 2.56 = 17.44

Answer is:

10% of the (lucky!) patients will now be seen faster than 17.44 minutes

(11)

50 % of the patients wait more than 20mins to be seen by the doctor

because 20 minutes coincides with mean which divides the area into 2 equal halves.