Question

Erinn must decide whether to add the cost of towing insurance to her car insurance policy. The additional cost is $50 annually. She has an older car that is not super reliable. She decides to analyze her expected annual towing costs and then she will decide whether it is cost effective for her to buy the towing insurance or not. She estimates the probability that her car will break down and need to be towed once during the year is 0.15. She also estimates that the probability that it will need to be towed twice during the year is 0.10. Lastly, she believes that it is extremely unlikely that it will need to be towed more that twice, so she assigns probability of 0 to having the car towed more than twice. The typical price of towing a vehicle in CT is approximately $225.00. Let the random variable X = Erinn’s towing charges annually. To begin her analysis she completes the table below:

	X= Amount Spent on Towing	Probability
Erinn’s car does not need to be towed	$0.00
Erinn’s car needs to be towed once during the year
Erinn’s car needs to be towed twice during the year

Calculate E(X). (Note: E(X) = her expected average annual towing charges.)

	a.	$78.75
	b.	$45.00
	c.	$240.00
	d.	$62.00
	e.	$24.00

Answer 1

Answer:

X = Amount spent on Towing	Probability
Erinn's car does not need to be towed	$0.0	0
Erinn's car needsto be towed once in a year	$225.00	0.15
Erinn's car needs to be towed otwice a year	$450.00	0.10

The completed table is given above:

X = Amount spent on Towing

E (X) = 0*0 +(225*0.15)+(450*0.10)

= $78.75

So since E(X) is greater than $50, hence, OPTION (a) IS CORRECT