Question

In: Math

A survey was conducted to determine, on average, how long patients have to wait to see...

A survey was conducted to determine, on average, how long patients have to wait to see a doctor. The number of patients that ended up waiting up to 2 hours to see a doctor, recorded in 10-minute intervals, is given in the table below.
a) Define and name the random variable associated with this data.
b) Find the probability distribution for the random variable and draw the corresponding histogram. Express all probabilities as a fraction in lowest terms.
c) What is the probability that a patient had to wait between 20 and 50 minutes?
d) What is the expected number of minutes a patient must wait? Give an exact answer. (2 marks

e) Calculate the standard deviation for this distribution. Explain what it means in the context of this problem.
2.

Waiting Time in minutes
10
20
30
40
50
60
70
80
90
100
110
120 Frequency of Occurrence
1
4
15
20
35
42
28
19
13
5
2
1
e) Calculate the standard deviation for this distribution. Explain what it means in the context of this problem.

Solutions

Expert Solution

a. Random variable= how long patients have to wait to see a doctor

b.

c. Probability that a patient had to wait between 20 and 50 minutes=0.0216+0.0811+0.1081+0.1892=0.4

d.

E(X)= expected number of minutes a patient must wait =60 min.

e. E(X2)=3992.34

Var(X)=E(X2) -E2(X)=3992.34-602=392.34

Standard deviation for this distribution= square root of Var(X)=19.8076 min.


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