Chapter #5 – Continuous Random Variables
Question #1: Let X be a random variable with the following probability density function
𝑐(1 − 𝑥 2 ) 𝑖𝑓 𝑥 ∈ (−1,1)
𝑓(𝑥) = {
. (a) What is the value of the constant c in the PDF given by
0
𝑜/𝑤
𝑓(𝑥)? (b) What is the cumulative distribution function (CDF) of X?
∞
1
1
a) To find constant, we have that 1 = ∫−∞ 𝑓(𝑥) 𝑑𝑥 = ∫−1 𝑐(1 − 𝑥 2 ) 𝑑𝑥 = 𝑐[𝑥 − 3 𝑥 3 ]1−1 =
1
1
1
1
2
4
𝑐 [(1 − 3) − (−1 + 3)] = 𝑐 (1 − 3 + 1 − 3) = 𝑐 (2 − 3) = 𝑐 3. Since we have found
3
3
that 1 = 𝑐 4, it follows that the constant 𝑐 = 4.
𝑥
3
𝑥
3
1
𝑥
b) We have that 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∫−∞ 𝑓(𝑡) 𝑑𝑡 = 4 ∫−1(1 − 𝑡 2 ) 𝑑𝑡 = 4 [𝑡 − 3 𝑡 3 ]−1
=
3
1
1
3
1
1
3
1
2
[(𝑥 − 3 𝑥 3 ) − (−1 + 3)] = 4 (𝑥 − 3 𝑥 3 + 1 − 3) = 4 (𝑥 − 3 𝑥 3 + 3). We can express
4
𝑖𝑓 𝑥 ∈ (−∞, −1]
0
3
1
2
the CDF as 𝐹(𝑥) = { 4 (𝑥 − 3 𝑥 3 + 3)
.
𝑖𝑓 𝑥 ∈ (−1,1)
𝑖𝑓 𝑥 ∈ [1, ∞)
1
Question #4: The probability density function of X, the lifetime of a certain type of electronic
10
𝑖𝑓 𝑥 > 10
device (measured in hours) is given by 𝑓(𝑥) = {𝑥 2
. (a) Calculate the 𝑃{𝑋 > 20}.
0 𝑖𝑓 𝑥 ≤ 10
(b) What is the cumulative distribution function (CDF) of X? (c) What is the probability that,
of 6 such types of devices, at least 3 will function for at least 15 hours?
∞
∞
𝑡
a) We find as follows 𝑃(𝑋 > 20) = ∫20 𝑓(𝑥) 𝑑𝑥 = 10 ∫20 𝑥 −2 𝑑𝑥 = 10 𝑙𝑖𝑚 ∫20 𝑥 −2 𝑑𝑥 =
𝑡→∞
1
10 𝑙𝑖𝑚 [− 𝑥]𝑡20
𝑡→∞
b) We
have
1
1
1
1
10
1
= 10 𝑙𝑖𝑚 (− 𝑡 + 20) = 20 = 2 = 0.5.
𝑡→∞
𝑥
that
10 (− 𝑥 + 10) = 1 −
𝑥
1
𝑥
𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∫−∞ 𝑓(𝑡) 𝑑𝑡 = 10 ∫10 𝑡 −2 𝑑𝑡 = 10[− 𝑡 ]10
=
10
. We can express the CDF as 𝐹(𝑥) = {
𝑥
𝑖𝑓𝑥 ∈ (−∞, 10]
0
1−
10
𝑥
𝑖𝑓 𝑥 ∈ (10, ∞)
.
c) Let 𝐴 be the event that a single device works for at least 15 hours. We thus have that
10
10
2
𝑃(𝐴) = 1 − 𝐹(15) = 1 − (1 − 15) = 15 = 3. If we let 𝐴3 denote the event that at least
2 𝑖
1 6−𝑖
three will work for at least 15 hours, then we have 𝑃(𝐴3 ) = ∑6𝑖=1(6𝑖) (3) (3)
.
Question #7: The density function of X is given by 𝑓(𝑥) = {
𝑎 + 𝑏𝑥 2 𝑖𝑓 𝑥 ∈ [0,1]
. (a) If we
0
𝑜/𝑤
3
are given that the expected value is 𝐸(𝑋) = 5, find the constants 𝑎 and 𝑏. (b) Find the CDF.
a) We will create a system of two equations in terms of 𝑎 and 𝑏 so we can solve for them.
∞
3
The first is obtained from the fact that we know 𝐸(𝑋) = 5 = ∫−∞ 𝑥𝑓(𝑥) 𝑑𝑥 =
1
𝑎
𝑏
𝑎
𝑏
𝑎
𝑏
3
∫0 𝑥(𝑎 + 𝑏𝑥 2 ) 𝑑𝑥 = [2 𝑥 2 + 4 𝑥 4 ]10 = 2 + 4. Thus, we have that 2 + 4 = 5. The second
equation is found from the fact the probability density function over its domain must
∞
1
𝑏
𝑏
sum to 1 = ∫−∞ 𝑓(𝑥) 𝑑𝑥 = ∫0 𝑎 + 𝑏𝑥 2 𝑑𝑥 = [𝑎𝑥 + 3 𝑥 3 ]10 = 𝑎 + 3, implying that we
𝑏
3
6
have 𝑎 + 3 = 1. These two equations allow us to solve for 𝑎 = 5 and 𝑏 = 5.
𝑥
𝑥3
b) We calculate that 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) = ∫−∞ 𝑓(𝑡) 𝑑𝑡 = ∫0
5
3
2
3
6
𝑖𝑓 𝑥 ∈ (−∞, 0]
0
3
6
+ 5 𝑡 2 𝑑𝑡 = [5 𝑡 + 15 𝑡 3 ]0𝑥 =
5
2
𝑥 + 5 𝑥 3 . Thus, we have that 𝐹(𝑥) = { 5 𝑥 + 5 𝑥 3
.
𝑖𝑓 𝑥 ∈ (0,1)
𝑖𝑓 𝑥 ∈ [1, ∞)
1
Question #10: Trains headed for destination A arrive at the train station at 15-minute
intervals starting at 7 AM, whereas trains headed for destination B arrive at 15-minute
intervals starting at 7:05 AM. (a) If a certain passenger arrives at the station at a time
uniformly distributed between 7 and 8 AM and then gets on the first train that arrives, what
proportion of time does he or she go to destination A? (b) What if the passenger arrives at a
time uniformly distributed between 7:10 and 8:10 AM?
a) 𝑃(𝐴) = 𝑃{(5 < 𝑋 < 15) ∪ (20 < 𝑋 < 30) ∪ (35 < 𝑋 < 45) ∪ (50 < 𝑋 < 60)} =
𝑃(5 < 𝑋 < 15) + 𝑃(20 < 𝑋 < 30) + 𝑃(35 < 𝑋 < 45) + 𝑃(50 < 𝑋 < 60) =
15−0
5−0
30
20
45
35
60
50
40
2
[(60−0) − (60−0)] + [(60) − (60)] + [(60) − (60)] + [(60) − (60)] = 60 = 3. Here, we
calculated each probability using the CDF of 𝑋~𝑈𝑁𝐼𝐹(𝑎, 𝑏), which is given by the
𝑥−𝑎
formula 𝐹(𝑥) = 𝑏−𝑎 for all 𝑥 ∈ (𝑎, 𝑏). Note that the CDF of 𝑋 is equal to zero for every
𝑥 ∈ (−∞, 𝑎] and is equal to 1 for every 𝑥 ∈ [𝑏, ∞). (Draw clock to get four intervals.)
b) The answer is identical to the one given above.
Question #13: You arrive at a bus stop at 10 o’clock, knowing that the bus will arrive at some
time uniformly distributed between 10 and 10:30. (a) What is the probability that you will
have to wait longer than 10 minutes? (b) If, at 10:15, the bus has not yet arrived, what is the
probability that you will have to wait at least an additional 10 minutes?
a) If we let 𝑋 be the random variable equal to the arrival time of the bus, then we have
30
that 𝑋~𝑈𝑁𝐼𝐹(0,30). Thus, 𝑃(𝑋 > 10) = ∫10
b) 𝑃(𝑋 > 25|𝑋 > 15) =
𝑃(𝑥>25 ∩ 𝑋>15)
𝑃(𝑋>15)
1
𝑥
2
30
𝑑𝑥 = [30]10
= 3.
30−0
30 1
𝑃(𝑥>25)
= 𝑃(𝑋>15) =
∫25 30𝑑𝑥
30 1
∫15 30𝑑𝑥
5/30
5
1
= 15/30 = 15 = 3.
Question #16: The annual rainfall (in inches) in a certain region is normally distributed with
𝜇 = 40 and 𝜎 = 4. What is the probability that, starting with this year, it will take over 10
years before a year occurs having a rainfall of over 50 inches?

If we let 𝑋 be the random variable equal to the annual rainfall in the region, then we
have that 𝑋~𝑁(40,4). The probability that rainfall is over 50 inches in any year is
𝑋−40
𝑃(𝑋 > 50) = 𝑃 (
4
>
50−40
10
4
4
) = 𝑃 (𝑍 >
) = 𝑃(𝑍 > 2.5) = 1 − Φ(2.5) = 0.0062.
If we let 𝐴 denote the event that the rainfall is less than 50 in ten consecutive years,
then we see that 𝑃(𝐴) = (1 − 0.0062)10 = 0.993810 .
Question #19: Let X be a normal random variable with mean 12 and variance 4. Find the
value of 𝑐 such that 𝑃{𝑋 > 𝑐} = 0.10.

𝑋−12
We have that 𝑋~𝑁(12,2), so 𝑃(𝑋 > 𝑐) = 𝑃 (
2
>
𝑐−12
𝑐−12
2
2
) = 𝑃 (𝑍 >
) = 0.1. From
the z-table, we have that 𝑃(𝑍 < 1.28) = Φ(1.28) = 0.9, which thus implies that
𝑃(𝑍 > 1.28) = 0.1, so we must solve the equation
𝑐−12
2
= 1.28, which gives 𝑐 = 14.56.
Question #22: The width of a slot of a duralumin forging is (in inches) normally distributed
with 𝜇 = 0.9 and 𝜎 = 0.003. The specification limits were 0.9 ± 0.005. (a) What percentage
of forgings will be defective? (b) What is the maximum value of 𝜎 that will permit no more
than 1 in 100 defectives when the widths are normally distributed with 𝜇 = 0.9 and 𝜎?
a) We
have
0.895−0.9
𝑃(
0.003
that
<
𝑊−0.9
0.003
𝑃(0.9 − 0.005 < 𝑊 < 0.9 + 0.005) = 𝑃(0.895 < 𝑊 < 0.905) =
<
0.905−0.9
0.003
0.005
0.005
) = 𝑃 (− 0.003 < 𝑍 < 0.003) = 𝑃(−1.67 < 𝑍 < 1.67) =
Φ(1.67) − Φ(−1.67) = 0.905. Thus, 1 − 0.905 = 0.095 = 9.5% are defective.
b) We start with 𝑃 (−
0.005
𝜎
<𝑍<
0.005
Thus, we must solve the equation
) = 0.99 when 𝑃(−2.575 < 𝑍 < 2.575) = 0.99.
𝜎
0.005
𝜎
= 2.575, which gives 𝜎 = 0.0019.
Question #31: (a) A fire station is to be located along a road of length 𝐴 < ∞. If fires occur
at points uniformly chosen on (0, 𝐴), where should the station be located so as to minimize
the expected distance from the fire? That is, choose 𝑎 so as to minimize 𝐸[|𝑋 − 𝑎|] when X
is uniformly distributed over (0, 𝐴). (b) Now suppose that the road is of infinite length —
stretching from point 0 outward to ∞. If the distance of a fire from point 0 is exponentially
distributed with rate 𝜆, where should the fire station now be located? That is, we want to
minimize 𝐸[|𝑋 − 𝑎|], where X is now exponential with rate 𝜆.
∞
𝑎
𝐴
1
1
a) We have that 𝐸[|𝑋 − 𝑎|] = ∫−∞(𝑥 − 𝑎)𝑓(𝑥)𝑑𝑥 = ∫0 (𝑎 − 𝑥) 𝐴 𝑑𝑥 + ∫𝑎 (𝑥 − 𝑎) 𝐴 𝑑𝑥 =
𝑎
𝑎
1
1
𝑎
𝐴
𝑎2
𝑎2
𝐴2
[𝐴 𝑥 − 2𝐴 𝑥 2 ] + [2𝐴 𝑥 2 − 𝐴 𝑥] = ( 𝐴 − 2𝐴) + ((2𝐴 −
0
𝐴2 −2𝐴𝑎−𝑎2 +2𝑎2
2𝐴
𝑎
=
2𝑎2 −2𝐴𝑎+𝐴2
2𝐴
𝑑
and set equal to zero: 𝑑𝑎 (
𝑎2
𝐴𝑎
) − (2𝐴 −
𝐴
𝑎2
)) =
𝐴
2𝑎2 −𝑎2
2𝐴
+
. Since we want to minimize this, we take the derivative
2𝑎2 −2𝐴𝑎+𝐴2
2𝐴
1
) = 2𝐴 (4𝑎 − 2𝐴) =
2𝑎
𝐴
𝐴
− 1 = 0 → 𝑎 = 2. Thus, we
can minimize the expected value by choosing the midpoint of the interval (0, 𝐴).
∞
b) Using integration by parts we can find that 𝐸[|𝑋 − 𝑎|] = ∫0 (𝑥 − 𝑎)𝑓(𝑥)𝑑𝑥 =
𝑎
∞
∫0 (𝑎 − 𝑥)𝜆𝑒 −𝜆𝑥 𝑑𝑥 + ∫𝑎 (𝑥 − 𝑎)𝜆𝑒 −𝜆𝑥 𝑑𝑥 = 𝑎(1 − 𝑒 −𝜆𝑎 ) + 𝑎𝑒 −𝜆𝑎 +
𝑎𝑒 −𝜆𝑎 +
𝑒 −𝜆𝑎
𝜆
𝑒 −𝜆𝑎
𝜆
1
−𝜆+
− 𝑎𝑒 −𝜆𝑎 . After differentiating and setting equal to zero, we discover that
1
2
𝑒 −𝜆𝑎 − 2 = 0, which gives the minimum value at 𝑎 = 𝑙𝑛 (𝜆).
Question #34: Jones figures that the total number of thousands of miles that an auto can be
driven before it would need to be junked is an exponential random variable with parameter
1
. Smith has a used car that he claims has been driven only 10,000 miles. a) If Jones
20
purchases the car, what is the probability that she would get at least 20,000 additional miles
out of it? b) Repeat under the assumption that the lifetime mileage of the car is not
exponentially distributed, but is (thousands of miles) uniformly distributed over (0, 40).
a) If 𝑋 is the continuous random variable equal to the total number of thousands of miles
that a car can be driven before it must be junked, then we know 𝑋~𝐸𝑋𝑃(20). The PDF
1
𝑥
𝑒 −𝜆
𝜆
of 𝑋 is 𝑓(𝑥) = {
0
𝑖𝑓 𝑥 ≥ 0
𝑖𝑓 𝑥 < 0
1
𝑥
𝑥
while the CDF of 𝑋 is 𝐹(𝑥) = ∫−∞ 𝑓(𝑡)𝑑𝑡 = 1 − 𝑒 −𝜆 .
1
We also know that 𝐸(𝑋) = 𝜆 while the 𝑉𝑎𝑟(𝑋) = 𝜆2 . By the memoryless property of
the exponential distribution, we have that 𝑃(𝑋 ≥ 𝑠 + 𝑡 | 𝑋 ≥ 𝑡) = 𝑃(𝑋 ≥ 𝑠). Thus, we
want to find 𝑃(𝑥 ≥ 20 + 10 | 𝑋 ≥ 10) = 𝑃(𝑋 ≥ 20) = 1 − 𝑃(𝑋 < 20) = 1 − 𝐹(20) =
20
1 − (1 − 𝑒 −20 ) = 𝑒 −1 = 0.37. Thus, the probability of getting 20 more miles is 37%.
b) We now have that 𝑋~𝑈𝑁𝐼𝐹(0,40) and wish to calculate the conditional probability
𝑃(𝑋 ≥ 30 | 𝑋 ≥ 10) =
𝑃(𝑋≥30 ∩ 𝑋≥10)
𝑃(𝑋≥10)
40 1
𝑃(𝑋≥30)
= 𝑃(𝑋≥10) =
∫30 40𝑑𝑥
40 1
∫10 40𝑑𝑥
=
40−30
40
40−10
40
10
1
= 30 = 3.
1
Question #37: If X is uniformly distributed over (−1,1), find (a) 𝑃{|𝑋| > 2} and (b) the
density function of the random variable |𝑋|.
1
1
1
a) We have that 𝑋~𝑈𝑁𝐼𝐹(−1,1) and wish to calculate 𝑃 (|𝑋| > 2) = 𝑃 (− 2 > 𝑋 > 2) =
𝑃 (𝑋 < −
1
2
1
1
1
1
− 1
2
2
2
2
11
∪ 𝑋 > ) = 𝑃 (𝑋 < − ) + 𝑃 (𝑋 > ) = ∫−12 𝑑𝑥 + ∫1
2
2
1
1
1
4
4
2
𝑑𝑥 = + = .
b) We begin by finding the CDF of 𝑌 = |𝑋|, where we have 𝐹𝑋 (|𝑋|) = 𝑃(|𝑋| ≤ 𝑥) =
𝑥 1
1
𝑃(−𝑥 ≤ 𝑋 ≤ 𝑥) = ∫−𝑥 2 𝑑𝑡 = [2 𝑡]
𝑥
−𝑥
=
2𝑥
2
= 𝑥 for all 𝑥 ∈ (0,1). We then differentiate
𝑑
𝑑
the CDF to obtain the PDF as 𝑓|𝑋| (𝑥) = 𝑑𝑥 𝐹𝑋 (|𝑋|) = 𝑑𝑥 (𝑥) = 1 for all 𝑥 ∈ (0,1).
Therefore, we have found that 𝑌 = |𝑋|~𝑈𝑁𝐼𝐹(0,1).
Question #40: If X is uniformly distributed over (0,1), find the density function of 𝑌 = 𝑒 𝑋 .

First we start with the cumulative density function 𝐹𝑌 (𝑦), since we want to find the
PDF for 𝑌. We have 𝐹𝑌 (𝑦) = 𝑃(𝑌 ≤ 𝑦) = 𝑃(𝑒 𝑋 ≤ 𝑦) = 𝑃(𝑋 ≤ 𝑙𝑛(𝑦)) = 𝐹𝑋 (𝑙𝑛(𝑦)),
which is the CDF with respect to 𝑋. After taking the derivative to get the PDF we obtain
𝑑
1
𝑓𝑌 (𝑦) = 𝑑𝑦 𝐹𝑋 (𝑙𝑛(𝑦)) = 𝑓𝑋 (ln(𝑦)) ∗ 𝑦 by the chain rule. But since 𝑋~𝑈𝑁𝐼𝐹(0,1), we
know that 𝑓𝑋 (ln(𝑦)) = 1 and the domain changes to 𝑒 0 < 𝑒 𝑋 < 𝑒 1 = 1 < 𝑦 < 𝑒, so we
1
have that the PDF of 𝑌 = 𝑒 is 𝑓𝑌 (𝑦) =
𝑋
{𝑦
0
𝑖𝑓 1 < 𝑦 < 𝑒
𝑜/𝑤.
.