Download IOE 265 Midterm II Winter

IOE 265
Midterm II
Winter ‘03
Name (printed): ________________________________
Student ID #: ____________________________
Lab Section # (circle): Sec 2
Sec 3
Sec 4
TU 12:30-1:30 TU 1:30-2:30 TU 2:30-3:30
Sec 5
TU 3:30-4:30
Sec 6
TU 4:30-5:30
Sec 7
W 9:30-10:30
Sec 8
W 10:30-11:30
“I have neither given nor received aid, nor have I concealed any
violation of the Honor Code.”
Signature: _____________________________
You are allowed to use a single crib sheet for this exam. The crib sheet
should be 8.5” x 11” and you can have information on one side of the sheet.
There are 10 written pages to this exam, including the cover page. Please
check now to make sure that you have the appropriate number of pages. Put
your name at the top of each page. Pace yourself on the exam. Good luck!
Part
1 – Multiple Choice
2 – Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
TOTAL
Total Points
Points Possible Your Points
30
15
15
15
15
15
105
Part I. Multiple Choice. (30 points).
Read each question carefully. Circle the answer – a,b,c, or d – which you feel best
answers the question. Each question is worth 5 points. No partial credit will be given for
questions 2 through 6. Partial credit will be given for question 1 (3 points for correct
choice and 2 points for additional statement).
1. A random sample of 20 observations was selected and the following probability plots
were constructed in Minitab. What is the most plausible distribution family from
which the given sample was selected? Please include a statement to support your
choice.
a. Exponential
b. Normal
c. Lognormal
d. Weibull
Points fall closest to a straight line on the lognormal probability plot
2. Which of the following is the best estimate for the variance of the sample mean
statistic X-bar?
n
 x
i1
a.
n 1
n
 x

 x
2
i
 x
2
i
i1
n
b.
n
x

i1
c.
 x
2
i
n 1
n
= S2/n is the correct choice
n
 (x  x)

i 1
d.
2
i
n2
n 1
3. Consider recording the number of defects per hour found by two quality inspectors,
1 and 2, respectively. The two populations of defect rates analyzed by the two
inspectors each follow a Poisson distribution with parameters, λ1 = 5 and λ2 = 4,
respectively. Two independent samples of sizes n1=35 and n2=45, respectively, are
selected from the given populations. What is the probability that the estimator for the
difference between the two sample means, X 1  X 2 is less than or equal to two
defects?
Use the Central Limit Theorem, both n > 30
2  ( 1   2 )
2  (5  4)
) = (
P( X 1  X 2  2 ) =  (
) =  (2.08) = 0.9812
2
2
5
4
1  2


35 45
n1
n2
a. 0.1774
b. 0.9812
c. 0.5997
d. 0.6293
4. A sample of n observations is selected from a population with known pdf f(x;)=e-x.
If the method of moments is used to find an estimate for , which of the following
equation(s) must be solved?
a.

b.

E X  
E X 
x
i
n
x

i
n
and
E X
2

x

2
i
n

x
c. E ( X )  
 xe dx
0
d. Both a and c
5. Assume the weights of individuals are independent and normally distributed with a
mean of 160 pounds and a standard deviation of 30 pounds. Suppose that 25 people try
to squeeze into an elevator. What is the design limit of this elevator such that only 0.02%
of the time the load (total weight) exceeds the design limit?
a. 4230.9 lbs
b. 4523.5 lbs
c. 3784.2 lbs
d. 4500 lbs
Design Limit = DL, T0 = load (total weight)
P(T0 > DL) = 0.0002, where T0 ~ N(nμ,σ*sqrt(n)) = N(4,000;150)
So DL is 99.98th percentile of T0 = 4,000 + 150 * 99.98th percentile of Z
= 4,000 + 150 * 3.49 = 4,523.5 lbs
6. Upon completion of a marketing campaign, an advertising agency claimed that 20%
of customers in a state had become aware of the newly advertised product. The product’s
distributor then surveyed a random sample of 1,000 of customers in the state. Assuming
that the ad agency’s claim is true, what is the probability that between 185 and 220
(inclusive) of the selected customers in the state had become aware of the newly
advertised product?
a. 0.8381
b. 0.6724
c. 0.0567
d. 0.1769
There is an underlying binomial distribution with n = 1000 and p = 0.20. However, since np
≥ 10 and
nq ≥ 10, we can use the normal approximation
x  .5  np 
p(X  x)  

npq 

Thus,
p(185  X  220)  p(X  220)  p(X  184)
220  .5  1000 .2  184  .5  1000 .2 
   
  
 
 1000.2.8   1000.2.8 

 





  1.62   1.23
 0.9474  0.1093  0.8381
Part II. Problems (15 points each)
Solve the following problems. Show all your work and clearly mark your answer.
1. The continuous random variables X and Y have the joint pdf:
c( x 2  y ) for 0  y  x  2
fxy(x,y) = 
0
otherwise
Also, it can be shown that fy(y) = 4y(1-y2) for 0≤y≤2 and E(Y) = 8/15.
Note: if you need to perform a double sum or integral, it may be more convenient to sum
or integrate over y before summing or integrating over x.
a. Find the value of c.
2 x
c   ( x 2  y)dydx  1
0 0
2
2
y2 x
x2
x4 x3 2
) 2 dx  c  ( x 3  )dx  c(  ) 0  1
2
2
4
6
0
0
c = 3/16 = 0.1875.
c ( x 2 y 
b. Find fx(x), the marginal distribution of X.
x
y2 x
x2
fx(x) =  0.1875( x 2  y)dy  0.1875( x 2 y 
) 0  0.1875( x 3  ) 0  x  2
2
2
0
c. Find E(X).
E(X) =
2 x
2
2
y2 x
x3
x5 x4 2
2
3
4
x
(
0
.
1875
(
x

y
))
dydx

0
.
1875
(
x
y

x
)
dx

0
.
1875
(
x

)
dx

0
.
1875
(
 )0
0 0
0
0
2 0
2
5
8
= 63/40 = 1.575
d. Are X and Y independent? Why or why not?
f xy ( x, y ) 0.1875( x 2  y)
f y x ( y) 

 f y ( y )  4 y(1  y 2 ) , so X and Y are not
2
f x ( x)
x
0.1875( x 3  )
2
independent.
e. Find the covariance of X and Y.
2 x
2
2
2
y3 x
x5 x4
2
3 y
E(XY) =   xy(0.1875( x  y)dydx  0.1875 ( x
 x ) 0 dx  0.1875 (  )dx
2
3
2
3
0 0
0
0
x6 x5 2
 0.1875(  ) 0  21 / 15  1.4
12 15
 xy  E ( XY )  E ( X ) E (Y )  21 / 15  (63 / 40)(8 / 15)  0.56
2. While taking an IOE 265 exam, previous students have reported an interesting
phenomenon. Students have noted that upon turning to each new page of the exam there
is the possibility of having a panic attack. Given a 7-page exam, the data below was
gathered on the number of panic attacks experienced by thirty students. (Assume that the
probability of having a panic attack is constant and is identical for each student.)
x1 =1 x2 =1 x3 =1 x4 =4 x5 =3 x6 =2 x7 =2 x8 =0 x9 =1 x10=1
x11=2 x12=2 x13=0 x14=2 x15=1 x16=2 x17=2 x18=1 x19=3 x20=2
x21=1 x22=0 x23=1 x24=3 x25=2 x26=0 x27=1 x28=1 x29=2 x30=2
30
x
30
 7  x   164
 46
i
i
i1
Hint: i1
a. What is the maximum likelihood estimate (MLE) of the probability of having a panic
attack on any given page? Show all steps involved in the MLE process.
n  x
nx


 p 1 p
x
The pmf for a binomial distribution is given by,  
Thus, the likelihood function is given as follows:
n
n
n   xi
nx 

i 1
Lp    p 1 pi 1 i

x
i1  i 
n    n

 n

n
*
ln Lp  L p     x i ln p   n  x i ln 1 p
x

i1

i1  i  i1
n

The maximum of this function is determined by taking the first derivative of the natural log of the
likelihood function and setting it equal to zero,

 x  n  x 
L  p


0
n
n
i
*
i
i1
p
i1
1 p
p
Thus,
n

pˆ 
x
i

i1
n
n
 n  x    x
i
i1
46
 0.219
164  46
i
i1
Note: These thirty data were randomly generated from a binomial distribution with n=7 and p=0.25
using Minitab.

b. What is the maximum likelihood estimate (MLE) of the probability of a student
having his/her second panic attack on page 5 of the 7-page exam? Name the principle
you used to find this MLE
Due to the invariance principle, we can simply use the negative binomial distribution with the mle
estimate of p determined above.
x  r 1 r
x
p(x)  
p 1 p
 r 1 
Thus,
 3  2  1
0.22 2 (1  0.22) 3  0.092
p( x  3)  
 2  1 
3. Suppose that the number of accidents occurring in an industrial plant is described by a
Poisson distribution with an average of 6 accidents per year. Let X denote the time (in
months) between successive accidents.
a.
Find the probability density function of X.
Poisson rate 6 accidents per year = 0.5 accidents per month … X ~ EXP(0.5)
Density f(x) = ½ exp(-x/2) for x≥0
b.
Using the pdf of X, find the expected time (in months) between successive
accidents. Is this result consistent with the initial statement given in this problem?
Why or why not?
E(X) = 1/λ = 1/0.5 = 2, consistent … average of 6 accidents for 12 months … expect 2
months btw successive accidents
c.
Find the probability that the time between successive accidents is more than 4
months.
P(T>4) = 1-FX(4)=1-(1-exp(-4/2))=exp(-2)=0.1353
d.
Find the probability that the time between successive accidents is between 1 and 3
months.
P(1<X<3)= FX(3)-FX(1)=(1-exp(-3/2))-(1-exp(-1/2))=exp(-1/2)-exp(-3/2)=0.3834
e.
Suppose that an accident occurs less than two weeks after the plant’s most recent
accident. Would you consider this event unusual enough to warrant a special
investigation? Explain why with supporting calculations.
P(X<0.5)= FX(0.5)=1-exp(-0.5/2)=0.2212, not a small probability, do not consider it
unusual event.
4. A five-year old study claims that selling prices of new homes/ condominiums in
the Ann Arbor/ Ypsilanti area have a normal distribution with mean  = $180,000 and
standard deviation  = $40,000.
a.
Assuming that the results of the study are still valid, what is the probability that a
randomly selected new house in the Ann Arbor/ Ypsilanti area will cost more than
$210,000?
P(X>210,000)=P(Z>0.75)=1-0.7734=0.2266
b.
A new couple decides to settle in this area. They are willing to pay between
$160,000 and $190,000 for a new home. What is the probability that they will
find a new house to purchase in the Ann Arbor/ Ypsilanti area that satisfies their
price range?
P(160,000<X<190,000)=P(-0.5<Z<0.25)=0.5987-0.3085=0.2902
c.
What is price below which the 10% of the lowest priced new homes in the Ann
Arbor/ Ypsilanti area are sold?
10th percentile of price = μ + 10th percentile of Z * σ = $180,000 – 1.28 * $40,000 =
$128,800
d.
Suppose a random sample of 4 new homes is selected to show to prospective
buyers. Find the probability that the sample mean selling price for these new
homes is between $185,000 and $220,000.
n=4, μ(X-bar)=μ=180,000, σ(X-bar)=σ/sqrt(n)=40,000/sqrt(4)=20,000
P(185,000<X-bar<220,000) = P( (185,000-180,000)/(40,000/sqrt(4)) < Z < (220,000180,000)/(40,000/sqrt(4))) = P(0.25<Z<2)=0.9772-0.5987=0.3785
5. Assume that the life of a roller bearing follows a Weibull distribution with
parameters α = 2 and β = 10,000 hours
a.
Determine the mean time until failure and the variance of the time until failure of
a roller bearing.
E(T)=μ=β*Γ(1+1/α)=10,000*Γ(1+1/2)=10,000*Γ(1.5)=10,000/2 * Γ(0.5) =
5,000*Γ(0.5)=5,000*sqrt(π)=8,862 hours
σ2=10,0002*[Γ(1+2/2)-Γ2(1+1/2)]=10,0002*[Γ(2)-Γ2(1.5)]=
10,0002*(1-π/4)=21,460,183.66
σ=4,632.5 hours
b.
Determine the probability that the bearing lasts at least 8,000 hours.
P(T>8,000)=1-FT(8,000)=1-(1-exp(-(8,000/10,000)2))=exp(-(8,000/10,000)2)=
exp(-0.82)=0.5273
c.
If 10 bearings are in use and failures occur independently, what is the probability
that all 10 bearings last at least 8,000 hours?
P(all 10 bearings last at least 8,000 hours)=(by independence)=
P(T>8,000)10=0.527310=0.00166
d.
If a second random sample of 64 bearings is selected, what is the probability that
their sample mean time until failure is less than 9,500 hours?
P(T-bar<9,500)=P(Z<(9,500-8,862)/(4,632.5/sqrt(64)))=P(Z<1.1)=0.8643