Question

A large medical clinic is working on improving waiting times for patients. While some delays are inevitable, for example due to doctors having to deal with urgent cases, having patients wait more than 15 minutes past their appointment time is considered excessive. Through patient surveys the clinic management has learnt that only 25% of the patients were able to see their doctor without waiting more than 15 minutes; 5% said they had to wait over 60 minutes. The clinic management assumes that waiting times roughly follow a Normal distribution.
(a) Calculate the mean and standard deviation for the waiting times at this clinic. For full marks, include all your working out and relevant MINITAB output that supports your answer.
(b) One way the clinic management could solve this problem would be to allow for waiting times longer than 15 minutes. What maximum waiting time should the clinic be aiming for if it is to be achieved with 60% of its patients? For full marks, include an appropriate probability statement and show MINITAB output that supports your answer.
(c) Wishing to maintain the ‘at most 15-minute wait’ claim, the clinic hopes that revising its scheduling procedures will allow them to improve the success rate to 60%. If the standard deviation stays the same, what new mean patient waiting time does the clinic need to achieve? For full marks, include an appropriate probability statement, your working out and MINITAB output that supports your answer.

Answer 1

Let us suppose that the waiting time of the patient to see the doctor is denoted with the variable X and is follows normal distribution with mean waiting time M and its standard deviation

From the given problem we observe that probability of waiting time less than 15 minutes is 0.25 and probability of waiting time more than 60 minutes is 0.05

i.e P(X < 15) =0.25 and P(X> 60) =0.05

by standard normal variate P(X<15) =P(Z,< -z1) =0.25 where z1= 15-M / ---------- (1)

   P(X>60) = P(Z>z2) = 0.05 and P(Z< z2 ) = 0.45 where z2 = 60 - M / ------ (2)

from normal tables z1 = 0.65 and z2 = 1.645 after substituting these values in equations 1 and 2

we will get M= 27.76 and = 19.60

a) Hence average waiting time M= 27.76 and its standard deviation = =19.60 minutes

b) maximum waiting time maintain doctor to meet 60 % of the patients is denoted with x

then P (X <x ) > 0.60

it is also same as P(Z < z ) > 0.60 or 0.5 + P( 0<Z < z ) =0.60 where z= x-M / and x= maximum waiting time of the patients

P(0<Z<z ) = 0.10

From normal tables the value of z= 0.255

x- M / = 0.255

x- 27.76 / 19.60 = 0.255

x= 32.76

Hence the maximum waiting time maintine to meet 60% of the patients = 32.76 minutes

c) the mean waiting time to maitine atmost 15 minutes waiting time to meet the success rate of 60 % if the standard deviation is same

P( X, <15 ) > 0.60

P(Z < z ) > 0.60 wher z= 15 - M / 19.60 M = new mean

0.5+ P(0<Z<z) .> 0.60

P(0<Z<z) = 0.10

from normal tables z = 0.255

15-M / 19.60 = 0.255

M= 10.002 minutes

the new mean time is 10 minutes to maintine 15 minutes waiting time to reach 60 % of the patients