251y0441 08/16/04
ECO 251 QBA1
FINAL EXAM
MAY 3, 2004
Name
KEY
Class ________________
Part I. Do all the Following (14 Points) Make Diagrams! Show your work! Exam is normed on 75 points.
There are actually 127 possible points.
x ~ N 10, 3 .
Remember z 
x
Comment: There is no excuse for

giving me a probability that is not between zero and one.
1. P0  x  13
0.3
f
13  10 
 0  10
 P
z
 P 3.33  z  1.00 
3 
 3
 P3.33  z  0  P0  z  1.00   .4496  .3413
 .8409
0.4
0.2
0.1
0.0
-4
-3
-2
-1
0
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
x
2. P1  x  9
0.4
0.3
f
9  10 
1  10
 P
z
 P 3.00  z  0.33 
3 
 3
 P3.00  z  0  P0.33  z  0  .4987  .1293
 .3694
0.2
0.1
0.0
-4
-3
-2
-1
0
x
0.4
0.3
f
14 .65  10 

3. Px  14 .65   P  z 
  Pz  1.55 
3


 Pz  0  P0  z  1.55   .5  .4394  .0606
Note that colors on the diagram are reversed.
0.2
0.1
0.0
-4
-3
-2
-1
0
x
4. F 19 .32  (Cumulative Probability)
19 .32  10 

 Pz 
  Pz  3.11
3


 Pz  0  P0  z  3.11  .5  .4991  .9991
0.4
f
0.3
0.2
0.1
0.0
-4
-3
-2
-1
0
x
1
251y0441 05/07/04
5. P6.4  x  18 .4
0.4
0.3
f
18 .4  10 
 6.4  10
 P
z
  P 1.20  z  2.80 
3
3


 P1.20  z  0  P0  z  2.80   .3849  .4974
 .8823
0.2
0.1
0.0
-4
-3
-2
-1
0
1
2
3
4
x
6. x.018 (Find z .018 first) Make a diagram! Make a diagram! x.018 is a point with 1.8% above it and
100%-1.8% = 98.2% below it. Since 50% of the distribution is below zero, we have
P0  z  z .018   48 .2% . The closest we can come is P0  z  2.10   .4821 , so z.018  2.10.
. Since x ~ N 10, 3 , we use the reverse of z 
x

, which is x    z , so x.018    z .018
 10  2.103  16.30.
Check:
16 .30  10 

Px  16 .30   P  z 
  Pz  2.10   Pz  0  P0  z  2.10   .5  .4821  .0179  .018 .
3


7. A symmetrical region around the mean with a probability of 35%. Make a diagram! Show a Normal
curve with a mean of zero in its center. If we split 35% in two, we get two areas, on either side of zero with
probabilities of .1750. The lower of these two points is  z .325 , since it has 32.5% below it and 17.5% +
50% = 100% - 32.5% = 67.5% above it. The upper of the two points is z .325 and has 32.5% above it. The
important measure, however, is P0  z  z.325   .1750 . The closest we can come is
P0  z  0.45   .1736 or P0  z  0.46   .1772
The first is a little closer than the second. So we can say z .325  0.45 and
x    z  10  0.453  10  1.35 or 8.65 to 11.35.
11 .35  10 
 8.65  10
z
Check: P 
  P 0.45  z  0.45   2 P0  z  0.45   2.1736   .3472  .35
3
3


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251y0441 05/07/04
II. (10 points+-2 point penalty for not trying part a .) Show your work!
x1
x2
x12
x 22
1
2
3
4
5
6
7
8
9
10
79.2
0.1
12.9
16.2
23.0
22.4
64.4
10.2
7.6
27.7
263.7
12.1
1.0
-4.2
11.0
2.5
4.4
11.8
2.3
1.2
2.4
44.5
6272.64
0.01
166.41
262.44
529.00
501.76
4147.36
104.04
57.76
767.29
12808.71
146.41
1.00
17.64
121.00
6.25
19.36
139.24
5.29
1.44
5.76
463.39
The data above represents returns of a sample of 10 low-risk mutual funds in 1999( x1 ) and the first quarter
of 2000( x 2 ). Calculate the following.
a. The sample standard deviation s x1 of x1 (2) and x 2 (1)
b. The sample covariance s x1x2 between x1 and x 2 . (3)
c. The sample correlation rx1 x2 between x1 and x 2 . (2)
d. Given the size and sign of the correlation, what conclusion might you draw on the relation
between x1 and x 2 ? (1)
e. Assume that the return on the funds were .1 higher in the first quarter of 2000. Find
x 2 , s x2 , s x1x2 and rx1 x2 . Use only the values you computed in a-c and rules for functions of x and y to
get your results. If you state the results without explaining why, or change x1 and x 2 and recompute the
results, you will receive no credit. (4).
f. Do a 95% confidence interval for the mean return in 2000. (4)
g. Was there a significant difference between the return on these funds in 1999 and 2000? A
relatively simplistic way to answer this is to check if the mean return in 1999 was in the confidence interval
for 2000. (2)[33]
x12  12808 .71 ,
Solution: The following have been given to us.
x  263.7 ,
x  44.5 ,

x
2
2
 463 .39 and n  10 . We need x1 
a. s x21 



x12
 nx1
2

n 1
x
1
n

1

263 .7
 26 .37 and x 2 
10
x
n

2

2
44 .5
 4.45 .
10
12808 .71  10 26 .37 2
 650 .549 , s x1  650.549  25.5059 ,
9
463 .39  10 4.45 2
 29 .4850 and s x2  29.4850  5.43001.
n 1
9
b. To go any further, we should repeat our data.
s x22
row
1
2
3
4
5
6
7
8
9
10
x1
79.2
0.1
12.9
16.2
23.0
22.4
64.4
10.2
7.6
27.7
x 22
x2
12.1
1.0
-4.2
11.0
2.5
4.4
11.8
2.3
1.2
2.4
 nx 2 2

x12
6272.64
0.01
166.41
262.44
529.00
501.76
4147.36
104.04
57.76
767.29
x 22
x1 x 2
146.41
1.00
17.64
121.00
6.25
19.36
139.24
5.29
1.44
5.76
958.32
0.10
-54.18
178.20
57.50
98.56
759.92
23.46
9.12
66.48
2097.48
We now have
x x
1 2
 2097 .48 ,
which means
x1 x 2  nx1 x 2 2097 .48  10 26 .37 4.45 
s x1x2 

n 1
9
 102 .668
x1 x 2 by
Do not compute

multiplying
x

1
by
x
2
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251y0441 05/07/04
c. rx1 x2 
s x1 x2

s x1 s x2
102 .668
25 .5059 5.43001 
 .7413
This must be between -1 and +1.
d. On a -1 to +1 scale, this may seem large, but strength is usually measured by squaring the
correlation and looking at the results on a zero to one scale. The square of the correlation is
.5495, which is neither strong or weak. The sign of the correlation indicates that the two variables
moved together, which means that the funds with the best gains in 1999 tended to have the best in early
2000.
e. From the syllabus supplement article:
“Let us introduce two new variables, w and v , so that w  ax  b , and v  cy  d , where
a, b, c, and d are constants. From the earlier part of this section we know the following:
 w2  Varw  a 2Varx  a 2 x2
 w  Ew  Eax  b  aEx   b
 v  E v   E cy  d   cE y   d
 v2  Var v   c 2Var  y   c 2 y2
To this we now add a new rule: Covw, v   wv  acCovx, y   ac xy
To find the correlation between w and v , recall that  wv 
 w2  a 2 x2
 wv 
 wv
. But since
 w v
and  v2  c 2 y2 , then
ac xy
a 2 x2 c 2 y2

ac xy
ac  x y

ac  xy
 signac xy .”
ac  x y
Since these rules work for sample statistics too, then w  x1 , v  x 2  .1 , so a  1,
b  0 c  1 and d  0.1. w  x1  26.37
s w2  12 s x21  s x21 , so
s w  650 .549  25 .5059 . v  x 2  .1  4.35 s v2  12 s x2x  s x22 , so
s v  29 .4850  5.43001 s wv  11s x1x2  102.668 .
Finally rwv 
acs x1x2

a 2 s x21 c 2 s x22
acs x1x2
ac s x1 s x2

ac s x1x2
 signacrx1x2  .7413
ac s x1 s x2
g. To do a 95% confidence interval for the mean, recall the formulas quoted in the problem
solutions.
When  is known   x  z  x , where  x 
2
x
n
, or  x 
x
N n
when the sample
N 1
n
is more than 5% of the population.
N n
when the sample
N 1
s
s
When  is unknown   x  tn1 s x , where s x  x , or s x  x
2
n
n
is more than 5% of the population.
In this case, the population variance is unknown, as is the population size, so we use s x 
Recall that s x1  25.5059 and that n  10 . s x 
25 .5059
10
sx
.
n
9
 8.0657 . t n1  t .025
 2.262.
2
  x  t n1 s x  26.37  2.2628.0657  26.37  18.24 or 8.13 to 44.61.
2
h. This does not include x 2  4.45 , so we can guess that the means are significantly different.
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251y0441 05/07/04
III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what
sections of the problem you are answering! If you are following a rule like E ax  aEx  please state it! If
you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think
are inappropriate - you might get partial credit. Choose the problems that you do carefully – most of us are
unlikely to be able to do more than half of the entire possible credit in this section!) This is not an opinion
questionnaire. Answers without reasons or supporting calculations or table references will not be
accepted!!!!
1. Suppose that you pick up 100 packages sent by a mailer. You believe that this mailer produces packages
with a population mean of 6 oz and a population standard deviation of 2.5 oz. Let x represent the sample
mean weight that you calculate on this particular day.
a. According to the central limit theorem, what distribution, with what mean and standard
deviation should apply to the sample mean weight of the packages?(2)
b. If you assume that the package weights are normally distributed, what is the probability that a
randomly picked package will weigh more than 7 oz? (2)
c. What is the probability that the sample mean of the weight of the packages is more than 7 oz?
(3)
d. What is the probability that the combined weight of all 100 packages is above 40 lbs? (640 oz.)
(3)
e. If you assume that the package weights are Normally distributed, what is the chance that three
randomly picked packages are all above the median weight? (2) [45]
f. How heavy does a package have to be to be heavier than 95% of the packages sent out by this
mailer? (2) [47]
Solution: a. If the parent population has a mean of 6 and a standard deviation of 2.5, The central limit
theorem says that the sample mean will have an approximate Normal distribution. So


 
2.5 
  N  6,
  N 6,0.25  .
x ~ N   ,



n
100 


7  6

 Pz  0.4  .5  .1554  .3446 .
b. x ~ N 6,2.5 and Px  7   P  z 
2.5 


7  6
2.5 
  N 6,0.25  Px  7   P  z 
 Pz  4.00   .5  .5000  0
c. x ~ N  6,

0.25 

100 

d. If the combined weight of all the packages is 640 oz., the sample mean of the 100 packages is
6.4  6 

640
 Pz  1.60   .5  .4452  .0548 .
 6.40 Px  6.40   P  z 
100
0.25 

e. For any continuous distribution, the probability that a randomly picked point is above the median is .5.
The probability for three such events is .5 3  .125
f. According to the bottom of the t table z .05  1.645 , so x.05    z.05  6  1.645 2.5  10.1125 .
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251y0441 05/07/04
2. I send out a survey to 200 people. The probability that each person returns the survey is 10%.
This is a Binomial problem.
a. What is the chance of between 10 and 30 returns. Answer the question by showing that you
can use the Normal distribution to solve this problem and solving it.(3)
b. Answer the same question by showing that you can use the Poisson distribution to solve it and
solving it. (3)
c. Let p represent the fraction of the surveys that are returned. Assume that we send out 103
surveys, what is the distribution (including the mean and standard deviation) of p . (2)
d. What is the probability that p is above 11% if (i) n  100 , (ii) n  1000 . (4)
e. What is the probability that the first survey returned is between the 20th and the 30th sent out?
(2) [61]
Solution: This is based on one of Dr. Tahany Naggar’s problems, which you were given as Problem M7.
a. First, test to see if we can substitute the Normal distribution in place of the binomial
distribution. I tested to see if np  5 and nq  5 . Since np  200 .1  20 , and
nq  200 .9  180 , np is large enough and we can use the Normal distribution with   np  20
and  2  npq  20.9  18. If we use the continuity correction,
 9.5  20
30 .5  20 

z
P10  x  30   PN 9.5  x  30.5  P

18 
 18
 P2.47  z  2.47   2P0  z  2.47   2.4932   .9864 .
b. Second, test to see if we can use the Poisson distribution in place of the binomial distribution. I
n
n 200
 500 . Since

 2000 is above 500, we can use the Poisson
tested to see if
p
p
.1
distribution with a parameter of m  20. From the Poisson table P10  x  30 
 Px  30   Px  9  .98653  .00500  .98153 .
c. From the outline p is approximately Normal with a mean of p  .10 and a standard deviation
of
pq
.1.9 

 .0296 .
n
103

d. p ~ N  p,


pq 
.
n 
pq
.11  .10 
.1.9 

 Pz  .33 

 .0009  .03 P p  .11  P  z 
.03 
n
100

=.5 - .1293 = .3707
pq
.1.9
.11  .10 

 Pz  1.05 

 .00009  .009487 P p  .11  P  z 
n  1000  p
.009487 
n
1000

= .5 - .3531 = .1469
e. This is a geometric distribution problem. P20  x  30   Px  30   Px  19 
n  100 ,  p






 F 30   F 19   1  q 30  1  q19  q19  q 30  q19 1  q11  .919 1  .911  .1350851  .313811
 .0927 .
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251y0441 05/07/04
3. A manufacturer is producing bolts with a nominal length of 5cm. A random sample of 10 bolts is taken
from a box containing a large number of bolts. From the sample we get a sample mean of 5.512 and a
standard deviation of .22754.
a. On the basis of long experience, we know that the standard deviation for the bolts is .2236.
Find a 95% confidence limit assuming that this population standard deviation is correct. (3)
b. Find a 95% confidence interval for the mean, assuming that the sample standard deviation is
correct. (4)
c. Find a 95% confidence interval for the mean assuming that the sample standard deviation is
correct and that the sample of 10 bolts was taken from a batch of only 50 bolts. (3)
d. On the basis of the tests in a-c, is a population mean of 5 cm. reasonable? (1)
e. Assume that the median length of the bolts is 5cm., and that all 10 bolts in the sample are
more than 5 cm. A p-value is the probability of obtaining a result as extreme or more extreme
than what actually happened. If the p-value is below 1% we strongly doubt that the median
length is 5cm. An event as extreme or more extreme than what actually occurred is either
getting all bolts longer than 5cm or all bolts shorter than 5cm.What is the probability of this
happening? (3)
f. Assume that a basketball team scores at an average rate of 1.5 points per minute. What is the
probability of scoring more than 22 points in a 10 minute interval? What is the probability of
scoring less than 10 points in a 10 minute interval? (2)
g. By eliminating scores in the bottom 5% or less of the distribution and in the top 5% of the
distribution, find an interval between scores with a probability of about 95%. This will look
something like P2  x  26   .95, where Px  2  .025 and Px  26   .025 and these two
numbers are picked so that the first number is the largest with probability below 2.5% and the
second number is the smallest with probability over 2.5%. (2)[76]
h. Try a cleaner version of g. Assume that the team plays for a half hour. Use the Normal
distribution to find a 95% symmetrical interval around the mean number of points scored. (3)
i. Re-do a) with a 96.4% confidence interval after looking at page 1(2). [81]
Solution: Remember!!!
When  is known   x  z  x , where  x 
2
x
n
, or  x 
x
n
N n
when the sample
N 1
is more than 5% of the population.
s
s
When  is unknown   x  tn1 s x , where s x  x , or s x  x
2
n
n
is more than 5% of the population.
a. Given: x  5.512 ,   0.2236 , n  10 , N unknown,   .05 .  x 
z 2  z.025
N n
when the sample
N 1
x

0.2236

.22757
 0.0707 .
n
10
 1.960 .   x  z 2  x  5.512  1.9600.0707  5.512  0.139 or 5.37 to 5.65.
b. Given: x  5.512 , s  0.22754 , n  10 , N unknown,   .05 . s x 
sx
n
 0.07196 .
10
9
t n1  t.025
 2.262   x  tn1 s x  5.512  2.262 0.07196   5.512  0.162 or 5.35 to 5.67.
2
2
c. Given: x  5.512 , s  0.22754 , n  10 , N  50 ,   .05 .
N n
40
9
 0.07186
 0.07186 .90351   .06493 t n1  t .025
 2.262
2
N

1
49
n
  x  tn1 s x  5.512  2.262 0.06493   5.512  0.147 or 5.36 to 5.66.
sx 
sx
2
d. It is not reasonable. None of these intervals contain 5.
7
251y0441 05/07/04
i. (Out of order) Since 1    .964 ,   .036 and

2
 .018 , in a) replace 1.960 with
z.018  2.10. You find this in Part I, Problem 6.
e. Since the probability of being above a median in a continuous distribution is .5, the probability
of all 10 bolts being above the median equals the probability of all 10 bolts being below the
median and is .510  .009766 . By the addition rule the probability of one or the other mutually
exclusive event occurring is 2.009677   .00195 . Since this probability is below .01, if we got
either all above 5 cm or all below 5 cm, we would strongly doubt that the median was actually 5
cm.
f. If the team scores 1.5 points per minute, it scores an average of 15 points in a 10 minute
period. Using a Poisson table with a parameter of 15, we find
Px  22   1  Px  22   1  .96726  .0327 and Px  10   Px  9  .06985 .
g. Since Px  24   1  .98054  .01946 is the largest probability below 2.5% on the top and
Px  7  .01800 is the largest probability below 2.5% on the bottom, the best we can do is
.98054  .01800  Px  23  Px  7  P8  x  23  .96254  .95.
h. This is a lot easier. If the team scores 1.5 points per minute, it scores an average of 45 points
in a 30 minute period. A Poisson table with a parameter of 45 is not available, but we can
approximate it with a Normal distribution with a mean of 45 and a standard deviation of 45 .
Since we already know that z .025  1.960 , a symmetrical interval around the mean with a 95%
probability is 45  1.960 45  45  13.15 , or, in integers 32 to 58.
8
251y0441 05/07/04
4a) Moe, Shemp and Curley work in a fast food restaurant. Moe fills 30% of the orders, Shemp fills 45% of
the orders and Curley fills 25% of the orders. Moe gets the order wrong 20% of the time, Shemp gets it
wrong 12% of the time and Curley screws up 5% of the time. If the order was filled correctly, what is the
probability that it was filled by each of the 3 workers? If the order was filled incorrectly, what are the
probabilities? (6)
Solution: This is the obligatory Bayes’ rule problem, and I was surprised at how easy it was.
Given: The probability that Moe gets the order is PM   .30, the probability that Shemp gets the order is
PS   .45, The probability that Curley gets the order is PC   .25 . If Moe gets the order, the probability


that he messes up is P W M  .20, If Shemp gets the order, the probability that he messes up is
PW S   .12, If Curley gets the order, the probability that he messes up is PW C   .05. The easiest way
to do this is a variant of the box method, but we can use the probabilities as they are stated. To start with
divide the orders between the three Stooges.
Moe
Shemp
Curley
Total
Wrong
Right
Total
.30
.45
.25
1.00
Moe gets 30 out of every 100 orders and messes up 20% of them. 20% of 30 is 6 or
PW  M   P W M PM   .20.30  .06 . Likewise PW  S   P W S PS   .12.45  .054 and
 

PW  C   PW C PC   .05.25  .0125. Fill in the ‘Wrong’ row.

Moe
Shemp
Curley
Total
Wrong
.06
.054
.0125
Right
Total
.30
.45
.25
1.00
Now add up the ‘Wrong’ row and fill in everything else so that it adds up.
Joint Probabilities
Moe
Shemp
Curley
Total
Wrong
.06
.054
.0125
.1265
Right
.24
.396
.2375
.8735
Total
.30
.45
.25
1.00
So 12.45% of the orders are wrong, and Moe’s bad orders are 6% of all the orders, so the probability that a
PW  M 
.06
bad order is Moe’s is PM W  

 .4743 . So to get the conditional probabilities divide
PW 
.1265
the first row by .1265 and the second row by .8735.
Conditional Probabilities
Moe
Shemp
Curley
Total
Wrong
.4743
.4269
.0988
1.0000
Right
.2748
.4533
.2719
1.0000
Total
9
251y0441 05/07/04
The following table gives probabilities for x and y.
y
x
1
2
1
.01
.03
2
.02
.06
3
.07
.21
3
.06
.12
.42
b. Find E x , E  y   x ,  y ,  xy , and  xy .(6)
Solution:
x
1
1
y
2
3
Px 





xPx 
x 2 Px 
.01
2
.03
.02
.06
.07
.21
.10
.30
 x2
.06 

.12 
.42 

.60
P y 
.10
yP y  y 2 P y 
0.10
0.10
.20
0.40
0.80
.70
2.10
6.30
1.00
2.60
7.20
0.10  0.60  1.80 
2.50
0.10  1.20  5.40 
6.70
 Px   1 ,   E x    xPx   2.50 , E x    x Px  6.70 ,  P y   1 ,
 E  y    yP y   2.60 and E y    y P y   7.20
 E x    6.70  2.50   0.45 and   Ey    7.20  2.60  0.44
2
To summarize
y
3
2
x
2
2
2
2
x
2
2
y
2
2
y
2
 x  0.45  .6708 and  y  0.44  0.6632 Now we compute the covariance,
unless we have noticed that x and y are independent.
 .0111 .0321 .06 31
E xy  
xyPxy    .02 12 .06 22 .12 32
  .07 13 .2123 .42 33
 0.01 0.06 0.18 
 0.04 0.24 0.72   6.50 .
 0.21 1.26 3.78 

 xy  Covxy  Exy   x  y  6.50  2.502.60  0.00
So that  xy 
 xy
 x y

0
0
.45 .44
c. x  y can take the values 2, 3, 4, 5 and 6. Find the probabilities for each of these values and show that
they belong to a valid distribution. (5)
Solution: Let us repeat our data. Sums are marked in the table. The fact that the Px  y  column sums to 1
shows that this is a valid distribution.
y
x
1
2
3
1
.01/2
.03/3
.06/4
2
.02/3
.06/4
.12/5
3
.07/4
.21/5
.42/6
10
251y0441 05/07/04
We get the following table from the table above.
Probabilities
x  y 
P x  y 
2
3
4
5
6
Sum
.01
.02+.03 =
.07 + .06 + .06 =
.21 + .12 =
.42
.01
.05
.19
.33
.42
1.00
x  y Px  y 
0.02
0.15
0.76
1.65
2.52
5.10
 x  y 2 P x  y 
0.04
0.45
3.04
8.25
15.12
26.90
d) Find Ex  y  and Var x  y  two different ways – getting the same values each time. (5)
 x  y Px  y   5.10
Ex  y 2   x  y 2 Px  y   26.90
From the table above E x  y  
Varx  y   E x  y 2  E x  y 2  26 .90  5.10 2  0.89
On the previous page we had  x  E x   2.50 ,  y  E y   2.60 ,  x2  0.45 ,  y2  0.44 and
 xy  0.00 .
Ex  y   Ex   E y   2.50  2.60  5.10
Varx  y   Varx   Var y   2Covx, y   0.45  0.44  0  0.89
Once more, we repeat the table.
y
x
1
2
3
1
.01
.03
.06
.10
2
.02
.06
.12
.20
3
.07
.21
.42
.70
.10
.30
.60
e) If Event A is x  3 and Event B is y  3 find P A and PB  . (0) Find P A  B  (1)
f) Find P A  B  (1)
 
g) Find P A B (1) [105]
Solution:
e) If Event A is x  3 and Event B is y  3, P A  .60 , PB   .70 and the table says
P A  B   .42 (1) Caution: P A  B  P A PB only because Event A and Event B are
independent. This is not generally true.
f) P A  B   P A  PB   P A  B   .60  .70  .42  .88 (1)
 
g) P A B 
P A  B  .42

 .60 or you could say that, because of independence, PA B  P A  .60 .
PB 
.70
11
251y0441 05/07/04
5. Assume that in any batch of soft drink bottles 25% are defective.
a) If I take a sample of 5 bottles from a case of 24, what is the chance that exactly 3 will be
defective? (2 – 3 with complete computation)
b) If I take a sample of 5 bottles from a case of 24, what is the chance that at least one will be
defective? (2 – 3 with complete computation)
c) If I take a sample of 5 bottles from a shelf with many, many bottles on it, what is the chance
that exactly 3 will be defective? (2)
d) If I take a sample of 5 bottles from a shelf with many, many bottles on it, what is the chance
that at least one will be defective? (2)
e) If I take a sample of 100 bottles from a shelf with many, many bottles on it, what is the chance
that at least 25 will be defective? (2)
f) (extra credit)If I take a sample of 100 bottles from a shelf with 500 bottles on it, what is the
chance that at least 25 will be defective? (2)
Solution:
6! 18!
6 18
C C
3!3! 16! 2!
a) Hypergeometric N  24 , n  5, M  np  24.25   6. P3  3 242 
24!
C5
19!5!
6  5  4 18 17
5  4  9 17
3060


 .0720
 3  2 1 2 1
24  23  22  21  20
24  23 11  7 42504
5  4  3  2 1
b) Hypergeometric N  24 , n  5, M  np  24.25   6. Px  1  1  P0  1 
C 06 C 518
C 524
6! 18!
18 17 16 15 14
1
6!0! 13! 5!
18 17  2 14
8568
5  4  3  2 1
 1
 1
 1  .2016  .7984
 1
 1
24!
24  23  22  21  20
24  23 11  7
42504
19!5!
5  4  3  2 1
c) Binomial p  .25 , n  5 . P3  Px  3  Px  2  .98437  .89648  ..08789 or
C35 .253.752  10.015625.5625  .08789 .
d) Binomial p  .25 , n  5 Px  1  1  P0  1  .23730  .7627 or 1  .75 5  .7627 .
e) Binomial p  .25 , n  100 ,   np  100 .25   25,  2  npq  100.25.75  25(.75)
 18 .75 .   4.3301  Because np and nq are above 5, use a Normal approximation.

24 .5  25 
PB x  25   PN x  24 .5  P  z 
  Pz  0.12   .5  .0478  .5478 It is also
18 .75 

possible to try PB x  25   1  PB x  24   1  PN x  24.5 .
 N n
f) Hypergeometric p  .25 , n  100 ,   np  100 .25   25,  2  
npq
 N 1 
 500  100 

100 .25 .75   .8016 25 (.75 )  15 .03   3.8769  Because np and nq are
 500  1 
above 5, use a Normal approximation to the binomial distribution with a finite population
correction.

24 .5  25 
PB x  25   PN x  24 .5  P  z 
  Pz  0.13   .5  .0517  .5517
15 .03 

12
251y0441 05/07/04
5
g) In an area of 5 square feet, we normally expect 2 paint blisters.
(i) What is the chance of no blisters in one square foot? (1)
(ii) What is the chance of at least two blisters in one square foot? (2)
(iii) 95% of surfaces of one square foot will have less than ? blisters. (1)
(iv) What is the chance of no blisters in 75 square feet? (2)
(v) What is the chance of at least 150 blisters in 75 square feet? (2)
(vi) 95% of 75 square foot surfaces will have less than ? blisters. (2) [127]
Solution: g-i) There are 2 5  0.4 blisters per foot, so use Poisson m  0.4 . P0  .67032 .
g-ii) Poisson m  0.4 Px  2  1  Px  1  1  .93845  .06155 .
g-iii) If we check the table for the Poisson distribution with a parameter of 0.4 Px  2  .99207
is the lowest value of the cumulative distribution with a value above .95. So 2 is the best answer
that we can give.
g-iv) If there are 0.4 blisters per square foot, the average 75 square foot area will have
m  750.4  30 blisters. We should use the Poisson distribution with a parameter of 30, but we
do not have the appropriate table. Since the parameter is above 20, we can use the Normal
distribution with a mean of 30 and a standard deviation of   m  5.4772 .

0.5  30 
PP x  0  PN x  0.5  P  z 
  Pz  5.39   Pz  0  P5.39  z  0
30 

 .5  .5000  0 .

149 .5  30 
g-v) Poisson m  30 PP x  150   PN x  149 .5  P  z 
  Pz  21 .82 
30 

 Pz  0  P0  z  21 .82   .5  .5000  1.0000 .
g-vi) From the bottom of the t table, we can read z .05  1.645 . We know that it is approximately
true that x ~ N 30, 5.4772  . So x.05    z.05  30  1.645 5.4772   30.01 .
13