AP Statistics
Random Variables
1. A commuter must pass through five traffic lights on her way to work and will have
to stop at each one that is red. She estimates the probability model for the
number of red lights she hits, as shown below.
X = # of red lights
0
1
2
3
4
5
P(X = x)
0.05
0.25
0.35
0.15
0.15
0.05
a. How many red lights should she expect to hit each day?
E(X) = 0(0.05) + 1(0.25) + 2(0.35) + 3(0.15) + 4(0.15) + 5(0.05)
E(X) = 2.25 red lights
b. What’s the standard deviation?
Var(X) = (0 – 2.25)2(0.05) + (1 -2.25)2(0.25) + (2 – 2.25)2(0.35) + (3 -2.25)2(0.15)
+ (4 – 2.25)2(0.15) + (5 – 2.25)2(0.05)
Var(X) = 1.5875
SD(X) = 1.26
2. A couple plans to have children until they get a girl, but they agree that they will
not have more than three children even if all are boys. (Assume boys and girls
are equally likely).
a. Create a probability model for the number of children they’ll have.
X = # of children
1
2
3
P(X = x)
1/2
1/4
1/4
b. Find the expected number of children.
E(X) = 1(1/2) + 2(1/4) + 3(1/4)
E(X) = 1.75
c. Find the expected number of boys they’ll have.
Y = # of Boys
0
1
2
P(Y = y)
1/2
1/4
1/8
E(Y) = 0(1/2) + 1(1/4) + 2(1/8) + 3(1/8)
E(Y) = 0.875
3
1/8
3. An insurance policy costs $100 and will pay policyholders $10,000 if they suffer a
major injury (resulting in hospitalization) or $3000 if they suffer a minor injury
(resulting in lost time from work). The company estimates that each year 1 in
every 2000 policyholders may have a major injury, and 1 in 500 a minor injury.
a. Create a probability model for the profit on a policy.
X = Profit
-$9900
-$2900 $100
P(X = x)
1/2000
1/500
399/400
b. What’s the company’s expected profit on this policy?
E(X) = -9900(1/2000) – 2900(1/500) + 100(399/400)
E(X) = $89
c. What’s the standard deviation?
Var(X) = (-9900 – 89)2(1/2000) + (-2900 – 89)2(1/500) + (100 – 89)2(399/400)
Var(X) = 67879
SD(X) = $260.54
4. Given independent random variables with means and standard deviations as
shown, find the mean and standard deviation of each of these variables:
a. 3X
Mean SD
E(3X) = 3E(X) = 3(10) = 30
X 10
2
SD(3X) = 3SD(X) = 3(2) = 6
Y 20
5
b. Y + 6
E(Y + 6) = E(Y) + 6 = 20 + 6 = 26
SD(Y + 6) = SD(Y) = 5
c. X + Y
E(X + Y) = E(X) + E(Y) = 10 + 20 = 30
Var(X + Y) = Var(X) + Var(Y) = 22 + 52 = 29
SD(X + Y) = 5.39
d. X – Y
E(X – Y) = E(X) – E(Y) = 10 – 20 = –10
Var(X – Y) = Var(X) + Var(Y) = 22 + 52 = 29
SD(X – Y) = 5.39
e. X1 + X2
E(X1 + X2) = E(X1) + E(X2) = 10 + 10 = 20
Var(X1 + X2) = Var(X1) + Var(X2) = 22 + 22 = 8
SD(X1 + X2) = 2.83
f. 3X – 2Y
E(3X – 2Y) = 3E(X) – 2E(Y) = 3(10) – 2(20) = –10
Var(3X – 2Y) =32Var(X) + 22Var(Y) = 9(22) + 4(52) = 136
SD(3X – 2Y) = 11.66
g. 2Y1 – 3Y2
E(2Y1 – 3Y2) = 2E(Y1) – 3E(Y2) = 2(20) – 3(20) = -20
Var(2Y1 – 3Y2) = 22Var(Y1) + 32Var(Y2) = 4(52) + 9(52) = 325
SD(2Y1 – 3Y2) = 18.03
5. Suppose a used car dealer runs autos through a two-stage process to get them
ready to sell. The mechanical checkup costs $50 per hour and takes an average
of 90 minutes, with a standard deviation of 15 minutes. The appearance prep
(wash, polish, etc.) costs $6 per hour and takes an average of 60 minutes, with a
standard deviation of 5 minutes.
a. What are the mean and standard deviation of the total time spent
preparing a car? (Assume that the two phases of the process are
independent)
X – Time for the mechanical checkup; Y – time for the appearance prep
E(X + Y) = E(X) + E(Y) = 90 + 60 = 150 minutes
Var(X + Y) = Var(X) + Var(Y) = 152 + 52 = 250
SD(X + Y) = 15.81 minutes
b. What are the mean and standard deviation of the total expense to prepare
the car?
Let CX = Cost of the mechanical checkup; CY = Cost of the appearance prep.
E(CX) = $50*90/60 = $75
E(CY) = $6*60/60 = $6
SD(CX) = $50*15/60 = $12.50
SD(CY) = $6*5/60 = $0.50
E(CX + CY) = 75 + 6 = $81
Var(CX + CY) = 12.52 + 0.502 = 156.5
SD(CX + CY) = $12.51
c. What are the mean and standard deviation of the differences in costs for
the two phases of the operation?
E(CX – CY) = 75 – 6 = $69
Var(CX – CY) = 12.52 + 0.502 = 156.5
SD(CX – CY) = $12.51
d. What is the probability that it will take longer to do the appearance prep
than the mechanical checkup? (We assume that each phase of the
process can be described by Normal Models)
If we take the difference of the two times, then when the appearance prep takes
longer we would get a negative value. So we need to find the probability that the
difference is less than zero.
E(X – Y) = 90 – 60 = 30 minutes
SD(X – Y) = 15.81 minutes (same as in a.)
N(30, 15.81)
0  30
z
 1.898
15.81
P(z < -1.898) = 0.0288
Probability is 2.88% that the appearance prep takes longer than the mechanical
checkup.