Question

5. According to Health Quality Ontario, the average wait time for first time assessment by a doctor in emergency department of a hospital is 1.6 hours as of February 2020. Source: https://www.hqontario.ca/System-Performance/Time-spent-in-emergency-departments. Assume that the emergency waiting times may be modelled by a normal distribution is applicable with a mean of 96 minutes and a standard deviation of 15 minutes.

                                                                                                           

5.1 The probability that a randomly selected emergency department patient will wait between 90 minutes and 105 minutes before first assessment by a doctor equals:                                                                                                                    

Calculator function(s):

____________________________________________      Answer: __________ (4 decimal pl)

                                                                                                                       

5.2 The probability that a randomly selected emergency department patient will wait more than 110 minutes equals:                                                                                             

Calculator function(s):

____________________________________________      Answer: __________ (4 decimal pl)

                                                                       

5.3 The waiting time exceeded by only 10% of emergency department patients equals:                     

                                                           

Calculator function(s):

_______________________________________      Answer: ____________ (nearest minute)

5.4 For a random sample of 25 emergency department patients what is the probability that the average waiting time will be between 88 minutes and 95 minutes:                                                 

                                                     

Calculator function(s):

____________________________________________      Answer: __________ (4 decimal pl)

Answer 1

Given:

= 96, = 15

a)

Find: P(90 < X < 105)

P(90 < X < 105) = P(-0.40 < Z < 0.60)

P(90 < X < 105) = P(Z < 0.60) - P(Z < -0.40)

P(90 < X < 105) = 0.7257 - 0.3446    ..............Using standard Normal table

P(90 < X < 105) = 0.3812

5.2) Find: P(X > 110)

P(X > 110) = P(Z > 0.93)

P(X > 110) = 1 - P(Z < 0.93)

P(X > 110) = 1 - 0.8247                       ..............Using standard Normal table

P(X > 110) = 0.1753

C) More than 10%

Find: 10th percentile

X = + *Z0.10

Where,

Z0.10 = -1.28       .                     ..............Using standard Normal table

Therefore,

X = + *Z0.10

X = 96 + 15*-1.28

X = 76.78 = 77 minutes

5.4)

      .                     ..............Using standard Normal table