Question

A car saleswoman has worked for the same car sales company for many years. Based on her track record, her selling prices of cars can be modelled by a normal distribution with mean $12,000AUD with standard deviation $2,000AUD. (a.) What is the probability that the car saleswoman sells a car that is more expensive than $13,500AUD? (b.) During a given month, 14 customers buy a car from her. What is the probability that at least 3 of them bought a car that is cheaper than $13,500AUD? Carefully state the assumptions of your solution. (c.) The car saleswoman has a bonus plan that entitles her to a bonus if she sells 4 cars that each costs $13,500AUD or more during a month. What is the probability that she receives the bonus when the 8th car of the month is sold?

Answer 1

a)

µ =    12000                  
σ =    2000                  
                      
P ( X ≥   13500.00   ) = P( (X-µ)/σ ≥ (13500-12000) / 2000)              
= P(Z ≥   0.750   ) = P( Z <   -0.750   ) =    0.2266   (answer)

b)

Probability of cheaper than 13500

µ =    12000      
σ =    2000      
          
P( X ≤    13500   ) = P( (X-µ)/σ ≤ (13500-12000) /2000)  
=P(Z ≤   0.750   ) =   0.7734

Now applying binomial

	X	P(X)
P ( X = 0) = C (14,0) * 0.773372647623132^0 * ( 1 - 0.773372647623132)^14=	0	0.0000000
P ( X = 1) = C (14,1) * 0.773372647623132^1 * ( 1 - 0.773372647623132)^13=	1	0.0000000
P ( X = 2) = C (14,2) * 0.773372647623132^2 * ( 1 - 0.773372647623132)^12=	2	0.0000010

P(X>=3) = 1 - P(X<=2)

= 0.9999

c)

Sample size , n =    7
Probability of an event of interest, p =   0.2266

	X	P(X)
P ( X = 3) = C (7,3) * 0.226627352376868^3 * ( 1 - 0.226627352376868)^4=	3	0.1457

Probability = 0.1457 * 0.2266

= 0.033

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