Will Landau
September 19, 2011
STAT 231
Problem Set 4 Solutions
Exercise 4.1.
I’ll walk you through how to generate one of the pmfs. Let’s say I want to
generate the pmf for Poisson(λ=4). My data looks ilke this so far:
I double-click on the title of the far-right column (the one I want to fill in) and
get this:
1
I go to Column Properties > Formula:
I get the following screen and click on the ”Edit Formula” button:
2
The formula-editing window comes up. I go to Discrete Probability > Poisson
Probability as I show below (Note: Poisson Distribution gives the cdf, not the
pmf):
And voila:
3
I enter ”4” where the formula says ”lambda” and click and drag ”x” from the
upper left area of the window to where the formula says ”k”. I click the
”Apply” button, and the column fills up like so:
There’s no built-in formula for the geometric distribution, so you’ll have to
write your own in a similar way. You’ve already used Graph Builder, so I
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didn’t walk you through that bit. Here are the plots:
5
6
7
Exercise 4.2.
Note: you can add rows by clicking on one of the red triangles in the upper
left-hand corner of a data table. Here are my simulated plots. They should
look like the analogous plots for the previous problem.
Exercise 4.3 (Devore 3.48, 3.49, 3.50, 3.51, 3.60).
3.48
a. P (X ≤ 2) =
P2
i=0
b(i; 25, .05) =
P2
b. P (X ≥ 5) = 1 − P (X ≤ 4) = 1 −
P4
i=0
8
25
i
i=0
(.05)i (.95)25−i = .873
b(i; 25, .05) =
P4
25
i
(.05)i (.95)25−i = 1 − .993 = .007
P4
P4
i
25−i
c. P (1 ≤ X ≤ 4) = i=1 b(i; 25, .05) = i=1 25
= .716
i (.05) (.95)
i
25−0
d. P (X = 0) = b(0; 25, .05) = 25
= .277
0 (.05) (.95)
1−
i=0
e. E(X) = np = (25)(.05) = 1.25
V (X) p
= np(1 − p) = (25)(.05)(.95) = 1.1875
σX = V (X) = 1.0897
3.49 Let X ∼ Bin(n = 6, p = .1), the number of goblets of six
randomly-selected ones that have cosmetic flaws.
a. P (X = 1) = b(1; 6, .1) = 61 .11 .95 = .3543
b. P (X ≥ 2) = 1 − P (X < 2) = 1 − P (X = 0) − P (X = 1) = 1 − b(0; 6, .1) −
b(1; 6, .1) = 1 − 60 .10 .96 − 61 .11 .95 = 1 − .3543 − .5314 = .1143
c. If we find four goblets that are not ”seconds” in at most five draws, then
one of the following two independent events must have occurred:
i. Four goblets are selected and none of them are seconds. Since
draws are independent and P(a random goblet is not a second) =
1-.1 = .9, the probability of this event is .94 = .6561.
ii. Select the first four goblets, one of which has a defect and the last
of
is good.
The probability of this event is
4which
0
4
(.1)
(.9)
·
(.9)
= .2624
0
Hence, we get a probability of .6561 + .2624 = .9185 for our event of
interest.
3.50 Let X be the number of incoming calls involving fax messages in a
random sale of 25 calls.
a. P (X ≤ 6) = B(6; 25, .25) = .561
b. P (X = 6) = b(6; 25, .25) = .183
c. P (X ≥ 6) = 1 − P (X ≤ 5) = 1 − B(5; 25, .25) = .622
d. P (X > 6) = 1 − P (X ≤ 6) = 1 − .561 = .439
3.51
a. E(X) = np = 25(.25) = 6.25
b. V ar(X) = np(1 − p) = 25(.25)(.75) = 4.6875, so SD(X) = 2.165
c. P (X > 6.25 + 2(2.165)) = P (X > 10.58) = 1 − P (X ≤ 10.58) =
1 − P (X10) = 1 − B(10; 25, .25) = .030
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3.60 Use X and h(X) as given in the hint.
h(X) = 1 · X + 2.25(25 − X) = 62.5 − 1.5X, so
E(h(X)) = 62.5 − 1.5E(X) = 62.5 − 1.5np = 62.5 − (1.5)(25)(.6) = $40.00
Exercise 4.4.
a. Y ∼ Geometric(p = .05). I take Y to be the number of boards inspected
before finding a defective one so that the pmf on page 126 of your book
will make sense to use.
b. Use the pmf on page 126. P (Y = 4) = (1 − p)4 p = .954 · .05 = 0.0407
P (1 ≤ Y ≤ 3) = P (Y = 1) + P (Y = 2) + P (Y = 3) =
.951 · .05 + .952 · .05 + .953 · .05 = 0.1355
1−p
.95
p = .05 =
1−p
.95
= p2 = .05
2
c. E(Y ) =
Var(Y )
19
= 380
Exercise 4.5.
a. W ∼ Geometric(p = .25). I take W to be the number of calls received
before getting a fax so that the pmf on page 126 of your book will make
sense to use.
b. P (W = 4) = (1 − p)4 · p = .754 · .25 = 0.0791
P (W > 4) = 1 − P (W ≤ 4) = P (W = 0) + P (W = 1) + P (W =
2) + P (W = 3) = .750 · .25 + .751 · .25 + .752 · .25 + .753 · .25 = 0.6835
1−p
.75
p = .25 =
.75
= 1−p
p2 = .252
c. E(W ) =
Var(W )
3
= 12, so SD(W ) =
√
12 = 3.4641
Exercise 4.6 (Devore 4.1, 4.3, 4.4, 4.11, 4.11a-e, 4.12, 4.13a-c).
4.1
a. Here is the pdf:
10
R∞
R5
Area under the curve = −∞ f (x)dx = 3 (.075x + .2)dx =
(.0375x2 + .2x |53 = .0375 · 52 + .2 · 5 − (.0375 · 32 + .2 · 3) = 1
b. P (X ≤ 4) = P (X < 4) since X is a continuous random variable.
R4
R4
P (X ≤ 4) = −∞ f (x)dx = 3 (.075x + .2)dx = (.0375x2 + .2x |43 =
.0375 · 42 + .2 · 4 − (.0375 · 32 + .2 · 3) = 0.4625
c. Similarly to the above,
R 4.5
R4
P (3.5 ≤ X ≤ 4.5) = 3.5 f (x)dx = 3.5 .5(.075x + .2)dx =
(.0375x2 + .2x |43.5 .5 = .0375 · 4.52 + .2 · 4.5 − (.0375 · 3.52 + .2 · 3.5) = .5
P (X < 4.5) =
0.2781
R∞
4.5
f (x)dx =
R5
4.5
(.075x + .2)dx = (.0375x2 + .2x |54.5 =
4.3
a. Here is the graph of the pdf:
11
R2
3
.09375(4 − x2 )dx = .09375(4x − x3 ) |20 = .5
R1
c. P (−1 < X < 1) = −1 .09375(4 − x2 )dx = .6875
b. P (X > 0) =
0
d. P (X < −.5 or X > .5) = 1 − P (−.5 ≤ X ≤ .5) =
R .5
1 − −.5 .09375(4 − x2 )dx = 1 − .3672 = .6328
4.4
a. f is continuous and clearly nonnegative. Now, we verify that it
integrates to 1:
Z
∞
Z
f (x; θ)dx =
−∞
0
∞
2
2
x −x2 /2θ2
e
dx = −e−x /2θ |∞
0 = 0 − (−1) = 1
θ2
R 200
b. P(X is at most 200) P (X ≤ 200) = −∞ f (x; 100)dx =
R 200 x −x2 /2θ2
2
2
dx = −e−x /2·100 |20 00 = −.1353 + 1 = .8647
1002 e
0
P(X is less than 200) = P(X is at most 200) because X is a continuous
random variable.
P(X is at least 200) = P (X ≥ 200) − 1 − P (X < 200) = .1353
R 200
R 200 x −x2 /2θ2
dx =
c. P (100 ≤ X ≤ 200 = 100 f (x; 100)dx = 100 100
2e
−x2 /2·1002 200
−e
|100 = .4712
d. For X > 0,
P (X ≤ x) =
Ry
−∞
f (y; θ)dy =
Ry
y −y 2 /2θ 2 x
|0 =
e
0 θ2
4.11a-e
a. P (X ≤ 1) = F (1) = 12 /4 = .25
b. P (.5 ≤ X ≤ 1) = F (1) − F (.5) =
3
16
12
= .1875
1 − e−x
2
/2θ 2
15
c. P (X > .5) = 1 − P (X ≤ .5) = 1 − F (.5) = 16
= .9375
√
2
d. .5 = F (e
µ) = µe4 , so µ
e2 = 2 ⇒ µ
e = 2 = 1.414
e. f (x) = F 0 (x) =
x
2
for 0 ≤ x < 2 and f (x) = 0 otherwise
4.12
a. P (X < 0) = F (0) = .5
b. P (−1 ≤ X ≤ 1) = F (1) − F (−1) =
11
16
= .6875
c. P (X > .5) = 1 − P (X ≤ .5) = 1 − F (.5) = 1 − .6836 = .3164
d. f (x) = F 0 (x) = 0 +
3
32 (4
−
3x2
3 )
=
2
.09375(4 − x ) for −2 ≤ x < 2 and 0 otherwise
e. By definition, µ
e is such that F (e
µ) = .5. We know F(0) = .5 from part a,
so µ
e= 0
4.13a-c
a. We need f (x) to integrate to 1, so
R∞
−3 ∞
1 = 1 xk4 dx = −k
|1 = 0 − (− k3 )(1) = k3 , so k=3
3 x
Rx
Rx
b. F (x) = −∞ f (x)dx = 1 x34 dx = −x−3 |x1 =
1 − x−3 for x > 1 and 0 otherwise
c. P (X > 2) = 1 − P (X ≤ 2) = 1 − F (2) = 1 − (1 − 1/8) = .125
d. P (2 < X < 3) = F (3) − F (2) = (1 − 1/27) − (1 − 1/8) = .088
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