Your first and last name:___________________________________ SSN:________________________
STAT 211
SPRING 2002
You have 50 minutes to complete this exam. You can only use your own calculator and there is penalty of
5 pt. if you separate your exam with any reason.
There is no partial credit on this exam
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If you did not mark your exam form on the scantron or did not place it in the right folder at the end of the
exam, your grade will not be changed.
If you caught cheating, you will get a grade of zero. Good Luck.
EXAM 2 - FORM A
Let X be a continuous random variable with the legitimate probability density function (pdf),
 3x 2
,

f ( x)   8
0,

0 x2
.
otherwise
Answer questions 1 to 3 using this information.
1.
Which of the following is the F(x) when 0  x  2 using the legitimate pdf?
(a) 0
2
(b) 3 x / 8
3
(c) x / 8
3
(d) x / 3
(e) 1
2.
Which of the following is the 50th percentile of the random variable X using the legitimate pdf?
(a) -4
(b) -1.5874
(c) -1.1547
(d) 1.1547
(e) 1.5874
3.
Which of the following is the expected value of 2X using the legitimate pdf?
(a) 3/8
(b) 3/4
(c) 1
(d) 3/2
(e) 3
Use the following choices to answer the next 3 questions.
(a) Binomial Distribution
(b) Hypergeometric Distribution
(c) Poisson distribution
(d) Exponential Distribution
(e) Geometric distribution
4.
Suppose that 20% of the applicants to an engineering school are women. An admissions committee
reviews applications in groups of 100. Which distribution would you use to find the probability that
the number of women in the next group is greater than 25?
5.
A computer time-sharing system receives teleport inquiries at an average rate of 0.1 per millisecond.
Which distribution would you use to find the probability that the number of inquiries in a particular 50
millisecond stretch will be between 5 and 12?
6.
A shipment contains 100 printed circuit boards. A sample of 10 boards will be tested. If 2 or fewer
defectives are found, the shipment will be accepted. Assuming that 10% of the boards in the shipment
are defective, Which distribution would you use to find the probability of accepting the shipment?
7.
Find the value of c for P(Z  c)=0.2236 where z is the standard normal random variable.
(a) -0.58
(b) -0.76
(c) 0.58
(d) 0.76
(e) 1
8.
Find the value of c for P(-c  Z  c)=0.5528 where z is the standard normal random variable.
(a) -0.58
(b) -0.76
(c) 0.58
(d) 0.76
(e) 1
9.
Find the value of c for P(|Z|  c)=0.4472 where z is the standard normal random variable.
(a) -0.58
(b) -0.76
(c) 0.58
(d) 0.76
(e) 1
When you toss a fair coin 4 times and count the number of tails (x), the legitimate probability distribution is
x
0
1
2
3
4
P(x)
1/16
1/4
6/16
1/4
1/16
Answer questions 10 to 12 using this information.
10. What is the probability that at least 2 tails are observed?
(a) 2/16
(b) 5/16
(c) 11/16
(d) 15/16
(e) The probability distribution is not legitimate to compute it.
11. Which of the following is the expected number of tails?
(a) 10/16
(b) 1
(c) 30/16
(d) 95/64
(e) 2
12. Which of the following is the variance of the number of tails?
(a) 10/16
(b) 1
(c) 30/16
(d) 95/64
(e) 2
13. How would you determine if the given data come from a normal distribution?
(a) I would graph data, x versus any f(x) to see if they make a symmetric graph.
(b) I would graph data, x versus any f(x) to see if they make a 45 line.
(c) I would graph the ordered data with their expected normality values to see if they make a
symmetric graph.
(d) I would graph the ordered data with their expected normality values to see if they make a 45 line.
(e) I would graph data, x versus exponential f(x) to see if they make a 45 line.
14. Which of the following is the answer for (5) ?
(a) 4
(b) 5
(c) 24
(d) 120
(e) 720
15. If time between failures are exponentially distributed with =1, what is the median time between
failures?
(a) -0.50
(b) -0.69
(c) 0.69
(d) 0.50
(e) 1
16. If time between failures are exponentially distributed with =1, what is the probability that the next
failure occur in less than 5 minutes?
(a) 0
(b) 0.0067
(c) 0.6321
(d) 0.9933
(e) 1
A joint pdf for x and y is defined by
 x  y,
f ( x, y )  
0,
if
0  x 1
otherwise
and
Answer questions 17 and 18 using this information.
17. Which of the following is the marginal pdf for X?
(a) x+0.5 for all x
(b) x, 0<x<1
(c) x+0.5, 0<x<1
(d) y, for all y
(e) y+0.5, 0<y<1
18. Which of the following is the f( y | x=0.5)?
(a)
y+0.5, 0<y<1
(b)
0.5+y, 0<x<1
(c)
x+y, 0<y<1
(d)
x+0.5, for all x
(e)
0.5+x, 0<x<1
0  y 1
19. If E(X)=5 and Var(X)=10, find the expected value of
(a)
(b)
(c)
(d)
(e)
X 2
3
.
2/3
4/9
1
3
5
20. Random sample of size three taken without replacement from the following student grade-point
averages: 2.0, 3.0, 4.0, 5.0 created the sample space as S={(2,3,4),(2,3,5),(2,4,5),(3,4,5)}. Which of
the following is the sampling distribution of the median (the middle data) for this situation?
(a) Median=3.5 with corresponding probability 1
(b) Median=3,4 with corresponding probabilities 1/2, 1/2
(c) Median=2,3,4 with corresponding probabilities 1/4, 1/2, 1/4
(d) Median=2,3,4,5 with corresponding probabilities 1/4, 1/4, 1/4, 1/4
(e) All of the above
Weight watcher's program has two parts to attend. For a randomly selected person, let X=the number of
points cumulated on the first part, Y=the number of points cumulated on the second part, and suppose that
the joint pmf of X and Y is given in the accompanying tabulation. (Hint: E(X)=7, E(Y)=5, E(XY)=35)
y
x
0
10
0
0.15
0.15
10
0.35
0.35
Answer questions 21 to 24 using this information.
21. Which of the following is the pmf of Y?
(a) x=0, 10 with corresponding probabilities 0.50, 0.50.
(b) x=0, 10 with corresponding probabilities 0.30, 0.70.
(c) y=0, 10 with corresponding probabilities 0.50, 0.50.
(d) y=0, 10 with corresponding probabilities 0.30, 0.70.
(e) all probabilities add up to 1 so it is a legitimate pmf.
22. If the total number of points, X+Y is recorded, which of the following is the P(X+Y=10)?
(a) 0.15
(b) 0.30
(c) 0.35
(d) 0.50
(e) 0.70
23. Are X and Y independent?
(a) Yes because p(x,y)=p(x)p(y) for all x and y
(b) No because p(10,10)p(x=10)p(y=10)
(c) Yes because p(0,0)=p(x=0)p(y=0)
(d) Yes because p(x,y) p(x)p(y) for all x and y
(e) It is not possible to determine
24. Which of the following is the covariance between X and Y?
(a) 0
(b) 1
(c) 5
(d) 7
(e) 35
25. If X1 and X2 are normally independently distributed random variables with the mean 10 and the
variance 16, What is the probability that 3 X 1  X 2 is less than 10 or greater than 30?
(a) 0.2148
(b) 0.4296
(c) 0.5704
(d) 0.7852
(e) 0.7906
Answer Key:
1.c
2.e
11.e
12.b
21.c
22.d
3.e
13.d
23.a
4.a
14.c
24.a
5.c
15.c
25.b
6.b
16.d
7.d
17.c
8.d
18.a
9.d
19.c
10.c
20.b
Formulas
Discrete probability distribution, p(x) is legitimate if
 0  p(x)=P(X=x)  1 for all x where X is a discrete random variable (r.v).

 p ( x)  1
all x
Cumulative Distribution Function, (CDF) for discrete X:
F ( x)  P( X  x) 
x
 P( X  y )
for all y
P(a  X  b)  F (b)  F (a  1) and P( X  a)  F (a)  F (a  1) where a and b are
The expected value of a discrete variable, X :
 x  p( x)
  E( X ) 
for all x
The variance of a discrete variable, X: 2 = Var(X) =
E( X ) 
2
x
2
E[( X   ) 2 ]  E ( X 2 )   2
where
 p ( x)
for all x
The standard deviation of numerical variable, X:  =
2
Bernoulli Distribution: It is based on Bernoulli trial (an experiment with two, and only two, possible
outcomes). A r.v. X has a Bernoulli(p) distribution if
P(X=x)=
p x  (1  p)1x ,
x  0,1 and 0p1. where E(X) = p and Var(X) = p(1-p)
Binomial Distribution: Approximate probability model for sampling without replacement from a finite
dichotomous population. X~Binomial(n,p).
 n fixed trials
 each trial is identical and results in success or failure
 independent trials
 the probability of success (p) is constant from trial to trial
 X is the number of successes among n trials
n
P( X  x)     p x  (1  p) n x ,
 x
E(X) = np
and
x  0,1,2,...., n
Var(X) = np(1-p)
Hypergeometric Distribution: Exact probability model for the number of successes in the sample.
X~Hyper(M,N,n)
 M  N  M 
 

x  n  x 

P( X  x) 
,
N
 
n 
max( 0, n  N  M )  x  min( n, M )
Let X be the number of successes in the sample, n be the sample size, N be the population size, and M be
the number of successes in the population
M
M
where
is the proportion of successes in the population.
N
N
N n
N n
M  M
 n   1   where
Var(X) =
is the finite population correction factor.
N 1
N 1
N 
N
E(X) = n 
Poisson Distribution: The probability of an arrival is proportional to the length of waiting time.
e   x
P( X  x) 
, x  0,1,2,3,...........,
x!
 0
: intensity parameter (mean rate, expected number of occurrences).
X : number of occurrences per given period

i
i 0
i!
e 

Note that
and E(X)=Var(X)=.
Continuous probability distribution, p(x) is legitimate
(i)
f(x)  0, for all x

(ii)
 f ( x)dx  1 = area under the entire graph of f(x).

x
Cumulative Distribution Function for continuous X (cdf) : F(x)= P ( X  x) 
 f ( y)dy.

F(-)=0, F()=1, P(a  X  b)  P(a  X  b)  F (b)  F (a ) , P(X>a)=P(Xa)=1-F(a)
Obtaining f(x) from F(x) : If X is a continuous r.v. with pdf f(x) and cdf F(x), then at every x at which the
derivative exists, F`(x)=f(x).
Percentile of a continuous distribution: Let p be a number between 0 and 1. the (100p)th percentile of
the
distribution
of
a
continuous
r.v.
X,
denoted
by
r(p),
is
defined
by
r ( p)
p  F (r ( p))  P( X  r ( p)) 
 f ( y)dy.
where P(Xmedian)=P(X>median)=0.50

Expected value for the continuous random variable, X:
  E( X ) 

 x  f ( x)dx.

Variance for the continuous random variable, X:
  Var ( X )  E ( X )  
2
2

2
where
E( X ) 
2
x
2
 f ( x)dx.

1
,
ba
X 
Uniform Distribution: X ~U[a,b] then f ( x) 
Normal Distribution: X ~ N ( , 2 ) and z 

axb
~ N (0 , 1)
Normal Approximation to the Binomial Distribution:
Let X be a binomial r.v. based on n trials with success probability p. Then if the binomial probability
histogram is not too skewed, X has approximately a normal distribution with  = np and
 x  0.5  np 
 (check if np10 and n(1-p)10 to use the

np
(
1

p
)


  np(1  p) then P( X  x)  
formula).
The Gamma Distribution:X~ Gamma (  ,  ) then (1 / 2) 

( )   x  1 e  x dx,
 , E ( X )   , Var ( X )   2
 0
0
 (  1)  (  1),
 (  1)!,
 1
 is any positive integer


If =1 then it is called standard gamma distribution.
When the random variable is a standard gamma r.v. then the cdf is called the incomplete gamma
function (Appendix Table A.4). F(x;,)=F(x/;)

If  = 1 then it is Exponential(=1/).
f ( x)    e    x , x>0 and P( X  x*)  1  e x* .
Discrete Data: If p(x,y) is the joint pmf for x and y, the marginal pmf for x can be computed as
 p( x, y) and the marginal pmf for y can be computed as  p( x, y) .
y
E(X)=
 x  p( x, y)   x  p( x) and E ( X
x
E(XY)=
y
)   x 2  p( x, y )   x 2  p( x)
x
x
 x  y  p( x, y)
x
x
2
and
E(h(X,Y))=
y
x
 h( x, y)  p( x, y)
x
p(x | y)=p(x,y) / p(y) where p(y)>0 and E(X|Y)=
y
y
 x  p( x | y )
x
Continuous Data: If f(x,y) is the joint pdf for x and y, the marginal pdf for x can be computed as
 p( x, y )dy and the marginal pdf for y can be computed as  p( x, y)dx .
y
E(X)=
x
  x  f ( x, y)dydx   x  f ( x)dx
x y
E(XY)=
and E ( X ) 
2
x
  x  y  f ( x, y)dydx
x
2
 f ( x, y )dydx   x 2  f ( x, y )dx
x y
and
E(h(X,Y))=
x y
x
  h( x, y)  f ( x, y)dydx
x y
f(x | y)=f(x,y) / f(y) where f(y)>0 and E(X|Y)=
 x  f ( x | y)dx
x
For both discrete and continuous,

Var(X)= E ( X )  E ( X )
2
2 , Cov(X,Y)=E(XY) -E(X)E(Y)
and
Corr(X,Y)=
Cov( X , Y )
Var ( X ) Var (Y ))
Random Sample:The random variables X1, X2, ….,Xn are said to form a random sample of size n if
(i) The Xi's are independent random variables.
(ii) Every Xi's has the same probability distribution.
If X1, X2, ….,Xn are said to form a random sample of size n with the mean  and the variance 2, the sampling
n
_
distribution of
x
has the mean  and the variance 2/n, the sampling distribution of
x
i 1
variance n2, and so on.
i
has the mean n and the