Question

The Heart Association plans t install a free blood pressure testing booth at the Palouse mall for a week. Previous experience indicates that, on average, 10 persons per hour request a test and the arrivals are Poisson distributed from an infinite population. Blood pressure measurements are exponential with an average time of 5 minutes per exam.

a) What is the average number of people in line (2 pts)?

b) What is the average number of people in the system (2 pts)?

c) What is the average amount of time a person can expect to spend in line (2 pts)?

d) What is the average amount of time it will take to measure a person’s blood pressure, including waiting in line (2 pts)?

e) What is the probability of having 3 people in line (2 pts)

Answer 1

Given are following data :

Arrival rate of patients = a = 10 per hour

Service rate ( i.e. blood pressure measurement @ 5 minutes per exam) =S = 12 per hour

1. Average number of people in the line

= a^2/ S x ( S – a )

= 10 x 10 / 12 x ( 12 – 10 )

= 100/( 12x 2 )

= 100/24

= 4.17 ( rounded to 2 decimal places)

1. Average number of people in the system

= average number of people in the line + a/S

= 4.17 + 10/12

= 4.17 + 0.83

= 5

1. Average amount of time a person can expect to be in the line

= a/S x ( s – a)

= 10/ 12 x ( 12 – 10) hour

= 10 / ( 12 x 2 ) hour

= 10 /24 hour

= 10 x 60 /24 minutes

= 25 minutes

1. Average amount of time it will take to measure one’s blood pressure , including waiting in line

a/S x ( S – a ) + 1 /S

= 10/24 + 1/12 hour

= 10/24 + 2/24 hour

= 3/24 hour

= 7.5 minutes

e)Probability of having 3 people in line

= ( a/s)^3 x ( 1 – a/s)

= ( 10/12)^3 x ( 1 – 10/12)

= ( 10/12)^3 x ( 2/12 )

= 2000/20736

= 0.964