Random Variables
• Expected values and standard deviation for a discrete random variable.
µ=
σ=
qX
X
xp(x)
(x − µ)2p(x) =
qX
x2p(x) − µ2
• (Linear combinations of r.v.’s) Suppose x1 and x2 are indpendent
random variables with means µ1, µ2 and standard deviations σ1 and
σ2.
If w = a + bx1, then
µw = a + bµ1 and σw2 = b2σ12
If w = ax1 + bx2, then
µw = aµ1 + bµ2 and σw2 = a2σ12 + b2σ22
Example. (Section 5.1, Exercise 6) The following data is for U.S.
family size
x 2
3
4
5
6 7
% 42% 23% 21% 10% 3% 1%
(a) Convert the percentages to probabilities and make a histogram
for the probability distribution.
(b) What is the probability that a family selected at random will have
only two members.
(c) What is the probability that a family selected at random will have
more than three members?
(d) Compute µ the expected family size. What does this mean?
(e) Compute the standard deviation σ.
Hints. (d)
µ = 2(.42) + 9(.23) + 4(.21) + 5(.10) + 6(.03) + 7(.01)
= 3.12
(e)
p
σ =
22 (.42) + 33 (.23) + 42 (.21) + 52 (.10) + 62 (.03) + 72 (.01) − 3.122
√
1.4456 = 1.202.
=
Example. Section 5.1, Exercise 10. Probability distribution for
x the number of prisoners out of five on parole who become repeat
offenders.
x
0
1
2
3
4
5
p(x) .237 .396 .264 .088 .015 .001
(a) Find the probability that one or more of the five parolees will be
repeat offenders.
(b) Find the probability that two or more of the five parolees will be
repeat offenders.
(c) Find the probability that four or more of the parolees will be
repeat offenders.
(d) Compute µ the expected number of repeat offenders out of five.
(e) Compute σ the standard deviation for the number of repeat offenders out of five.
Hints. (a) Why is this 1 − .237?
(b) Compute .264 + .088 + .015 + .001.
(c) .016
(d)
µ = .396 + 2(.264) + 3(.088) + 4(.015) + 5(.001)
= 1.253.
(e) σ is computed as
p
.396 + 22(.264) + 33(.088) + 42(.015) + 52(.001) − 1.2532
√
which is .938881 = .969.
Example. Section 5.1, Exercise 16. Random variables x1 and x2
represent, respectively, the length of time in minutes for a company
to check the computer and the length of time for a company to repair
the computer. For these random variables it is know that
x1 : µ1 = 28.1 σ1 = 8.2
x2 : µ2 = 90.5 σ2 = 15.2
(a) Let W = x1 + x2. Compute the mean, variance and standard
deviation for W .
(b) Suppose it costs $1.50 per minute to examine the computer and
$2.75 per minute to repair the computer. Then W = 1.50x1 + 2.75x2
represents the service cost. Compute the mean, variance and standard
deviation for W .
(c) Suppose L = 1.5x1 + 50. Compute the mean, variance and standard deviation for L.
Hints. (a) µW = µ1 + µ2 = 28.1 + 90.5 = 118.6.
2
= σ22 + σ22 = 8.22 + 15.22 = 298.28
σW
√
σW = 298.28 = 17.27.
(b) Compute µW = 1.5µ1 + 2.75µ2 = 293.01 and
Thus σW
2
σW
= 1.52σ12 + 2.752σ22
= (1.5)2(8.2)2 + (2.75)2(15.2)2 = 1898.53
√
= 1898.53 = 43.47.
(c) Compute µL = 1.5µ1 + 50 and σL2 = 1.52σ12 and so µL = 92.15
and σL2 = 151.29 and σL = 12.30.