Question

There are 12 agents in one office of a certain real estate firm. Much of the business in this office is conducted by taking a prospective customer out to view a particular property. The time to drive to and view a property is normally distributed and averages 47.3 minutes with a standard deviation of 11.7 minutes. Past studies have shown that​ twenty-six (26) percent of customers that visit properties with an agent will eventually buy a property being shown by that agent. On average each agent visits 3.6 properties a​ day, with a standard deviation of 1.9 visits.

a) What is the probability that an agent will visit at least 7 properties over the next three​ days? 

b)What is the probability that the average time of the next 50 property viewings will be less than 45​ minutes?

c) A particular agent currently has 17 prospective customers that are being shown properties for sale. What is the probability that less than 4 of these customers will eventually buy a property being shown by the​ agent?  

d) Eighty-two (82) percent of property visits will take what range of​ time, centered at the mean.

Answer 1

let x be the time to drive to and view a property and it is normally distributed with mean µ = 47.3 and standard deviation σ =11.7

and y be the number of visit properties with an agent it is also normally distributed with mean µ = 3.6 and standard deviation σ =1.9

p is the percent of customers that visit properties with an agent will eventually buy a property being shown by that agent = 0.26

a) We have mean µ = 3.6 and standard deviation σ =1.9 per day ,so for 3 days mean µ = 3*3.6 = 10.8 and standard deviation σ = = 3.2909

P( y ≥ 7 ) =   = P( z ≥ -1.15 ) = 1- P( z ≤ -1.15 ) = 1 - 0.1251 ---- ( from z score table, value of -1.15 )

P( y ≥ 7 ) = 0.8749

b) We have mean µ = 47.3 and standard deviation σ =11.7 , so average time   of the next 50 property viewings follows approximately normal distribution with mean µ = 47.3 and standard deviation = = =  1.6546

P( x <45 ) =   = P( z < -1.39 )

P( x <45 ) = 0.0823

c) For this part we have to use binomial distribution with n = 17 and p = 0.26

P( less than 4 customers out of 17 will buy the property )

= P(0) +P(1) + P(2)+P(3) = 0.0060 + 0.0357 + 0.1005 + 0.1765 ---from binomial distribution table for n=17 and p =0.26

= 0.3187

d) We have to find y₁ and y₂ such that area between them is 82%, therefore area outside y₁ and y₂ is 1- 0.82 = 0.18

Therefore area less than y₁ and area greater than y₂  = 0.18/ 2 = 0.09

So we have to find z score corresponding to area 0.09 using the z score table , it is -1.34

So y₁ = z*σ + µ = ( -1.34*11.7) + 47.3

y₁ = 31.622

y₂ = z*σ + µ = ( 1.34*1.9) + 3.6

y₂ = 62.978

82% percent of property visits will take 31.622 minute to 62.978 minutes.