MSA 8190 Statistical Foundations

Assignment 4 Pages

Georgia State University

MSA

Contributed by

Prajwal Hallale

s
r
s
r
Both increase.(e)
788867..2
=
2
30.0
30.0−1
=
2
p
p−1
=σ3333 and.= 3
30.0
1
=
p
1
) =X(E30, then.= 0pIf(d)
544212..= 1
2
471.0
471.0−1
=
2
p
p−1
=σand
123142.= 2
471.0
1
=
p
1
) =X(E471. Therefore.= 0phas a geometric distribution with
Xbe the number of women to sample before selecting a married woman. ThenXLet(c)
1044871..= 0
3
471.0(b)
1318051..471 = 0.0×471).0−(1×471).0−woman is married which is equal to (1
This is equal to the probability that the ﬁrst two women are not married and the third(a)
Solution:
event aﬀect the mean and standard deviation of the wait time until success?
Based on your answers to parts (c) and (d), how does decreasing the probability of an(e)
expect to sample before selecting a married woman? What is the standard deviation?
If the proportion of married women was actually 30%, how many women would you(d)
woman? What is the standard deviation?
On average, how many women would you expect to sample before selecting a married(c)
What is the probability that all three randomly selected women are married?(b)
third woman selected is the only one who is married?
We randomly select three women between these ages. What is the probability that the(a)
over are married.
The 2010 American Community Survey estimates that 47.1% of women ages 15 years and
Problem 1
Solution for Assignment 2
Fall 2022
MSA 8190 Statistical Foundations

Page 1
2
Problem 2
A not-so-skilled volleyball player has a 15% chance of making the serve, which involves
hitting the ball so it passes over the net on a trajectory such that it will land in the opposing
team’s court. Suppose that her serves are independent of each other.
(a) What is the probability that on the 10th try she will make her 3rd successful serve?
(b) Suppose she has made two successful serves in nine attempts. What is the probability
that her 10th serve will be successful?
(c) Even though parts (a) and (b) discuss the same scenario, the probabilities you calcu-
lated should be diﬀerent. Can you explain the reason for this discrepancy?
Solution:
(a) Let X be the number of trials until she get the 3rd successful serve. Then, X follows
a negative binomial distribution with p = 0.15 and r = 3. Then P(X = 10) =
dnbinom(10 − 3, 3, 0.15) = 0.03895012.
(b) Since the each trial is independent of the other trials, this probability is 0.15.
(c) Let’s assume E is the event of have a successful serve on the 10th trial. Moreover, let
F be the event of have 2 successful serves in the ﬁrst 9 trials. Part (a) is asking about
the probability of the intersections of events E and F which is
P(E ∩E) = P(E)P(F ) = dbinom(2, 9, 0.15) × 0.15 = 0.03895012,
where the ﬁrst quality follows by the independence of E and F . We can also see
that P(X = 10) = P(E ∩ E) = 0.03895012. But, in part (b) we are looking for the
conditional probability of P(E|F ) = P(E) = 0.15, where again the ﬁrst quality follows
by the independence of E and F .
Problem 3
In a particularly delicate process for manufacturing a particular type of experimental silicon
chip, the probability of successfully producing a marketable chip is 70%. The company needs
to produce 12 marketable chips, but only has the budget to manufacture 15 chips.
(a) What is the probability that they will be able to produce 12 marketable chips in at
most 15 attempts?
(b) How many chips described in this problem need to be manufactured so that the prob-
ability of getting at least 12 marketable chips is 95%.
Solution:
Let X be the number of chips to be produced until 12 of them are marketable. Then, X has
a negative binomial distribution with size 12 and p = 0.70
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3
(a) P(X ≤ 15) = pnbinom(15 − 12, 12, 0.7) = 0.2968679
(b) qnbinom(0.95, 12, 0.7)+12=22
Problem 4
A coﬀee shop serves an average of 75 customers per hour during the morning rush.
(a) Which distribution we have studied is most appropriate for calculating the probability
of a given number of customers arriving within one hour during this time of day?
(b) What are the mean and the standard deviation of the number of customers this coﬀee
shop serves in one hour during this time of day?
(c) Would it be considered unusually low if only 60 customers showed up to this coﬀee
shop in one hour during this time of day?
(d) Calculate the probability that this coﬀee shop serves 70 customers in one hour during
this time of day?
Solution:
(a) A Poisson distribution with λ = 75.
(b) E(X) = λ = 75, σ =
√
λ =
√
75 = 8.66.
(c) P(X ≤ 60) = ppois(60, 75) = 0.0433398, which seems a low probability.
(d) P(X = 70) = dpois(70, 75) = 0.04016033
Problem 5
Suppose f (X) = 1.5(x −1)
2
for 0 < x < 2. Determine the cumulative distribution function,
mean and variance of X.
Solution:
Note that
R
x
0
1.5(u − 1)
2
du = 0.5(u − 1)
3
|
x
0
= 0.5(x − 1)
3
+ 0.5. Then,
F (X) =



0 x < 0
0.5(x − 1)
3
+ 0.5 0 ≤ x < 2
1 2 ≤ x
E(X) =
Z
2
0
xf(x)dx =
Z
2
0
x × 1.5(x − 1)
2
dx = 1.5
Z
2
0
(x
3
− 2x
2
+ x)dx
= 1.5

x
4
4
−
2x
3
3
+
x
2
2





2
0
= 6 − 8 + 3 = 1
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4
E(X
2
) =
Z
2
0
x
2
f(x)dx =
Z
2
0
x
2
× 1.5(x − 1)
2
dx = 1.5
Z
2
0
(x
4
− 2x
3
+ x
2
)dx
= 1.5

x
5
5
−
x
4
2
+
x
3
3





2
0
=
48
5
− 12 + 4 =
8
5
= 1.6
=⇒ Var(X) = E(X
2
) − [E(X)]
2
= 1.6 − 1
2
= 0.6
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