251x0144 12/29/01
ECO251 QBA1
FINAL EXAM
DECEMBER 15, 2001
Name KEY
Class ________________
For problem references, see 251y0141.
Part I. Do all the Following (20 Points) Make Diagrams! Show your work!
A. z ~ N (0,1)
1. P1.50  z  1.50   2P0  z  1.50   2.4332   .8664
2. P3.34  z  1.93  P3.34  z  0  P1.93  z  0  .4996  .4732  .0264
3. P1.23  z  3.04   P0  z  3.04   P0  z  1.23  .4988  .3907  .1081
4. Pz  2.85   Pz  0  P0  z  2.85   .5  .4978  .9978
Note that probabilities cannot be above 1 or below 0. A large fraction of the errors made on this or
any other exam are exactly what you were reminded of in "Things you should Never do on an
exam…." Look at it before every exam!
z .27 is the 73rd percentile of the distribution, the point with a probability of .27 above it or .73
below it. Since 73% is below this point and 50% is below zero, from the diagram P0  z  z.27   .2300 .
5. z .27
the closest we can come to this probability using the Normal table is P0  z  0.61  .2291, though 0.62 is
almost as good. So z.27  0.61 .
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251y0144 12/03/01
B. x ~ N 1.5, 3 .
1.50  1.5 
  1.50  1.5
z
1. P1.50  x  1.50   P
  P 1.00  z  0  P1.00  z  0  .3413
3
3


 1.93  1.5 
  3.34  1.5
z
2. P3.34  x  1.93  P
  P1.61  z  1.14 
3
3


 P1.61  z  0  P1.14  z  0  .4463  .3729  .0734
3.04  1.5 
 1.23  1.5
z
3. P1.23  x  3.04   P
  P 0.09  z  0.51
3
3


 P0.09  z  0  P0  z  0.51  .0359  .1950  .2309
2.85  1.5 

4. Px  2.85   P z 
  Pz  0.45   Pz  0  P0  z  0.45   .5  .1736  .6736
3


.
5. x.27 - x.27 is the 77th percentile of the distribution, the point with a probability of .27 above it or .73
below it. Since 1.5 is the mean and median of this distribution, x.27 must be above it. On the previous page
we found z.27  0.61 . So x.27    z.17  1.5  0.613  3.33 .
3.33  1.5 

To check to see if this is correct: Px  3.33  P z 

3


 Pz  0.61  Pz  0  P0  z  0.61  .5  .2291  .2709  .27.
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II. (4 points-2 point penalty for not trying .) Show your work!
We are investigating the cost of remodeling a kitchen. Contractors are asked for
estimates, with the following results:
contractor
1
2
3
4
5
6
Estimate($thousands)
20.4
22.1
25.3
28.7
19.9
17.1
Compute the sample standard deviation, s , for the contractors' estimates (not the number of the
contractor!). Show your work. (4)
Solution:
x
1
20.4
2
22.1
3
25.3
4
28.7
5
19.9
6
17.1
Total 133.5
 x  133 .5 ,  x
x
2
 3056 .77 and n  6 . So
 x  133 .5  22.25
n
6
x2
416.16
488.41
640.09
823.69
396.01
292.41
3056.77
sx2 
x
2
 nx 2
n 1

3056 .77  622.25 2 86.395

 17.279
5
5
sx  17 .279  4.1568 .
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III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what
sections of the problem you are answering! If you are following a rule like E ax  aEx  please state it! If
you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think
are inappropriate - you might get partial credit.
1. Using the sample mean and sample standard deviation from the sample of 6 on the previous page (and
assuming that a confidence interval is a valid thing to do under the circumstances):
a. Compute a 90% confidence interval for the average estimate. (4)
b. Compute a 90% confidence interval for the average estimate assuming that the sample of 6 was taken in a
town where there are only 31 kitchen remodelers. (4)
c. Compute a 90% confidence interval for the average estimate using the sample mean that you found from
the sample of 6 on the previous page but assuming that there are only 31 kitchen remodelers in town and
that you know that the population standard deviation is 11.4 ($thousands) (4)
d. What must be true about the distribution from which the data on the previous page comes for a
confidence interval to be valid for such a small sample? (1)
x
x 2  nx 2
Solution: From the previous page x 
 22 .25 , sx2 
 17 .279 and
n
n 1


sx  17 .279  4.1568 .
For all these sections the confidence level is 1    .90 and the significance level is
  1  .90  .10 . n  6 .
Repeat after me! " z goes with  (sigma - population variance); t goes with s (sample variance)!"
 .10
5
a) The degrees of freedom are n  1  6  1  5 .

 .05 . From the t-table, tn1  t.05
 2.015 ,
2
2
2
s
4.1568
sx  x 
 1.697 .
n
6
Putting this all together   x  t n1 s  22.25  2.015 1.697   23.25  3.419 or 19.831 to 26.669.

2
x
More formally, P19.831    26.669   .90 .
b) The population size, N  31 , is less than 20 times the sample size n  6 , so sx 
sx
n
N n
N 1
31  6
5
 1.697 0.9129   1.549 . It is still true that tn1  t.05
 2.015 , so
2
31

1
6
  x  tn1 s x  22.25  2.015 1.549   22.25  3.122 or 19.128 to 25.379. More formally,

4.1568
2
P19.128    25.379   .90 .
x
N  n 11.4 31  6

 4.654 0.9129   4.249 . From the t-table z 2  z.05  1.645 .
N 1
6 31  1
  x  z 2 x  22.25  1.645 4.249   23.25  6.989 or 16.761 to 30.239. More formally
c)  x 
n
 P16.761    30.239   90% .
d) for a confidence interval based on z or t to be valid for a sample with n  30 , the underlying
distribution must be Normal.
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251y0144 12/03/01
2. A manufacturer claims that less than 1% of the jorcillators it sells you are defective.. To be charitable, in
the problem assume that the probability is exactly 1% (Remember, 'at least 4' means 4 or more.)
a. Assume that you buy 50 jorcillators and that 4 are defective.
(i) Without using a table find the probability that exactly 4 are defective.(3)
(ii) To check your answer, find the same probability using a binomial table (2)
(iii) Find the probability that at least 4 are defective. (2)
b. Assume that you buy 150 and that the 1% probability is still valid. Show that you can use the Poisson
distribution and find
(i) The probability that exactly four are defective (1)
(ii) The probability that at least 4 are defective (2)
c Assume that you buy 550 and that the 1% probability is still valid. Show that you can use the Normal
distribution. and find the probability that at least 4 are defective. (3)
d. Suppose that you buy 50 jorcillators, but that they come from a batch of 500 that is 1% defective. Find
the probability that at least 4 are defective. You may leave your answer in factorial form. (3)
e. Suppose that you buy a large number of jorcillators and that they are 1% defective. What is the
probability that the first defective one you find will be among the first five that you inspect? (2)
f. EXTRA CREDIT - The big question here is do we believe that 1% or fewer are defective. The test is to
take a sample and see how many are defective. If k items are defective, we figure out the probability that k
or more items are defective if the probability is 1%. If this probability is below 1% we say we strongly
doubt that less than 1% are defective. On the basis of what you found here if 5 were defective, do we
strongly doubt that less than 1% are defective in a), b) and c)? In each case tell why. (1.5)
Solution: a) (i) Binomial: Px  Cxn p xqn x . Px  4 when n  50 and p  .01 ,
q  1  p  1  .01  .99 so
 50! 
50  49  48  47
.014 .99 46 
.00000001 .629824   .0014505 . .
P4  C450 p 4 q 46  
4  3  2 1
 46!4! 
(ii) Using the table, we get Px  4  Px  4  Px  3  .99985  .99840  .00145 .
(iii) Px  4  1  Px  3  1  .99840  .00160 .
b) This is still a binomial probability problem. p  .01 , q  1  p  1  .01  .99 , n  150 . If we test to
see if the Poisson Distribution can be used, we find
n 150

 15000 . Since this is above 500,
p .01
we can use the Poisson Distribution. The mean is   np  150 .01  1.5.
(i) Using the Poisson1.5 table, we find that Px  4  .047067 . .
(ii) Using the Poisson1.5 table, we find that Px  4  1  Px  3  1  .93436  .06564 .
c) This is still a binomial probability problem. p  .01 , q  1  p  1  .01  .99 , n  550 . If we test to
see if the Normal Distribution can be used, we find np  550 .01  5.5 and
nq  n  np  550  5.5  544.5. Since these are both above 5, we can use the Normal
Distribution. The mean is   np  550 .01  5.5 and the variance is
 2  npq  5.5.99  5.445 . (   5.445  2.33345.) We want Px  4 , but if we use the
continuity correction we expand the bottom of the interval downward by
1
2
to get

3.5  5.5 
  Pz  0.43 
Px  3.5  P z 
5.445 

 P0.43  z  0  Pz  0  .1664  .5  .6664 .
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251y0144 12/03/01
d) Because the total possible defectives are now limited by the size of the population to 1% of 500, which
is 5, and the population size is less than 20 times the sample size n  50  , we must use the
Hypergeometric Distribution. Px  
C xM C nNxM
M  Np  5 . Px  4  P4  P5
C Nn
, N  500 ,
  495!   495!  
 5
  1

   449!46!   450!45!  




 . Whee!
 
500!
 



450!50!


e) Because we are now asking about the probability of the first success, we can use the Geometric
Distribution with p  .01 , q  1  p  1  .01  .99 , F c  Px  c  1  q c . So
495
 C45C46

 C 500
 50
5 495
C51
C45
500
C50
Px  5  1  q5  1  .99 5  1  .95099  .04901 .
f) Since in a), Px  4 is below 1%, we can say that we strongly doubt the hypothesis that the
defect rate is 1% or lower. For the larger samples, the probabilities rise, as we should expect.
Obviously, as the sample gets larger, the probability of 4 or more defective items goes up, so the
results in b) or c) would not lead to any suspicions.
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251y0144 12/03/01
3. Businesses of the type that use my industrial park produce an average of 980 gallons of sewage per day
with a standard deviation of 210 gallons. There are 49 businesses in my park.
a. If , on one day, I compute a sample mean for these 49 businesses, what is the standard error (standard
deviation of the sample mean)? (2)
b. Using the standard error that you found in a), find
(i) A symmetrical interval about the mean with a probability of 72% (2)
(ii) The probability that the sample mean is between 930 and 1030 gallons. (2)
(iii) If my sewer lines can take no more than a total of 51000 gallons per day, what is the probability that
they will overflow on a given day? (3)
c. Assume that I do not know the population mean above, but that I find that the total amount of sewage on
a given day by all 49 businesses is 44278 gallons, make a confidence interval for the average sewage for an
individual business with a confidence level of 72%. (3)
d. Assume again that the population mean is 980 gallons with a standard deviation of 210. I charge each
business a daily maintenance charge of $12 plus $0.075 for each gallon of sewage. What is the mean and
standard deviation of the daily charge for an individual business? (3) What is the probability of a daily
charge above $100.00? (Assume Normal distribution) (2)
Solution: The problem says   980,   210 and n  49.

210

 30 .
n
49
b) (i) We wish to find a symmetrical interval about the mean with a probability of 72%. Make a diagram!
The diagram for z will show a central area with a probability of 72%. It is split in two by a
vertical line at zero into two areas with probabilities of 36%. The tails of the distribution each
have a probability of 50% - 36% = 14%. From the diagram, we want two points z.14 and  z.14
so that P z.14  z  z.14   .7200 . The upper point, z.14 will have
a) The formula for the standard error is  x 
P0  z  z.14  
72 %
 .3600 . From the interior of the Normal table the closest we can come to
2
.3600 is P0  z  1.08   .3599 . So z.14  1.08 . The interval for x can then be written
x    z.18 x  980  1.0830   980  32.4 , or 947.6 to 1012.4.
1012 .4  980 
 947 .6  980
z
To check this P947 .6  x  1012 .4  P 

30
30


 P 1.08  z  1.08   2P0  z  1.08   2(.3599 )  .7198  .72 .
1030  980 
 930  980
z
(ii) P930  x  1030   P 
 (Make a diagram!)
30
30


 P 1.67  z  1.67   2P0  z  1.67   2.4525   .9050
(iii) P
1040 .82  980 


 x  51000   P x  51000
  Px  1040 .82   P  z 

49 
30



 Pz  2.03  Pz  0  P0  z  2.03  .5  .4788  .0212 . (Make a diagram!)
c) x 
442788
 903 .63 . Note that the significance level is   1  .72  .28 . We have already found out
49
that z  z.14  1.08 . The formula for a two-sided confidence interval when  is
2
known is   x  z  x  903 .63  1.0830.0  903 .63  32.40 or 871.23 to 936.03.
2
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251y0144 12/03/01
d) If x is the amount of sewage, the formula for the daily charge for an individual firm is
w  0.075 x  12.00.
(i) We know that if w has the form w  ax  b ,  w  Eax  b  aEx   b and Varw   w2 
Varax  b  a 2Varx In this case a  0.075 and b  12 .00 . So  w 
 0.075 980   12.00  85.50. , and Var y    y2  0.075 2 210 2 or  y  0.075 210   15 .75. .
(ii) There are two ways to do this problem. We can solve w  0.075 x  12.00  100 and find that
1172 .33  980 

at 100, x  1172 .33 . We can then find that Px  1172 .33  P  z 

210


 Pz  0.92   Pz  0  P0  z  0.92   .5  .3212  .1788 . Or we can realize that w has the
Normal distribution with a mean of 85.50 and a standard deviation of 15.75. We can then compute
100  85 .50 

 Pz  0.92   Pz  0  P0  z  0.92   .5  .3212  .1788 .
Pw  100   P  z 
15 .75 

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251y0144 12/03/01
4. A survey of top executives reveals that 30% of them read Time Magazine and 40% read US News and
World Report. 10% read both. Let U be the event that an executive reads US News, U be the event that
the executive does not, T be the event that an executive reads Time and T that the executive does not
read Time.
a. If these are the only newsmagazines that executives read, what is the probability that a randomly picked
executive reads at least one newsmagazine? (2)
b. What is the probability that an executive who reads Time also reads US News? (2)
c. Use your result in b), PU  , PT  and Bayes' rule to find the probability that an executive who reads US
News also reads Time. (2)
d. Make a joint probability table for the 4 events. (2)
e. Assuming that PT   .30 and PU  =.40, what would P U  T have to be if T and U were mutually
exclusive? (Hint: Use a new joint probability table.) (2)
f. Using the probabilities given at the beginning of the problem, assume that Time costs $2.55 a week and
US News costs $2.40 a week, find the distribution of the amount a randomly picked executive spends on
newsmagazines weekly and the mean and variance of the amount. (5)


Solution: PT   .30, PU   .40, PT U   PU  T   .10 .
a) PT U   PT   PU   PT U   .30  .40  .10  .60.
  PU  T   .10  .3333 .
PT 
.30
PU T PT  .3333 .30 

 .25 . Most of you seem to have forgotten the
c) By Bayes' Rule PT U  
PU 
.40
b) By the Multiplication Rule P U T 
difference between a joint and a conditional probability!
U U
U U
.
10
.
30
T .10 .20
T
d) We start with
and fill in
T .30 .40
T
.40 .60
.40
1.00
e) We start with
T
U U
0
.30
T
.40
1.00
f) The probabilities come from d).
Magazine
Time only
US News only
Both
Neither
So  x  E x  
and fill in
 x  Px  1.625
Cost
x
2.55
2.40
4.95
0
T
T
.30
.70
1.00
U U
0 .30
.30
.40 .30
.40 .60
.70
1.00
Probability
P x 
.20
.30
.10
.40
1.00
 

 

So P U  T  P T  U  .30 .
x Px 
x 2 Px 
0.510
0.720
0.395
0.000
1.625
1.30050
1.72800
1.95525
0.00000
4.98375
 x2  Varx   E x2   x2  4.98375  1.625 2  0.656875 .
9
251y0144 12/03/01
5. The following data gives occupancy rates x (in per cent) and room rates y (in $) for a randomly picked
sample of 10 US hotel markets.
Market
occ
rate
1
2
3
4
5
6
7
8
9
10
79
70
73
62
68
69
66
72
69
70
room
rate
109
97
107
69
67
86
72
85
76
101
According to Minitab the mean occupancy rate is 69.8000, the variance for occupancy rates is 19.9553, the
mean for room rates is 86.9000 and the variance for room rates is 250.544. Do not compute any means
and variances in this problem!
a.
The sample covariance between the occupancy rate and the room rate (4)
b.
The sample correlation between the occupancy rate and the room rate (2)
c.
Interpret the correlation (1)
Answers to questions d) and e) must be based on the results in questions a-c and the variances given by
Minitab. Do not recompute the answers after changing x or y ! Show how you got your answers!
d.
If, due to recession, occupancy rates fall by 5 (i.e. to 74, 65, 68 etc.), but room rates are
unchanged, what would the variance for occupancy rates and the covariance and
correlation computed above become? (2)
e.
If, instead, occupancy rates fall by 5, but all room rates rise by 6% (i.e. they become
112.36, 102.82, 113.42 etc.), what would the variance of room rates and the covariance
and correlation computed in a), b) and c) above become? (3)
Solution: a) From the computations below s xy  56.6445
b) From the computations below rxy  .8011.
c) Since the sign is positive, x and y tend to move together. Since rxy2  .642 , on a zero to one scale, this
is a middling relationship. Note that r 2  1 is impossible.
10
251y0144 12/03/01
x
x2
79 6241
70 4900
73 5329
62 3844
68 4624
69 4761
66 4356
72 5184
69 4761
70 4900
698 48900
Note that only the
obs
1
2
3
4
5
6
7
8
9
10
xy
y
y2
109 11881 8611
97 9409 6790
107 11449 7811
69 4761 4278
67 4489 4556
86 7396 5934
72 5184 4752
85 7225 6120
76 5776 5244
101 10201 7070
869 77771 61166
xy column was needed.
x
 x  698  69.8
n

10
48900  10 69 .82
n 1
9
 19 .9556 Slight discrepancy 
s x2 
y
s 2y 
x 2  nx 2

 y  869  86.9
n
y
10
2
 ny 2
n 1
 250 .544

77771  10 86 .9 2
9
 x  698 ,  x  48900 ,  y  869 ,  y  77771 and  xy  61166
 xy  nx y  61166  1069.886.9  509 .8  56.6444 so
s 
2
xy
rxy 
n 1
sxy
sx s y

2
9
9
56 .6444
19 .9556 250 .544
 0.8011
d) Occupancy rates fall by 5, so x is replaced by x  5. Recall the following formulas:
Varax  b  a 2Varx
Varcy  d   c 2Var y 
Cov(ax  b, cy  d )  acCov( x, y)
and if w  ax  b and v  cy  d , rwv  signacrxy .
In this case a  1, b  5, c  1 and d  0.
Var1x  5  12Varx  19.9553.
Cov(1x  5,1y  0)  11Cov( x, y)  56.6444 .
rwv  sign11rxy  .8011.
e) In this case x is replaced by x  5. and y is replaced by 1.06 y , so a  1, b  5, c  1.06 and d  0.
Var1.06 y  0  1.062Var y   1.062 250.544  281.511.
Cov(1x  5, 1.06 y  0)  11.06 Cov( x, y)  1.06 56.6444  60.0431 .
rwv  sign11.06rxy  .8011.
11
251y0144 12/03/01
6. A jorcillator has two components, a phillinx and a flubberall. As long as both are working the
jorcillator will not be junked. If either fails the jorcillator must be junked.
a. If a phillinx has a lifespan approximated by a uniform distribution between 12 and 24 years,
find:
(i)
The mean and standard deviation of its life. (2)
(ii)
The probability that it will last less than 10 years (2)
(iii)
The probability that it will last between 10 and 20 years (2)
(iv)
The probability that it will last more than 20 years (1)
b. If a flubberall has a lifespan approximated by a Normal distribution with a mean of 18 years and
a standard deviation of 5 years, find
(i)
The probability that it will last less than 10 years (2)
(ii)
The probability that it will last between 10 and 20 years (2)
(iii)
The probability that it will last more than 20 years (1)
c. Now, put them together and find
(i)
The probability that the jorcillator will last less than 10 years. (2)
(ii)
The probability that the jorcillator will last between 10 and 20 years. (3)
(iii)
The probability that the jorcillator will last more than 20 years. (2)
Solution:
a) For the Phillinx: c  12, d  24 Make a diagram. Show a box between 12 and 24 with a height of
f x  1d c   12412  112 . Shade the areas between 12 and 20 and between 20 and 24.
0
10
12
20
24
c  d 12  24
d  c2  24  12 2  12.00   12.00  3.4641

 18
2 
2
2
12
12
(ii)
Let A be the event that the Phillinx lasts 0 - 10 years. There is no probability below 12.
P A  P0  x  10   0 .
(iii)
Let B be the event that the Phillinx lasts 10 - 20 years. There is no probability below 12.
1
PB   P10  x  20   20  12   .66667 .
12
(iv)
Let C be the event that the Phillinx lasts more than 20 years. Since there is no probability above
1
24, PC   P20  x  24   24  20   .33333 .
12
Note that these three probabilities must add to 1.
(i)

12
251y0144 12/03/01
b) For the Flubberall: x ~ N 18, 5 . Make a diagram. Show normal curve with a center at 18 divided into
three areas by vertical lines at 10 and 20. Remember z 
x
. We need the following probabilities:

Px  10  , P10  x  20  and Px  20  . Though these can be found directly, it is more economical to use
the cumulative distribution.
10  18 

F 10   Px  10   P z 
  Pz  1.60   Pz  0  P1.60  z  0  .5  .4452  .0548 .
5 

20  18 

F 20   Px  20   P z 
  Pz  0.40   Pz  0  P0  z  0.40   .5  .1554  .6554 .
5 

(i) Let D be the event that the Flubberall lasts 0 - 10 years. PD   F 10   .0548
(ii) Let E be the event that the Flubberall lasts 10 - 20 years.
PE   P10  x  20   F 20   F 10   .6554  .0548  .6006 .
(iii) Let F be the event that the Flubberall lasts more than 20 years.
PF   Px  20   1  F 20   1  .6554  .3446 . These probabilities also add to 1.
D
0
E
0
F
0
0
c) Here then, is our joint probability table. B
.03653
.40040
.22973
.66667
C
.01827
.20020
.11487
.33333
.0548
.6006
.3446
1
A
Event
Fails in
0-10
Fails in
10-20
Component Joint Events
P A  D   A  E    A  F 
 B  D  C  D 
PB  E   B  F   C  E  
000
Probability
.0548
 .03653  .01827 
.40040  .22973  .20020 
.83033
Fails in
.11487
PC  F  
.11487
10+
Sum
1.0000
There is another way to do this, but caution is recommended, since it is hard to extend this to more than
three periods or different failure rules. Let u represent the life of the jorcillator.
Pu  10   P A  D  P A  PD  P A  D  0  .0548  0  .0548
Pu  20   PC  F   .11487
P10  u  20   1  P A  D  PC  F   1  .0548  .11487  83033 .
13
251y0144 12/03/01
6. d. Extra Credit The time between landings at an airport follows an exponential distribution with a
mean of 60 seconds. i) What is the probability of less than a 2-minute gap between landings (2) ? ii) What is
the probability of between a 15 and a 45 second gap between landings? (2) iii) What is the standard
deviation? (1)
In the exponential distribution from the outline F x 1  ecx , when the mean time to a success is
  
1
c
1
.
c
Note that this is only for x  0 . There is no probability below zero.
a) Since   60, c 
1


1
120
. x  120 so cx 
 2.
60
60
Px  120  F 120  1  e 2  1  .13534  .86466.
b) For 15 cx 
15
45
 0.25 , for 45 cx 
 0.75 .
60
60


P15  x  45  F 45  F 15  1  e 0.75  1  e 0.25  e 0.25  e 0.75  .77880  .47237  .30643.
c) Since    
1
1
.
, 
c
60
.
14