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Name&Surname:
28 Nov 2016
STAT 240 FALL 2016
APPLIED STATISTICS AND PROBABILITY FOR ENGINEERS
Duration of this exam is 120 minutes. Turn off your cellular phones. All notes are closed.
Formulas are at the back side of this page.
1. (40 points) Consider the joint probability distribution, fXY(X,Y) for variables X and Y
shown in table below. Variable X has three values (1,2,3) and variable Y has two values
(0,1).
Variable X
1
Variable Y
2
3
0 0.12 0.2 0.08
1 0.18 0.3 0.12
a. Calculate Px(X): marginal probability distribution of X.
Px(X=1)=0.3, Px(X=2)=0.5, Px(X=3)=0.2
b. Calculate PY(Y): marginal probability distribution of Y.
PY(Y=0)=0.4, PY(Y=1)=0.6
c. Calculate 𝜇𝑋 : mean of X.
1*0.3+2*0.5+3*0.2=1.9
d. Calculate 𝜇𝑌 : mean of Y.
0*0.4+1*0.6=0.6
e. Calculate 𝜎𝑋2 : variance of X.
(1-1.9)20.3+(2-1.9)20.5+(3-1.9)20.2=0.49
f. Calculate 𝜎𝑌2 : variance of Y.
(0-0.6)20.4+(1-0.6)20.6=0.16
g. Calculate cov(X,Y): covariance of X and Y.
For each cell: (X-𝜇𝑋 )(Y-𝜇𝑌 )P(X=x,Y=y) is calculated and summed = 0
h. Calculate 𝜇𝑋|𝑌=1 : conditional mean of X given Y=1.
Same as (c)
2
i. Calculate 𝜎𝑋|𝑌=1
: conditional variance of X given Y=1.
Same as (e)
j. Prove or disprove that X and Y are independent.
P(X,Y)=P(X)P(Y) for each cell.
2. (10 points) Consider a medical test for a disease. The test has a probability of 0.95
of currently (or positively) detecting an infected person (sensitivity). It has a
probability of 0.90 of identifying a healthy person (specificity). In the population 3%
have the disease. What is the probability that a person testing positive is actually
infected.
We are given P(Infected)=0.03. This implies P(Healthy)=0.97.
What does detecting an infected person mean? Test must be positive for that person. So
for 95% of the infected population test will be positive. This means 0.95 x 0.03 =
0.0285.
What does identifying healthy person mean? Test must be negative given that the person
is healthy, right? So P(N|H)=0.90. So of healthy people 90% will be tested as negative.
Then we write 0.90 x 0.97 = 0.873.
This info is enough to fill-out the table. See below.
The question is P(Infected|Positive)=0.0285 / 0.1255 = 0.227092
Positive Negative
Infected
0.0285
0.0015 0.03
Healthy
0.097
0.873 0.97
0.1255
0.8745
1
3. (5 points) The following circuit operates only if there is a path of functional devices
from left to right. The probability that each device functions is shown on the graph.
Assume that devices fail independently. What is the probability that the circuit operates?
(1-(1-0.9)(1-0.9)) (0.9) (1-(1-0.9)(1-0.9))=0.88209
4. (10 points) The probability that a student passes from a course is 85%.
a. From a group of 12 students find the probability that 9 of them will pass from
the course. (Hint: Bi)
12
𝑃(𝑋 = 9) = ( ) 0.859 0.153 = 0.171976
9
b. Among a group of 200 students what is the probability that 180 or less will pass?
(Hint: Use normal approximation)
np=200*0.85=170 & n(1-p)=200*0.15=30 are both more than 5.
√200 ∗ 0.85 ∗ 0.15 = 5.049752
180.5 − 170
𝑧=
= 2,07931~2.08
5,049752
From the table, P(z<2.08)=0.9812
5. (10 points) The average number of traffic accidents per hour in a town is 4. (Hint: Po)
a. What is the probability that there are 3 or fewer accidents in one hour?
Use the formula; P(X=0)+P(X=1)+P(X=2)+P(X=3)
= 0.018315639+0.073262556+0.146525111+0.195366815 = 0.43347012
b. What is the probability that there are exactly 9 accidents in two hours?
Change lambda to 8 use the formula: P(X=9)= 0.124076917
6. (15 points) Let 𝑋~𝑁(10,36). (Hint: 𝜎 2 = 36)
a. Find (i)P(X=9)=0, (ii)P(X<8)=P(z<-0.33)=1-0.6293=0.3707,
(iii)P(0<X<11)=P(-1.67<Z<0.17)=(1-0.5675)+(1-0.9525)=0.48
(iv)P(X>3) = P(Z>1.17)=0.8790
b. Find x if (i) P(-x<X<x)=0.95 >> z=1.96 so X=1.96*6+10 = 21.76
(ii)P(X>x)=0.90 >> z=-1.28 so X=-1.28*6+10 = 2.32
7. (10 points) In a large corporate computer network, user log-ons to the system can be
modeled as a Poisson process with a mean of 25 log-ons per hour. What is the probability
that there are no log-ons in the next 6 minutes (0.1 hour)? (Hint: Expo)
∞
𝑃(𝑋 > 0.1) = ∫ 25𝑒 −25𝑥 𝑑𝑥 = 𝑒 −25(0.1) = 0.082
0.1
Name&Surname:
28 Nov 2016
𝑃(𝐻∩𝐸)

Conditional Probability:

Bayes’ Formula: Pr(𝐻|𝐸𝑖 ) =

Binomial Distribution: Pr(𝑋 = 𝑥) = 𝑓(𝑥) = (𝑛𝑥)𝑝 𝑥 (1 − 𝑝)𝑛−𝑥
𝑃(𝐻|𝐸) =
𝑃(𝐸)
Pr(𝐸𝑖 |𝐻 )Pr(𝐻)
Σ𝑗 Pr(𝐸𝑗 |𝐻 )Pr(𝐻)
𝜇 = 𝐸[𝑋] = 𝑛𝑝 and 𝜎 2 = 𝑉𝑎𝑟(𝑋) = 𝑛𝑝(1 − 𝑝)
𝑒 −𝜆 𝜆𝑥

Poisson Distribution: Pr(𝑋 = 𝑥) = 𝑓(𝑥) =




Exponential Distribution: 𝑓(𝑥) = 𝑒 −𝜆𝑥 , 𝜇 = 1/𝜆 and 𝜎 2 = 1/𝜆2
Mean of a random variable: 𝜇𝑋 = 𝐸(𝑋) = ∑ 𝑥 ∙ 𝑃(𝑋 = 𝑥)
Variance of a random variable: 𝜎 2 = 𝐸[(𝑋 − 𝜇𝑋 )2 ] = ∑(𝑥 − 𝜇𝑋 )2 ∙ 𝑃(𝑋 = 𝑥)
Covariance of two variables:
𝑥!
𝜇 = 𝜎2 = 𝜆
𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸[(𝑥 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 ) = ∑ ∑(𝑥 − 𝜇𝑋 )(𝑌 − 𝜇𝑌 )𝑃(𝑋 = 𝑥, 𝑌 = 𝑦)
Standard Normal Table for P(Z<z) and z>0.