Stat 330
Exam II Practice Questions
Spring 2015
1. Continuous Distributions
The length of time (in hours) taken by a student to complete a one hour exam is a continuous random
variable X with probability density function fX defined by fX (x) = cx2 + x, 0 < x ≤ 1, for some
constant c, and zero otherwise.
(a) Find the value of c.
The two conditions for a probability density function are, that the function has only positive values
(which is true for all positive c in this problem) and the integral over R is 1:
1
Z ∞
Z 1
c 1
c
1 Setting 1 =
f (x)dx =
(cx2 + x)dx = x3 + x2 = +
3
2
3
2
−∞
0
0
⇒ c = 1.5.
For c = 1.5 the above function is a probability density function.
(b) By integration, find the distribution function of X, FX .

Z t
Z t
 0
0.5t3 + 0.5t2
FX (t) =
f (x)dx =
(cx2 + x)dx =

−∞
0
1
t<0
0≤t≤1
t>1
(c) Find the probability that a student completes the exam in less than half an hour.
Let X be the time the student needs for the exam, X ∼ FX . Then
P (X < 0.5) = FX (0.5) = 0.5 · 0.53 + 0.5 · 0.52 = 0.1875.
(d) Given that a student takes longer than fifteen minutes to complete the exam, find the probability
that he/she requires at least half an hour.
P (X > 0.5 | X > 0.25) =
P (X > 0.25 ∩ X > 0.5)
P (X > 0.5)
1 − FX (0.5)
1 − 0.1875
=
=
=
= 0.8455.
1 − P (X < 0.25)
1 − P (X < 0.25)
1 − FX (0.25)
1 − 0.039
(e) In a class of sixty students, find the probability that at least half the students complete the exam
in fewer than fifty minutes.
The probability for each student to finish in fewer than fifty minutes is p = P (X ≤ 5/6) = 0.64
Let Y be the number of students finishing in fewer than fifty minutes. Y then has a binomial
distribution with n = 60 and p = 0.64.
Need P (Y ≤ 30) where Y ∼ Bin(60, 0.64). By central limit theorem Y is distributed approximately as N (60 · 0.64, 60 · 0.64 · (1 − 0.64)), i.e as N (38.4, 13.824)
P (Y ≤ 30)
=
30 − 38.4
P (Z ≤ √
)
13.824
Φ(−2.26)
=
1 − Φ(2.26) = 1 − .0119 = .9881
≈
2. Binomial, Exponential, and Erlang Distributions
Suppose that the times till failure of circuit packs of a particular design are exponential with rate
λ = 0.4 (per year).
(a) Find the probability that a particular single circuit pack lasts at least 2.0 years.
Let Y be the time until failure (in years); then Y ∼ Exp(0.4)
P (Y > 2) = 1 − P (Y ≤ 2) = 1 − (1 − e−0.4·2 ) = e−0.8 = 0.45.
(b) Suppose that a switching system uses 20 of these circuit packs (simultaneously and independently).
Let X be the number of the original circuit packs that are still installed and working in the
switching system, 2.0 years into the life of the system. Use your answer to a) and find the
1
probability that P (X > 18). What is the distribution of X? (Give both a name and appropriate
parameter value(s).)
Since each circuit works independently, we have 20 independent experiments with binary outcomes:
still working/ failed. X is therefore a binomial variable with n = 20 and p = 0.45.
P (X > 18) = P (X ≥ 19) = 20 · 0.4519 · 0.55 + 0.4520 = 2.95 · 10−6 .
(c) In a particular application, only one of these circuit packs is in service at a time. In addition
to the in-service circuit pack, there are 2 back-up packs. (The first back-up is placed in service
when the original pack fails and the second when the first replacement fails.) Let T be the total
operating time of the original and 2 back-up packs. What are the mean and variance of T ?
T is the sum of three exponential variables, T is therefore an Erlang variable wih stage parameter
k = 3 and λ = 0.4 (in years).
1
= 7.5 (years)
0.4
1
= 18.75(years2 )
V ar[T ] = 3 ·
0.42
P (T > 5) = 1 − P (T ≤ 5) = 1 − FY (5)
E[T ]
=
=
3·
1 − (1 − FX (2))
= FX (2)
=
where Y ∼ Erlang(3, 0.4)
where X ∼ P oi(0.4 · 5)
where X ∼ P oi(2)
0.677.
(d) If 100 of these circuit packs are purchased by a telecommunications company and placed into
service (in independent applications), approximate the probability that the mean life of these 100
packs exceeds 2.2 years.
The expected life of each individual circuit is 1/0.4 = 2.5 years with a variance of 1/0.42 = 6.25.
The average life until failure of all these circuits X has (according to the central limit
theorem) approximately a normal distribution with mean = 2.5 and variance 6.25/100.
P (X > 2.2) = 1 − P (X ≤ 2.2) ≈ 1 − Φ(
2.2 − 2.5
) = Φ(1.2) = 0.8849.
0.25
3. Normal Distribution
The average monthly balances on checking accounts of the ISU Bank are normally distributed with
mean $1000 and standard deviation $500. There is a monthly fee if the average monthly balance is
below $1500.
(a.) What is the probability that a customer can avoid the monthly fee?
Let X = average monthly balance of an account at ISU Bank. It is given that X ∼ N (1000, 5002 )
P (avoid fee)
=
=
=
P (X > 1500)
1500 − 1000
P Z>
= P (Z > 1)
500
1 − Φ(1) = 1 − .841 = .159
(b.) What is the probability that a customer can avoid the fee given that the balance on his/her
account exceeds $1000?
P (X > 1500 | X > 1000)
=
=
=
P (X > 1500 ∩ X > 1000)
P (X > 1000)
P (X > 1500)
P (X > 1000)
1 − Φ(1)
1 − .841
=
= .318
1 − Φ(0)
1 − .5
2
(c.) What is the probability that the balance of a customer is between $ 900 and $1100?
900 − 1000
1100 − 1000
P (900 ≤ X ≤ 1100) = P
≤Z≤
500
500
= P (−.2 ≤ Z ≤ .2) = Φ(.2) − Φ(−.2)
=
Φ(.2) − (1 − Φ(.2)) = 2Φ(.2) − 1 = 2 × .579 − 1 = .158
On the other hand, AMES Bank customers have a normally distributed balances with mean $1200 and
variance 90000. The customer account balances at each bank are independent.
(d.) What is the distribution of the difference between the balances of ISU Bank and AMES Bank
customers? Give an appropriate distribution and parameter(s).
Let Y = average monthly balance of an account at AMES Bank. It is given that Y ∼ N (1200, 3002 )
2
Let D = X − Y . Then D ∼ N (µD , σD
) where
µD
=
µX − µY = 1000 − 1200 = −200
2
σD
=
=
2
σX
+ σY2
Cov() term =0 as X and Y are independent
2
2
500 + 300 = 340, 000
(e.) What is the probability that a customer at ISU Bank has more money on his account than a
customer from AMES Bank? (use your answer to (d))
P (X > Y )
=
P (X − Y > 0)
=
P (D > 0)
= P
0 − (−200)
Z>p
(340, 000)
!
= P (Z > .343) = 1 − Φ(.343) = 1 − .633 = .367
4. Central Limit Theorem
(a.) Suppose 49% of students are in favor of a new administrative decision. The Iowa State Daily asks
1000 randomly selected students if they are in favor of the new decision. Assume the responses are
iid Bernoulli trials. Approximate the probability that more than 51% of the students interviewed
are in favor of the new decision.
Let X = # of students in favor out of 1000. Thus X ∼ Bin(1000, .49)
By the CLT, X ∼
˙ N (1000(.49), 100(.49)(.51)) ; that is X ∼
˙ N (490, 249.9). Thus,
510 − 490
P (X > 510) ≈ P (Z > √
)
249.9
= P (Z > 1.27)
=
1 − Φ(1.27) = 1 − .898 = .102
(b.) A psychologist wants to estimate the mean intelligence quotient (IQ) of the students of a university.
To do so, she takes a sample of size n of the students and measures their IQs. Then she finds the
average of these numbers. She believes that the IQs of these students are iid random variables with
a variance of 170. How big should she choose the sample so that the (approximate) probability
that her estimate differs from the true mean by more than 1 is .02?
Let X n = sample average IQ and let the unknown mean IQ be µ
By the CLT, we know that, X n ∼
˙ N (µ, 170/n). We want to find n so that P (|X n − µ| > 1) = .02
P (|X n − µ| > 1)
= P (X n − µ < −1) + P (X n − µ > 1)
!
!
Xn − µ
−1
Xn − µ
1
= P p
<p
+P p
>p
(170/n)
(170/n)
(170/n)
(170/n)
!
!
−1
1
≈ P Z<p
+P Z > p
(170/n)
(170/n)
p
p
= Φ(− n/170) + 1 − Φ( n/170)
p
= 2(1 − Φ( n/170))
3
Set this equal to .02 and
p
2(1 − Φ( n/170))
p
Φ( n/170)
p
n/170
n
solve for n:
=
.02
=
.99
=
2.33
=
√
(2.33 170)2 ≈ 923
4