251y0132 11/19/01
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ECO251 QBA1
THIRD EXAM
NOVEMBER 20, 2001
NAME: ______KEY________
SECTION ENROLLED: MWF TR 10 11 12:30
(Circle both days and time separately or just write down days and time)
Throughout this exam 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!
Part I. Do all the Following (10+ Points) Penalty for not doing question 1. Show your work!
(Keller, Warrack simplified) A manufacturing firm has recorded its total machine time (in thousands of hours) ( x ) and its
electrical power costs (in $1000s) ( y ) over 12 months.
Month Hours Costs
x
1
2
3
4
5
6
7
8
9
10
11
12
Note that
7
10
9
8
11
11
6
8
7
10
9
12
108
y
0.74
0.98
0.87
0.86
1.05
1.01
0.68
0.89
0.73
0.93
0.85
1.09
10.68
y  0.890000 , s 2y  0.0168000 . Please don’t do these again!
Compute the following:
1.
The sample variance for the hours (4).
2.
The sample covariance between the hours and the cost (3)
3.
The sample correlation between the hours and the cost (2)
4.
Interpret the correlation (1)
Answers to questions 5) and 6) must be based on the results in questions 1-4. Do not recompute the answers after
changing x or y !
y column rise by 15% (i.e. to
5.
If there is a retroactive energy cost rise of 15%, so that all the numbers in the
6.
.851 , 1.127, 1.0005, 0.989 etc.), what would the variance, covariance and correlation computed above be?
(1.5)
If, instead, a monthly charge of $200 ($0.2 thousand) is added to all the numbers in the y column (i.e. they
become 0.94, 1.18, 1.07, 1.06 etc.), what would the variance, covariance and correlation computed above be?
(1.5)
Solution:
1) From the computations below s x2  3.45455 (so that s x  1.85864 ).
2) From the computations below s xy  0.233637
3) From the computations below rxy  .9698 .
4) Since the sign is positive, x and y tend to move together. Since rxy2  .94 , on a zero to one scale, this is
a strong relationship. We would be justified in saying that hours of use enable us to predict costs very well.
Note that r 2  1 is impossible as is a negative variance.
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251y0132 11/19/01
obs x
x2 y
1 7 49 0.74
2 10 100 0.98
3 9 81 0.87
4 8 64 0.86
5 11 121 1.05
6 11 121 1.01
7 6 36 0.68
8 8 64 0.89
9 7 49 0.73
10 10 100 0.93
11 9 81 0.85
12 12 144 1.09
108 1010 10.68
y2
0.5476
0.9604
0.7569
0.7396
1.1025
1.0201
0.4624
0.7921
0.5329
0.8649
0.7225
1.1881
9.6900
xy
5.18
9.80
7.83
6.88
11.55
11.11
4.08
7.12
5.11
9.30
7.65
13.08
98.69
 x  108 ,  x
 1010 ,
 y  10 .68 ,  y
2
x
s x2
 x  108  9.000
n
x

12
2
 nx 2
n 1

1010  12 9.000 2
11
 3.45455
y
 y  10.68  0.8900
s 2y 
n

12
y 2  ny 2
n 1

9.6900  12 0.8900 2
11
 0.01680
2
 9.69 and
 xy  98.69
2
Note that the y and y columns were not needed! I can't think of a bigger waste of effort except
recomputing y in parts 5 and 6.
s xy 
rxy 
 xy  nx y  98.69  129.0000 0.8900   2.5700  0.233637
n 1
s xy
11
so
0.233637
 0.9698
3.45455 0.016800
5) There is a retroactive energy cost rise of 15%, so that all the numbers in the y column rise by 15%.
sx s y

11
Varcy  d   c 2Var y  , so Var 1.15 y  0  1.15 2 Var  y   1.15 2 0.0168   0.0222 . Also,
Cov(ax  b, cy  d )  acCov( x, y) , so Cov( x  0, 1.15 y  0)  11.15 Cov( x, y)  11.15 0.233637   0.2687
and if w  ax  b and v  cy  d ,  wv  signac xy . signac  sign11.15    , so the correlation
between the two is 0.9698.
6) Instead, a monthly charge of $200 ($0.2 thousand) is added to all the numbers in the y column.
Varcy  d   c 2Var y  , so Var  y  0.2  12 Var  y   12 0.0168   0.0168 . Also,
Cov(ax  b, cy  d )  acCov( x, y) , so Cov( x  0, y  0.2)  11Cov( x, y)  110.233637   0.233637
and if w  ax  b and v  cy  d ,  wv  signac xy . signac  sign11   , so the correlation
between the two is 0.9698. In short, adding 0.2 to y has no effect.
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Part II. Do the following problems ( do at least 40 points ). Show your work!
1.
The following table represents the joint probability of x and y .
x
6 12
0 .1
.1 0
0
 3  .1

y
9  0
11  .1

0
18


.1
.2 

.1
a.
b.
Fill in the missing number. (1)
Are x and y independent? Why?(2)
c.
Compute
 xy , the covariance of x
and
y , and interpret it. (3)
d.
Compute
 xy , the correlation of x
and
y , and interpret it. (3)
e.
Find the probability that
f.
(i) Find the distribution of
x  y is less than 11. (1)
x  y . (2)
x  y .(3)
Solution: a) The missing number must be .2 because the numbers inside the table must add to 1.
b) Check for independence: First you need to find Px  and P y  . Look at the upper left hand probability
(ii) Using only the results of a)-d), find the mean and variance of
below. Its value is .1 and it represents Px  0   y  3 . If x and y were independent , we would
have Px  0   y  3  Px  0 P y  3  .4.2  .  8 . Since this is not true, we can say that x
and y are not independent. A faster way to do this is to note that the zeroes are not parts of rows or
columns of zeroes.
c) Below, we find that  xy  Covxy  Exy    x  y  60 125  0 - meaning: see d)
d) Below, it says that  xy 
 xy
 x y
0

 0 . This means that there is no linear relationship
50 .4 43 .2
between x and y .
x
3
y
9
11
Px 
xPx 
x 2 Px 
E xy  
P y 
.4
0
6 12
 .1 0 .1

 0 .1 0
 .1 0 .1

.2 .1
.2
18
.2 

.1
.2

.5
0
0.6
9.0
12 .0
0
3.6 28 .8 162
194 .4

2.4
.2
yP y  y 2 P y 
 1.2
3.6
1.8
16 .2
.4
4.4
48 .4
1.0
5.0
68 .2
 Px   1 ,
  E x    xPx   12 .0 ,
E x    x Px   194 .4 ,
 P y   1 ,
  E  y    yP y   5.0 and
E y    y P y   68 .2
To summarize
x
2
2
2
2
y
 .10 3  06 3  .112  3  .218  3
xyPxy     009  .169
 012 9
 .118 9
 .1011  0611  .112 11  .218 11
 0  3.6  10 .8
 0
  0  5.4
 0  16 .2  60 .0
 0
 0  13 .2  39 .6
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251y0132 11/19/01
 xy  Covxy  Exy   x  y  60 125  0 ,
 
 y2  E y 2   y2  68.2  52  43.2 .
(  x  7.0993 ,  y  6.57287 )
 
 x2  E x 2   x2  194 .4  12 2  50 .4 and
So that  xy 
 xy
 x y
0

 0.
50 .4 43 .2
Note that  2  1 is impossible as is a negative variance.
e) Repeat the joint probability table, and, in a second table, show the sums of x and y . Note that 15 is
repeated, so that the probability of 15 is Px  y  15   .2  .1  .3.
x
x
0

3
.1
.

y
9  0
11  .1

6 12
0 .1
.1
0
18
.2

.1
.2

0
.1
0
6 12

3
3 3 9

y
9  9 15 21
11  11 17 23

18
15 

27 
29 

It is easiest to do f(i) first. If you look at the table below Px  y  11  P3  P3  P9  .1  0  .1  .2
f(i))
x y
x y
Px  y 
Px  y 
-3
3
9
11
15
.1
0
.1
.1
.3
17
21
23
27
29
0
0
.1
.1
.2
It would be a good idea to take out the zero probabilities and get the table below. Note that you were not
supposed to do the last 2 columns, but that they serve as a check on f(ii)), verifying that E x  y   17 and


Var x  y   E x  y 2  E x  y 2  382 .6  17 2  93.6.
x y
Px  y 
x  y  Px  y 
-3
9
11
15
23
27
.1
.1
.1
.3
.1
.1
-0.3
0.9
1.1
4.5
2.3
2.7
0.9
8.1
12.1
67.5
52.9
72.9
29
.2
1.0
5.8
17.0
168.2
382.6
 x  y 2  P x  y 
f(ii)) Ex  y   Ex  E y    x   y  12  5  17 and
Var x  y    x2   y2  2 xy  Var x   Var  y   2Covx, y   50.4  43.2  20  93.6
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2. A warehouse will accept a lot of jorcillators if a randomly picked box of 50 jorcillators contains no more
than two defective jorcillators. Assume that, in fact, 4% of jorcillators in the lot are defective.
a) Without using a table, figure out the probability that no jorcillators in a box of 50 will be
defective. (2)
b) Without using a table, figure out the probability that no more than two jorcillators in a box of 50
will be defective. (3)
c) Using our test for use of the Poisson distribution to show that it is appropriate to use the Poisson
distribution, find an approximate answer to a) using a Poisson table. (2)
d) Find an approximate answer to b) using a Poisson table. (2)
e) Assume instead that jorcillators come in boxes of 625. What is the probability that the number
of defective jorcillators in a box will be between 11 and 20? (3)
f) Again consider a box of 625 jorcillators. We will reject the hypothesis (that the defect rate for
jorcillators is less than or equal to 4%) if the number, x 0 , of defective jorcillators in a box is so
large that Px  x 0   .05 . Using a table and the mean that you used is e), what is the minimum
number of defective jorcillators in a box that would lead to rejection of the hypothesis? (3).
Solution: a) Binomial (because we are taking 50 items in succession with a constant probability of
n!
getting a defective item.) p  .04, n  50 . Px  C xn p x q n x , q  1  p , C xn 
. Both
n  x ! x!
q and p must be positive and not above 1.
P0  C 050 p 0 q 50 
50 !
.04 0 .96 50  11.12989   .12989
50 ! 0 !
b) Px  2  P0  P1  P2  C050 p 0 q 50  C150 p1q 49  C250 p 2 q 48
 .12989 
50 !
.04 1 .96 49  50 ! .04 2 .96 48
49 !1!
48 ! 2 !
 .12989  50.04 .135298   502149 .0016 .140935 
 .12989  .27060  .27623  .67672
n 50

 2500 is above 500, use the Poisson distribution table with parameter
c) Since
p .02
m  np  50.04   2. Px  0  .13534 .
d) From the same table Px  2  .67668 .
e) Since
n 625

 15625 is above 500, use the Poisson distribution table with parameter
p .04
m  np  625 .04   25. P11  x  20   Px  20   Px  10   .18549  .00059  .1849 .
f) Again m  25. We want the minimum number x 0 , so that Px  x 0   .05 . This is the same as
saying Px  x 0   .95 . If we check the Poisson table we find that Px  33   .95022 is the first
value above 95%. So. If there are 33 or more defective jorcillators in the box, we reject the
hypothesis.
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3. A company produces camcorders. In any batch of camcorders 80% will last five or more years without a repair, 20%
will need repairs.
a. If I take 5 camcorders from a batch of 25, what is the probability that exactly four will need a repair before 5
years pass? (3)
b. If I take 5 camcorders from a batch of 25, what is the probability that at least one will need a repair before 5
years pass. (2)
c. What is the expected value and standard deviation of the number that will need repairs before 5 years pass.
(2)
d. Assume instead that the batch from which I take the 5 camcorders is a batch of 1200. Now find the
probability that exactly four will need a repair before 5 years pass without using the Hypergeometric distribution. (1)
e. Assume that the batch from which I take the 5 camcorders is a batch of 1200. Now find the probability that
at least one will need a repair before 5 years pass without using the Hypergeometric distribution. (2)
f. What is the mean and variance of the number in e) that will need repairs before 5 years pass? (2)
g. In d) and e) I said not to use the Hypergeometric distribution. This was to make your computations easier.
Explain what criterion I had to satisfy to be able to do this. (1)
h. Assume that the batch of camcorders is 10000. How many camcorders must I take (I took 5 before) to be
able to use the Poisson distribution to find the probability that at least one needs a repair? (2)
Solution: This problem is basically a hypergeometric problem (because we are taking a sample of
5 items from a population of only 25, so that the probability of a defective item keeps changing).
C M C N M
 N n
Px   x Nn  x , where M  Np ,   np and  2  
 npq .
 N 1 
Cn
a) N  25 , n  5, M  Np  25.20   5 .
5!
20 
5  4  3  2 1
1!4!
P4 

 520 
 .00188 .
25
25!
25  24  23  22  216
C5
20!5!
 20! 

1
5 20
15!5! 
C0 C5
20  19  18  17  16

b) Px  1  1  P0  1 
1
1
 1  .29181  .70819 .
25
25
!
25
 24  23  22  21
C5
20!5!
C 45 C120
25  5
20
 N n
0.8  .6667 .
5.20 .80  
c)   np  5.20   1 and  2  
 npq 
N

1
25

1
24


  0.6667  .8165 .
d) Use the binomial distribution when N is large compared with n . p  .2, n  5 .
Px  C xn p x q n x , q  1  p , C xn 
n!
. From the table
n  x ! x!
P4  Px  4  Px  3  .99968  .99328  .0064 . Or
P4  C 45 p 4 q 1  5.24 .8  .0064 .
e) With or without the table Px  1  1  P0  1  .85  1  .32768  .67232 .
f)   np  5.20   1 and  2 npq  5.20.80  0.800 or
1200  5
1195
 N n
0.8  0.7973 . So   .89 .
5.20 .80  
 npq 
1200  1
1199
 N 1 
2 
g) The condition to replace The Hypergeometric by the bimomial is N  20 n  20 5  100 . Since
N  1200 , we have satisfied this condition.
n
n
 500 to get n  500 .2  100 We can use the
 500 . If p  .2, we can solve
h) We need
.2
p
Poisson distribution if the sample size is above 100, as long as it is not above 5% of the
population.
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251y0132 11/19/01
4. I sell camcorders for $300 apiece. My variable cost to produce a camcorder is $150 a unit and my fixed
costs are $2000 a month. Thus, if I sell 20 camcorders, my revenue will be $300 (20) = $6000 and my
costs will be $150 (20) + $2000 = $5000 and I will have profits of $1000. Assume that the mean number of
units sold per month are 20 and that sales follow the Poisson distribution . Let x represent my sales.
a) If it takes a month to make a camcorder, what is the minimum number of camcorders that I
should keep on hand to be sure that I have a probability of no more than 1% of running out before
I can produce more? (2)
b) What is the probability that I will sell between 20 and 30 camcorders in a month? (2)
c) What is the probability that I will sell less than 10 in a month? (1)
d) Write an equation for my costs in terms of x and find the mean and standard deviation of my
costs. (3)
e) Using an equation for revenue as well as your equation for costs find the covariance and
correlation between revenues and costs. (4)
f) Using only your means and variances for costs and revenues and the covariance you found, find
the mean and variance of my profits. Remember, profits are revenues minus costs. (3)
Solution: For a) - c) all we need is a Poisson table with a parameter of 20.
a) The first number we can use on the table is 31, since Px  31  .99191 , and 31 is the smallest number
with a cumulative probability above 99%. This means that Px  31  1  .99191  .00809 (below 1%).
b) The table says P20  x  30   Px  30   Px  19   .98653  .47026  .51627 .
c) Px  10   Px  9  .00500 .
d) As explained in the problems, the Outline says that if w  ax  b and v  cy  d , E w  aEx   b,
E v   cE y   d , Varw   w2  Varax  b  a 2Varx, Varv   v2  Varcy  d   c 2Var y  ,
Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y) and Corr u, v   signacCorr x., y  . In problem K3 we
found Covx, x   Varx  and Corr x, x   1 .
For costs let v  150 x  2000 , so that c  150 and d  2000 . Remember that Varx   E x    x  20 ,
since the mean and variance are equal in the Poisson distribution.. Then, if we replace y by x in the
formulas, Ev   cEx   d  150 20   2000  5000 and
Varv    v2  Varcx  d   c 2Varx   150 2 20   450000 . the standard deviation is
 
 v  450000  150 20  670 .82 .
e) For revenue, we need w  300 x, so that a  300 and b  0. Using the formulas above or the results of
problem K3, we see that
Cov(w, v)  Cov(ax  b, cx  d )  acCov( x, x)  acVarx   300 150 20   900000 .
Corr u, v   signacCorr x., x   sign300 150 1  sign45000 1  1.
f) Using the formulas above, Ew  E300 x   300 Ex   300 20   6000 and
Var w  300 2 20   1800000 From the outline or the syllabus supplement, we know that we can add and
subtract means and that Var(ax  cy)  a 2Var( x)  c 2Var( y)  2acCov( x, y) This means that
E w  v   E w  E v   6000  5000  1000 and
Var (w  v)  Var (w)  Var (v)  2Cov(w, v)  1800000  450000  2900000   450000 . To confirm this try
Pr ofit  300x  150x  2000  150x  2000. So EPr ofit   150 20   2000  1000 and
Var Pr ofit   150 2 20   450000 .
7
251y0132 11/19/01
5.
a. Identify the distribution in the problem, give and explain its mean and give its standard
deviation. Do not answer the question in the problems in part a.!
Example: An average of 200 cars go through a toll booth each hour, what is the
probability that over 150 go through the toll booth in a half hour? Solution: Poisson (with
a parameter of 100) . The mean is   100 , the average number of cars that go through
the booth in a half hour when the mean is 200 per hour.   100  10. Do not tell me
the probability that over 150 go through the booth!
(i) 18 customers enter my store. Each one has a 27% chance of buying something. What is the chance that at
least one buys something? (3)
(ii) Each customer that enters my store has a 27% chance of buying something. What is the chance that the 6 th
customer is the first to buy? (3)
(iii) In a package of 60 condoms, 5% are defective. If I sell 3, what is the chance that 2 are defective? (3)
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. From here on you are actually
doing the problems!
b. If a phillinx has a lifespan approximated by a uniform distribution between 2 and 12 years, find:
(i)
The mean and standard deviation of its life. (2)
(ii)
The probability that it will last between 0 and 5 years (2)
(iii)
The probability that it will last between 5 and 10 years (1)
(iv)
The probability that it will last more than 10 years (1)
c. If a flubberall has a lifespan approximated by a uniform distribution between 5 and 20 years, do
(i) b(ii)-b(iv) above for the flubberall (2) and then (assuming that the lifespan of the two
components is independent) find:
(ii)
The probability that the jorcillator will last between 0 and 5 years. (2)
(iii)
The probability that the jorcillator will last between 5 and 10 years. (3)
(iv)
The probability that the jorcillator will last more than 10 years. (2)
Solution: a)
(i) 18 customers enter my store. Each one has a 27% chance of buying something. What is the
chance that at least one buys something? (3)
Binomial Distribution with p  .27 , n  18 . q  1  p  1  .27  .73 .
  np  18.27   4.86 , the average number out of 18 customers that will buy
something.  2  npq  4.86.73  3.5478 .   npq  3.5478  1.8836 .
(ii) Each customer that enters my store has a 27% chance of buying something. What is the chance
that the 6th customer is the first to buy? (3)
1
1
 3.70370 , the number of the
Geometric Distribution with p  .27 ,   
p .27
average customer who buys first - in other words, on the average the fourth or fifth
q
.73
 10 .0137 so
customer to enter the store will actually buy something.  2  2 
p
.27 2
  10.0137  3.1644.
(iii) In a package of 60 condoms, 5% are defective. If I sell 3, what is the chance that 2 are
defective? (3)
Hypergeometric Distribution with p  .05 , n  3 , N  60 .
  np  3.05   0.15, the average number that will be defective out of 3.
 N n
 npq 
 N 1 
2  
 60  3 
 60  1  0.15 .95   0.1377 so   0.1377  0.3710 .


8
251y0132 11/19/01
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. From here on you are actually doing the problems!
b. If a phillinx has a lifespan approximated by a uniform distribution between 2 and 12 years, find:
(v)
The mean and standard deviation of its life. (2)
(vi)
The probability that it will last between 0 and 5 years (2)
(vii)
The probability that it will last between 5 and 10 years (1)
(viii)
The probability that it will last more than 10 years (1)
c. If a flubberall has a lifespan approximated by a uniform distribution between 5 and 20 years, do (i) b(ii)-a(iv) above for
the flubberall (2) and then (assuming that the lifespan of the two components is independent) find:
(ii)
The probability that the jorcillator will last between 0 and 5 years. (2)
(iii)
The probability that the jorcillator will last between 5 and 10 years. (3)
(iv)
The probability that the jorcillator will last more than 10 years. (2)
Solution:
b) For the Phillinx: c  2, d  12 Make a diagram. Show a box between 2 and 12 with a height of
1
1
d c   122  0.1 . Shade the areas between 2 and 5, between 5 and 10 and between 10 and 12.
c  d 2  12

7
2
2

(i)
2 
d  c2
12

12  22
12
 8.3333   8.3333  2.8868
Let A be the event that the Phillinx lasts 0 - 5 years. P A  P0  x  5  5  2
1
3

.
10 10
1
5
(iii)
Let B be the event that the Phillinx lasts 5 - 10 years. PB   P5  x  10   10  5 
.
10 10
(iv)
Let C be the event that the Phillinx lasts more than 10 years.
1
2
PC   Px  10   12  10  
10 10
Note that these three probabilities must add to 1.
(ii)
c) (i) For the Flubberall: c  5, d  20 . Make a diagram. Show a box between 5 and 20 with a height of
1
1
d c   205  15 . Shade the areas between 3 and 5 (There is none!), between 5 and 10 and between 10
and 20.
Let D be the event that the Flubberall lasts 0 - 5 years. PD   P0  y  5  0
1
Let E be the event that the Flubberall lasts 5 - 10 years. PE   P5  y  10   10  5
1 1
 =.3333
15 3
1 2
Let F be the event that the Flubberall lasts more than 10 years. PF   P y  10   20  10  
15 3
=.6667
(ii) - (iv) A joint probability table follows on the left with the unions of joint events that make up the
events that were specified summed to the right. Note that the probabilities sum to one. Note also that these
probabilities are not the same as on the 1998 exam.
D E
F
Event
Component Joint Events
Probability
Fails
in







P
A

D

A

E

A

F
3
6
3
3
6
9
A 0
0

00 
 .3000
30
30
10
0-5
30 30
30
 B  D  C  D 
10
5
B 0 5
30
C
0 2
0
30
1
3
30
4
30
2
3
10
2
10
1
Fails in
5-10
Fails in
10+
Sum
PB  E   B  F   C  E  
PC  F  
5 10 2



30 30 30
4

30
17
 .5667
30
4
 .1333
30
1.0000
9
251y0132 11/19/01
Extra Credit: The main computer we use for payroll crashes on the average 156 times a year. It has just
crashed. This means that the average waiting time before the next crash will be 1156 of a year. Assume that
a year consists of 52 weeks (A week is 152 year.)
a. What is the probability that the time before the next crash will be less than 1 week? (3)
b. What is the probability that the time before the next crash will be between 2 and 3 weeks? (2)
c. What is the standard deviation for this distribution? (1)
Solution: You were told in advance that this section would use the exponential distribution. In the
exponential distribution distribution from the outline F x 1  e cx , when the mean time to a success is
1
1
  
.
c
c
Note that this is only for x  0 . There is no probability below zero.
1
1
1
 156 . Px  152  F  152  1  e156 52  1  e3  1  .0498  .9502
a) Since   1156 , c  

b) P2 52  x 
3
52
c) Since    
1
156
  F 2 52  F 3 52  1  e156
1
,
c
2


  1  e156352  e6  e9  .00248  .00012  .00236
52
1
156
10