Download Test 2 and Solutions

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MATH 461 - D13, Test 2, Fall 2014
October 31, 2014
Books, notes and extra papers are not allowed on this test
Please show all your work and explain all answers to qualify for full credit
1. (16 points) Let X be a random variable whose distribution functions is given by

0,
x < −1




x/4
+
1/4
,
−1
≤x<0

1
,
0≤x<1
F (x) =
2


x/12 + 5/6 , 1 ≤ x < 2 .



1,
2≤x
Find (a) P(X < 1), (b) P(X = 1), (c) P(0 ≤ X < 2), (d) P(X > 1/2), (e) P(1 < X ≤
5).
Solution:
(a) P(X < 1) = F (1−) =
1
2
(b) P(X = 1) = p(1) = F (1) − F (1−) =
11
12
(c) P(0 ≤ X < 2) = F (2−) − F (0−) = 1 −
−
1
2
=
1
4
=
3
4
(d) P(X > 12 ) = 1 − P(X ≤ 12 ) = 1 − F ( 12 ) = 1 −
(e) P(1 < X ≤ 5) = F (5) − F (1) = 1 −
11
12
=
5
12
1
2
=
1
2
1
12
2. Independent trials, each of which results in a success with probability 1/3, are performed 4 times. Let X be the total number of successes, and let Y = cos( π2 X).
(a) (6 points) Find P (X ≥ 3).
(b) (6 points) Find the expectation of Y .
(c) (6 points) Find the variance of Y .
Solution: X is a binomial random variable with parameters (4, 31 ).
(a)
3 4
4
1
2
4
1
P(X ≥ 3) = P (X = 3) + P(X = 4) =
+
3
3
3
4
3
2
1
1
= 4·
+
=
81 81
9
(b)
4
X
π
π
E[Y ] = E[cos( X)] =
cos( j) P (X = j)
2
2
j=0
= 1 · P (X = 0) + 0 · P (X = 1) + (−1)P (X = 2) + 0 · P (X = 3) + 1 · P (X = 4)
0 4 2 2 4 0
4
1
2
4
1
2
4
1
2
=
−
+
0
3
3
2
3
3
4
3
3
16 24
1
7
=
−
+
=−
81 81 81
81
(c)
π
16 24
1
41
E[Y 2 ] = E[cos2 ( X)] =
+
+
=
2
81 81 81
81
2
41
7
− −
Var(Y ) = E[Y 2 ] − (E[Y ])2 =
81
81
3. (16 points) Consider a roulette wheel consisting of 38 numbers 1 through 36, 0, and
double 0. If Smith always bets that the outcome will be one of the numbers 1 through
12, what is the probability that
(a) Smith will lose his first 5 bets;
(b) His first win will occur on his fourth bet?
(c) Find the expected number of bets until his first win (including the winning bet).
Solution: First note that the probability of winning in one bet is 12/38 = 6/19.
13 5
= probability of losing 5 times in a row
(a) 19
3 6 (b) 13
= probability of losing in the first three bets and then winning
19
19
(c) Let X denote the number of bets until the first win. Then X has geometric
distribution with parameter p = 12/38 = 6/19. Therefore, EX = 19/6.
4. (16 points) Suppose that X is uniformly distributed over the interval (−1, 1). Find the
probability density function of Y = log(1 − |X|).
Solution: X ∈ (−1, 1) =⇒ |X| ∈ [0, 1) =⇒ 1 − |X| ∈ (0, 1] =⇒ log(1 − |X|) ≤ 0. Thus,
for a ≤ 0,
P(Y ≤ a) = P(log(1 − |X|) ≤ a) = P(1 − |X| ≤ ea ) = P(|X| ≥ 1 − ea )
1
= 2P(X ≥ 1 − ea ) = 2 (1 − (1 − ea )) = ea .
2
It follows that
fY (a) =
ea , a ≤ 0
0, a > 0
5. (16 points) The time (in hours) required to repair a machine is an exponential random
variable with parameter λ = 13 . Find (a) the probability that the repair exceeds 3
hours; (b) the conditional probability that the repair time is at least 9 hours, given
that its duration exceeds 7 hours.
Solution: Let X = the time in hours required to repair a machine. Then X is exponential with parameter λ = 1/3.
(a) P(X > 3) = e−3λ = e−1
(b) P(X > 9 | X > 7) = memoryless property = P(X > 2) = e−2/3
6. (16 points) A certain basketball player knows that on average he will make 80 percent
of free throw attempts. Use normal approximation to find the probability that in 100
attempts he will be successful at least 90 times.
Solution: Let X = the number of free throw attempts the player will make. Then X
is binomial with parameters (100, 0.8). We approximate X with the normal variable
with parameters (100(0.8), 100(0.8)(0.2)) = (80, 16).
89.5 − 80
X − 80
≥ √
)
P(X ≥ 90) = (continuity correction) = P(X ≥ 89.5) = P( √
16
16
9.5
) = P(Z ≥ 2.375) = 1 − Φ(2.375)
≈ P(Z ≥
4
5
≈ Φ(2.37) +
(Φ(2.38) − Φ(2.37)) = (normal table)
10
= 1 − 0.9912 = 0.0088 .
Without continuity correction one gets
90 − 80
X − 80
≥ √
)
P(X ≥ 90) = P( √
16
16
9
≈ P(Z ≥ ) = P(Z ≥ 2.5) = 1 − Φ(2.5) = (normal table)
4
= 1 − 0.99379 = 0.00621 .
Two points were taken off for solutions without continuity correction.