Download Stat/Math 418 Exam 2

Stat/Math 418
Exam 2 - Practice problems
1. Let the density of the random variable X be given by
2
f (a) = e−π(1−a) ,
−∞ < a < ∞.
(a) Identify the distribution of X. Write down the mean and variance of X.
(b) Find P (X < 1).
2. The number of years a radio functions is exponentially distributed with parameter λ = 0.5. If Jones
buys a used radio, what is the probability that it will be working after an additional four years?
3. Suppose X is uniformly distributed over [−1, 3] and Y = X 2 . Find the cumulative distribution function
(c.d.f) and the probability density function (p.d.f.) of Y .
4. Suppose that the height, in inches, of a 25-year-old man is a normal random variable with mean µ = 71
and variance σ 2 = 6.25. What percentage of 25-year-old men are over 6 feet tall? What percentage of
men in the 6-footer club are over 6 feet 5 inches?
5. A bus travels between the two cities A and B, which are 100 miles apart If the bus has breakdown, the
distance from the breakdown to city A has a uniform distribution over (0, 100). Three service stations
are located at a distance of 25, 50, and 75 miles respectively, from A. Find the expected distance the
bus needs to be towed.
6. Consider the function f (x) = C(2x2 −x3 ) if 0 ≤ x ≤ 2 and 0 otherwise. Could f be a probability
2
density function? If so, determine C. Repeat if f were given by f (x) = C(x − 1)ex for 0 ≤ x ≤ 2
and 0 otherwise.
7. Suppose X has normal distribution with mean 0 and variance 1. Find E(X 5 ), P (X 3 > 0), and write
down the density of Y = 2X − 1.
8. If X is uniformly distributed on (0, 1), then find the distribution of Y = − log X. Also find its hazard
rate.
9. The joint probability density function of X and Y is given by
3
cxe−x if − x < y < x,
fX,Y (x, y) =
0
otherwise,
x>0
for some c > 0. Are X and Y independent? Justify your answer. Find the marginal density of X.
10. Let J and K be independent random variables with probability mass functions

(

if j = 1, 3,
0.2
0.5
if k = −1, 1,
PJ (j) = 0.6
PK (k) =
if j = 2,

0
otherwise.

0
otherwise,
Find the probability mass function pJ+K of J + K.
Solutions to practice problems
1. Solution:
(a) Since f (a) = √
1
2
1
p
e− 2(1/2π) (a−1) for all a, the mean is 1 and variance is
2π
1/2π
1
2π .
(b) Find P (X < 1) = 0.5.
2. Solution: By the memoryless property of the exponential distribution, the required probability is
e−4λ = e−2 .
3. Solution: For y > 0, the c.d.f. is given by
√
√
FY (y) = P (X ≤ y) = P (− y ≤ X ≤ y) =
2
Z
√
y
√
− y
fX (x)dx.
R √y
R √y
√
√
For 0 ≤ y ≤ 1, FY (y) = −√y 14 dx = 12 y. As −1 ≤ X ≤ 3, we have FY (y) = −1 14 dx = 14 (1 + y)
for 1 ≤ y ≤ 9. Further FY (y) = 0 for y < 0, and FY (y) = 1 for y ≥ 9. The p.d.f. of Y is obtained by
differentiating FY ,

√

1/(4 y) if 0 ≤ y ≤ 1
√
FY0 (y) = fY (y) = 1/(8 y) if 1 < y ≤ 9


0
otherwise.
4. Solution: Let X denote the height, and Z denote a standard normal random variable.
X −µ
72 − 71
P (X > 72) = P
>
= P (Z > 0.4) = 1 − 0.6554 = .3446
σ
2.5
X −µ
77 − 71
P (X > 77|X > 72) = P
>
/0.3446 = P (Z > 2.4)/0.3446
σ
2.5
= (1 − 0.9918)/.3446 = .0082/.3446 = .0238
5. Solution: Let X denote distance from A to the breakdown site, and let L denote the distance from
the breakdown site to the nearest service station. So,


25 − X,
if 0 < X < 25




X − 25,
if 25 ≤ X < 37.5

L = |X − 50|
if 37.5 ≤ X < 62.5



75 − X
if 62.5 ≤ X < 75




X − 75,
if 75 ≤ X ≤ 100.
E(L) =
R2
2
100
Z
25
xdx +
0
4
100
Z
12.5
xdx =
0
25 12.5
+
= 9.375.
4
4
4
3
6. Solution: As 1 = 0 C(2x2 − x3 )dx = C( 16
3 − 4) = 3 C, it is a probability density function if C = 4 .
As f (0) = −C and f (2) = Ce4 , there is no value of C for which f (x) is always non-negative. So f
cannot be a probability density function.
1 2
7. Solution: As the density of X is symmetric around zero, P (X 3 > 0) = P (X > 0) = 21 . Since x5 e− 2 x
is an odd function of x, E(X 5 ) = 0. Y has normal distribution with mean E(2X − 1) = −1 and
1
1
2
variance V ar(2X − 1) = 4. So the density of Y is given by fY (x) = √ e− 8 (x+1) .
2 2π
8. Solution: fX (x) = 1 if 0 < x < 1 and 0 otherwise. Let g(x) = − log x. y = − log x implies x = e−y ,
d −1
so g −1 (y) = e−y for y > 0. Use the formula fY (y) = fX (g −1 (y))| dy
g (y)| if y = g(x) for some x and
−y
0 otherwise, to get fY (y) = e if y > 0 and zero otherwise. So Y is distributed as exponential with
λ = 1. The hazard rate is (fY (y)/P (Y > y)) = 1.
9. Solution: X and Y are not independent as the range of Y depends on X. The marginal density fX
of X is given by
Z x
3
3
fX (x) =
cxe−x dy = 2c x2 e−x ,
−x
if x > 0 and zero otherwise. As
Z
1=
∞
Z
fX (x) dx =
o
0
∞
2
3
2c x2 e−x dx = c,
3
c = 1.5.
10. Solution: By independence of J, K, we have for any integer n,
pJ+K (n) =P (K = 1, J = n − 1) + P (K = −1, J = n + 1)
=pK (1)pJ (n − 1) + pK (−1)pJ (n + 1)
=0.5 [pJ (n − 1) + pJ (n + 1)].
So


0.5 pJ (n + 1)
pJ+K (n) = 0.5 [pJ (n − 1) + pJ (n + 1)]


0.5 pJ (n − 1)
for n = 0, 1
for n = 2
for n = 3, 4.
Hence PJ+K (0) = PJ+K (4) = 0.1, PJ+K (1) = PJ+K (3) = 0.3 and PJ+K (2) = 0.2.