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500y0711 12/26/06
ECO251 QBA1
FIRST EXAM
ECO500 Business Statistics
Name: _____________________
Student Number : _____________________
Remember – Neatness, or at least legibility, counts. In all questions an answer needs a calculation or
explanation to count. Show your work!
Part I. (10 points)
The following numbers are a sample of 20.
22, 11, 22, 23, 17, 25, 17, 20, 26, 14, 23, 15, 34, 12, 15, 11, 19, 16, 34, 24
Compute the following: Show your work!
a) The Median (1)
b) The Standard Deviation (3)
c) The Interquartile Range (3)
d) The Coefficient of variation (1)
e) A measure of skewness (2)
Solution: The table computations which could be needed for this problem follow. Obviously, no one needs
to do all of these.
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
x
22
11
22
23
17
25
17
20
26
14
23
15
34
12
15
11
19
16
34
24
400
x in order x 2
11
11
12
14
15
15
16
17
17
19
20
22
22
23
23
24
25
26
34
34
x  x  x  x 2 x  x 3
x3
484 10648
121
1331
484 10648
529 12167
289
4913
625 15625
289
4913
400
8000
676 17576
196
2744
529 12167
225
3375
1156 39304
144
1728
225
3375
121
1331
361
6859
256
4096
1156 39304
576 13824
8842 213928
2
-9
2
3
-3
5
-3
0
6
-6
3
-5
14
-8
-5
-9
-1
-4
14
4
0
4
81
4
9
9
25
9
0
36
36
9
25
196
64
25
81
1
16
196
16
842
8
-729
8
27
-27
125
-27
0
216
-216
27
-125
2744
-512
-125
-729
-1
-64
2744
64
3408
 x 400
 x  400,  x  8842,  x  213928 so x  n  20  20 . If we use
definitional formulas  x  x   0 (a check),  x  x   842 and  x  x   3408.
2
To summarize
3
2
3
a) Median: position  pn  1  0.521  10.5 ; x1 p  xa  .bxa 1  xa  so
x.5  x10  .5x11  x10   19  .520  19   19 .5 .
b) Standard Deviation: s
s2 
  x  x 2
n 1

2
x

2
 nx 2
n 1

8842  20 20 2 842

 44 .3158 or
19
19
842
 44 .3158 . So s  44.3158  6.6570.
19
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c) Interquartile range: To find the first quartile position  pn  1  0.2521  5.25 .
x1 p  x a  .bx a 1  x a  so x.75  x 5  .25x 6  x 5   15  .2515  15   15
To find the third quartile position  pn  1  0.7521  15.75 . x1 p  x a  .bx a 1  x a  so
x.25  x15  .75x16  x15   23  .7524  23   23 .75 . IQR  x.25  x.75  23.75  15  8.75.
d) Coefficient of variation: C 
e) Measure of skewness:
n
k 3
x 3  3x
(n  1)( n  2)

s
6.6570

 3.3285
x
20
x
2

 2nx 3 

20
213928  320 8842  220 20 3
1918 

20
213928  530520  320000   20 3408  199 .2984 or
1918 
1918 
k
n
x  x 3  20 3408  199 .2984 or g1  33  199 .29843  0.6756 or
k 3
(n  1)( n  2)
19 18 
s
6.6570
3mean  median 320  19 .5

 0.225 . All these measures of skewness indicate that the
SK 2 
6.6570
std .deviation
distribution is skewed to the right. The last two tell us that the skewness is relatively small.


2
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Part II. (At least 40 points. Parentheses give points on individual questions. Brackets give cumulative
point total.) Exam is normed on 50 points.
1. For the sample above, find the two numbers that would be at the limits of an interval of 2.5 standard
deviations from the mean x  2.5s  . If these were population parameters instead of sample statistics, what
would be the minimum fraction of the data between these two numbers according to Tchebyschev's rule?
The Empirical rule says that about 98.8% of the data should be between these two points. How does the
fraction of the data actually between these two points compare? (5)
Solution: We have x  20 and s  44.3158  6.6570, so the interval in question is x  2.5s 
 20  2.56.6570   20  16.64 or 3.36 to 36.64.
The numbers in order are
11 11 12 14 15 15 16 17 17 19 20 22 22 23 23 24 25 26 34 34
All are between 3.36 and 36.64. The Tchebyschev Rule says that the proportion onside the interval
1
1

 16 % , which means at least 84% should be in the interval. There is
should be no more than
2
6
.
25
2.5
no reason why it shouldn’t be more than 84%. As far as the empirical rule is concerned, though the percent
in the interval seems close to 98.8%, the rule does not apply since we have already shown that the
distribution is not quite symmetrical.
2. (Most of the following problems are due to Mansfield and Schmidt) As you leave a restaurant the
attendant gives you a randomly selected coat from the coatroom. If 10% of the coats are worth $200, 40%
are worth $100 and 50% are worth $50, show that you have a valid distribution and compute the expected
value and standard deviation of the value of the coat you receive. (5)
Solution: Computations are below.
Row
1
2
3
Px  xPx  x 2 Px 
x
200
100
50
0.1
0.4
0.5
1.0
20
40
25
85
4000
4000
1250
9250
x   x  x   x Px x   x 2 Px 
115
15
-35
11.5
6.0
-17.5
0.0
1322.5
90.0
612.5
2025.0
We have shown that this is a valid distribution because all probabilitites are between zero and one and their
xPx   85 and E x 2 
x 2 Px   9250 . This means
sum is 1. The next two columns show  x 
that the (population) variance is  x2

 E x  
2
  
2
x
 9250  85 2  2025 . Thus the (population) standard
deviation is  x  2025  45 . The last three columns are totally unnecessary but show that
 x   Px   0 , which must be true if the mean is correct and, using the definitional formula, that
   x    Px  2025. This serves as a check on our work.
x
2
x
2
x
3. If the gross profit on each calculator sold is $5 and the fixed cost is $100000 per year, so that if the firm
sells 50000 calculators, annual profit is 550000   100000  150000 , what is the expected value and
standard deviation of the firm’s annual profit? (4) ( I should have said that average sales are $75000 units
with a standard deviation of $30000.)
[14]
Solution: The formula relating profit, w, to sales is w  5x  100000 . The formula table says
E ax  b  aEx   b and Varax  b  a 2Varx . So E5x  100000   5Ex   100000 and
Var5x 100000  5 2 Varx , so that the standard deviation is 5 times the standard deviation of x . If you
were around when I finally remembered the numbers, this gives
Ew  E5x  100000   575000   100000  275000 Varw  Var5x  100000   5 2 30000 2 . The
standard deviation is thus 530000   15000 .
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4. If a population has a mean weight of 140 pounds and a standard deviation of 30 pounds, what is the mean
and standard deviation of the total weight of 3 randomly selected people? (3)
[17]
Solution: The section on functions of random variables in the Syllabus Supplement says the following.
“ If x1 , x2 , x3 , xn are random variables, then Ex1  x 2  x3   x n 
 E x1   E x 2   E x3    E x n  ,” and “If x1 , x 2 , x3 ,  x n are independent random variables, then
Varx1  x 2  x3   x n   Varx1   Varx 2   Varx3   Var x n  . So, if the individuals weights are
x1 , x 2 and x3 , we have E x1  x 2  x3   Ex1   Ex 2   Ex3   3(140 )  420 .
Var x1  x 2  x3   Var x1   Var x 2   Var x3   330 2  3900   2700 , so that Std.Devx1  x 2  x3 
 2700  51.962 .
5. Two cards are drawn at random from a deck of cards. Identify the following events: A1 an ace on the


first draw and A2 an ace on the second draw. Find the following: (a) P A1  , (b) P A1  A2  , (c) P A2 A1 ,


(d) P A2 A1 , (e) P A1  A2  , (g) P A2  . Can you explain why P A1  A2   P A1   P A2  ? Repeat the
exercise for the following events H 1 , a head on the first flip and H 2 , a head on the second flip. By
contrasting P A1  A2  and PH 1  H 2  , show that only one of these pairs of events are independent. (8)
Solution: The difference between A1 and A2 on one hand and H 1 and H 2 on the other is that the first
pair is not independent and the second one is. Since there are for aces in the deck, we can write that
4
 .07692 . If we get an ace on the first draw, there are only 3 left in the deck so we have
(a) P A1  
52
3
4
 .05882 . (d) But P A2 A1 
 .07843 . By the multiplication rule P A1  A2 
(c) P A2 A1  
51
51
3 4
 .00452 The easiest way to do P A1  A2  is to note that there
(b) P A2  A1   PA2 A1 P A1  
51 52
47 48
 .85068 and that (e) P A1  A2 
are 48 non-aces in the deck, so that P A2  A1  P A2 A1 P A1 
51 52
47 48
 1  P A2  A1  1  P A2 A1 P A1  1 
 1  .85068  .14932 . It is also probably worth noting
51 52
4!
43
C 24 C 048
2! 2!
that P A2  A1  

 2 1  .00452 . (g)But we still haven’t found P A2  . There are
52
52
!
52  51
C2
50! 2!
2 1
several ways to do this, all requiring some thought. P A1  A2   P A1   P A2   P A1  A2  , which is, of






 
 
 
course why P A1  A2   P A1   P A2  . Perhaps this is best explained by saying that A1 and A2 are not
mutually exclusive . So .14932  .07692  P A2   .00452 and P A2   .14932  .07692  .00452  .07692

 
or P A2   PA2 A1 P A1   P A2 A1 P A1 
3 4
4 48 12  192
4



 .07692 . If you tried to work
51 52 51 52
51  52
52
A1 A1
A2
this with a joint probability table, you would have
A2
3 4
51  52
48  4
51  52
4
52
4  48
51  52
47  48
51  52
48
52
or
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500y0711 12/26/06
A1
A2
A2
A1
.0045249
.0723982
.0769231
.0723982
.8506787
.9230769
.0769231
..9230769
1.00000
. Somehow, it all adds up. But we can see clearly that
43
4 4
. So it seems that A1 and A2 are not mutually exclusive and
 P A2  A1   P A1 P A2  
52  51
52 52
not independent.
Because PH 1   PH 2   PH 2 H 1   P H 2 H 1  .5 we can say



PH 2  H 1   PH 2 H 1 PH 1  

11 1
 . S we have PH 1   .5 , PH 1  H 2   .25 , P H 2 H 1  .5 ,
22 4
P H 2 H 1  .5 , PH 1  H 2   .5  .5  .25  .75 , PH 2  . PH 1  H 2   PH 1   PH 2  and these two


events are not mutually exclusive but are independent.
6. Your significant other is supposed to pick you up at 5pm. The probability that said other remembered to
buy gas is 60%. If said other remembered there is a 90% probability that you will be picked up on time. If
buying gas was forgotten the probability that you will be picked up on time is 60%. No one shows at 5pm.
What is the probability that your significant other forgot to buy gas? To keep us all together on this, let G
be the event that gas is remembered and T be the event that you are picked up on time. There is partial
credit her for finding the probability of events like T  G or T  G . (5)
[30]
 
 
 
Solution: We are given that PG   .6, P T G  .9 and P T G  .6 . We are asked for P G T .We could
    , or we could do a box solution. If we can talk about 100 dates, SO
 
PT G PG
write Bayes’ rule P G T 

PT
 
will have remembered 60 times and forgotten 40. Since P T G  .9 , on the 60 times that SO remembers,
 
SO will be on time 54 times. Since P T G  .6 , on the 40 times SO forgets, SO will be on time .640   24
times. Lets work this out.
G
T T
54
60
G
24
40
. If we just fill in the blanks we get
G
T T
54 6
G
24 16
100
gives us
G
G
T T
54 6
60
24 16 40
78 22 100
60
40
100
. Adding
. So On the 22 occasions that SO is not on time there are 16 on which gas was
forgotten. The probability is thus
 
 
16
 .72727 . Lets try this formally P T G  .6 , so P T G  1  .6  .4
22
  
   
  
 
that PT G  1  .9  .1 . So PT   .1.6  .4P.4  .06  .16  .22 . So Bayes’ rule says
PT G PG  .4.4 .16
PG T  


 .72727
.22  .22
PT 
P T  P T  G  P T  G  P T G PG   P T G P G . We know that P G  1  PG  1  .6  .4 and
5
500y0711 12/26/06
7. The formula for the probability of x successes in n tries is Px  C xn p x q n x , where p is the
probability of a success on 1 try and q is the probability of failure. For example, if your probability of
success on 1 try is .20 and the probability of failure is .80, the probability of 3 successes on 7 tries is
7!
.20 3 .80 4 . This may help you in the following: You roll one die and then toss a coin the
C 37 p 3 q 4 
3!4!
number of times indicated by the die; for example if you roll a two you toss the coin twice, Let y be the
number of heads you get. Find the possible values of y and their probabilities. Show that y has a valid
distribution and find the probabilities of each value of y . Find E  y  and  y . (6)
[36]
Solution: I guess we could start with the probabilities we need. y will range from 1 to 6. Let w be our
result from the die. For example if y  3 , we can get three heads in 4 ways: w  3 and we get 3 heads in 3
tries, w  4 and we get 3 heads in 4 tries; w  5 and we get 3 heads in 5 tries; w  6 and we get 3 heads in
1
1
1
1
6 tries, The probability for this is C 33 .53  C 34 .54  C 35 .55  C 36 .56
6
6
6
6


1
5!
3
4
.55  6! .56   1 .12500  .25000  .31250  .31250   1 1.0000   .1667 .
1.5  4.5 
6
2! 3!
3! 3!
6
 6
We can read the probabilities in the brackets from table E.6 in the text. Let w represent the amount on the
face of the die.
w
Sum
P0
P 1
P2
P3
P4
P5
P6
.5000
.5000
1.0000
1
.2500
.5000
.2500
1.0000
2
.1250
.3750
.3750
.1250
1.0000
3
.0625
.2500
.3750
.2500
.0625
1.0000
4
.0312
.1562
.3125
.3125
.1562
.0312
0.9998
5
.0156
.0937
.2344
.3125
.2344
.0937
.0156 0.9999
6
0.9843 1.8749
1.5469
1.0000
0.4531
0.1249
0.0156 5.9997
Sum
There is a slight rounding error in these sums, but they should be good enough. The probabilities on the
table are, as far as the present problem is concerned, conditional probabilities. For example,
P 1 head w  3  .3750 . To allow for the die, multiply the column sums by 1 6 .


Row
y
1
2
3
4
5
6
7
0
1
2
3
4
5
6
P y 
0.1641
0.3125
0.2578
0.1667
0.0755
0.0208
0.0026
1.0000
yP y 
y 2 P y 
0.0000
0.3125
0.5156
0.5001
0.3020
0.1040
0.0156
1.7498
0.0000
0.3125
1.0312
1.5003
1.2080
0.5200
0.0936
4.6656
y   y  y   y Py y   y 2 P y
-1.7498
-0.7498
0.2502
1.2502
2.2502
3.2502
4.2502
-0.287142
-0.234313
0.064502
0.208408
0.169890
0.067604
0.011051
0
0.502441
0.175688
0.016138
0.260552
0.382287
0.219727
0.046967
1.603800
We have shown that this is a valid distribution because all probabilitites are between zero and one and their
yP y   1.7498 and E y 2 
y 2 Px   4.6656 . This
sum is 1. The next two columns show  y 
means that the (population) variance is  y2

 E y  
2
  
2
y
 4.6656  1.74982  1.6038 . Thus the
(population) standard deviation is  y  1.6038  1.2664 . The last three columns are totally unnecessary
but show that
formula, that
 y   P y   0 , which must be true if the mean is correct and, using the definitional
   y    P y   1.603800.
y
2
y
2
y
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500y0711 12/26/06
8. As everyone knows, a Veeblefetzer (Actually, it should have been a Jorcillator) has two components, a
Phillinx and a Flubberall. Depending on the design of the Veeblefetzer, the Veeblefetzer will work a) as
long as both components work or b) as long as either component works. Let us assume that rule b) is in
effect and that the life of a phillinx, x1 , is a random variable with a continuous uniform distribution
between 0 and 4, so that the probability of it failing in the first year is P0  x1  1 . The life of a
Flubberall, x 2 , is a random variable with a continuous uniform distribution between 0.5 and 2.7. What is
the probability that the Veeblefetzer fails in Year 1? Year 2? After Year 2? If you will lose $2million if it
fails in the first year, break even if it fails in the second year and make $5 million of it lasts beyond the
second year, what is the standard deviation of your earnings? (5)
[41]
Solution: For the first part of the problem find the uniform distribution probabilities. Define the following
events.
Period
Phillinx
Flubberall
Fails
Fails
Make a diagram for the Phillinx. Show a
1
rectangle that goes from zero to 4, with a
A1
B1
height of .25 Shade the areas from zero to 1, 1
2
A2
B2
to 2, and above 2.
3
A3
B3
A1
P A1   1  0.25  .25
P A2   2  1.25  .25
A2
A3
P A3   4  2.25  .50
Note that these add to 1.
Make a diagram for the Flubberall. Show a rectangle that goes from 0.5 to 2.7, with a height of
1
1

 .45455 Shade the areas from 0.5 to 1, 1 to 2, and above 2.
2.7  0.5 2.2
PB1   1  0.5.45455  .22727
B1
B2
B3
PB2   2  1.45455  .45455
PB3   2.7  2.45455  .31818 Note that these add to 1.
Now we need a joint probability table.
Event
B1
B2
A1
A2
A3
Sum
Sum
.05682
.05682
B3
.11364
.11364
.11364
.22728
.45455
.07955
.07955
.15909
.31818
.25
.25
.50
1.0000
.22727
Assume that the Veeblefetzer will work as long as one component works. Fill in the table with the period in
which the Veeblefetzer will fail.
Event
Sum
B1
B2
B3
A1
A2
A3
Sum
Period 1
Period 2
Period 3
Period 2
Period 2
Period 3
Period 3
Period 3
Period 3
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500y0711 12/26/06
Use these two tables to figure out the probability that the Veeblefetzer will fail in each period.
Period
1
Component Joint Events
A1 B1
Probability
.05682
2
A1 B2 , A2 B1 , A2 B 2
3
A1 B3 , A2 B3 , A3 B1 , A3 B2 ,
A3 B3
.05682 + .11364 + .11364 =
.28410
.11364 + .22728 + .07955 +
.07955 + .15909 = .65911
These probabilities add to one, So we can figure out our return as in the previous problems.
Computations are below.
Row
6
7
8
-2
0
5
xPx 
x 2 P x 
x   x 
-0.11364
0.00000
3.29555
3.18191
0.2273
0.0000
16.4778
16.7050
-5.18191
-3.18191
1.81809
P x 
x
0.05682
0.28410
0.65911
1.00003
 
x   x Px x   x 2 Px 
-0.29444
-0.90398
1.19832
0
1.52574
2.87639
2.17866
6.58078
 x2  E x 2   x2  16 .7050  3.18191 2  6.5805 . Thus the (population) standard deviation is
 x  6.58078  2.565 .
A1
B1
9. The following joint probability represents returns for two different stocks. B 2
B3

a)Find the following: P A1  B2  , P A1  B2  , P A1 B2
A2
A3
.05 .05 .20 


.05 .30 .05 
.20 .05 .05 



The returns on stock 1 are A1 x1  10 % , A2 x1  20% and A3 x1  30%
The returns on stock 1 are B1 x 2  5% , B 2 x 2  10% and B 3 x 2  15%
b) Find E x1  , E x 2  ,  x21 ,  x22 ,  x x ,  x x
1 2
1 2
c) If your return is x1  x 2 , find the mean and standard deviation of the return.
d) If your return is P1 x1  P2 x 2 and P1  P2  1 , find the proportions P1 and P2 that will lead to a
minimum risk. (9)
Solution: a) Complete the joint probability table as follows.
A1 A2 A3 sum
B1 .05 .05 .20  .30


From this we see P A1  B2   .05 ,
B 2 .05 .30 .05  .40


B3 .20 .05 .05
.30


.30 .40 .30 1.00
P A1  B2   P A1   PB2   P A1  B2   .30  .40  .05  .65
PA1 B2  
P A1  B2  .05

 .125
P B 2 
.40
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500y0711 12/26/06
b)

E x1 x 2  
 .00025
 .00050
.00300
 .05 .10 .05  .05 .20 .05  .20 .30 .05 
x1 x 2 Px1 , x 2    .05 .10 .10  .30 .20 .10  .05 .30 .10 
 .20 .10 .15  .05 .20 .15  .05 .30 .15 
.00050
.00300 
.00150   .01850 .
.00225 
.00600
.00150
We can now use the following tableau to compute the means and variances of x1 and x 2 .
x1
.05
y
.10
.15
Px1 





.10
.05
.20
.05
.05
.30
.20
.05
.30
.40
.30
.20 

.05 
.05 

.30
x1 Px1  .0300  .0800  .0900 
x12 P
x 
Px 2  x 2 Px 2  x 22 Px 2 
.30
.015
.00075
.40
.040
.00400
.30
.045
.00675
1.00
.100
.01150
.200
.0030  .0160  .0270 
.046
 Px   1 (a check),   Ex    x Px   0.200 , E x    x
 Px   1 ,   E x    Px   0.100 and E x    x Px   0.01150
To summarize
2
2
1
x2
x1
2
1
1
2
2
2
2
2
1
2
2
2
2P
x 2   0.046 ,
2
 x1x2  Covx1 x2   Ex1 x2    x1  x2  0.01850 0.2000.100  0.0015
 
 Ex  
 x21  E x11   x21  0.046  0.2002  0.006 and
 x22
 x1x2 
2
2
 x1x2
 x1  x2
2
x2

 0.01150 0.1002  .0015 . So that
 0.0015
0.006 0.0015

0.0015 2
  0.2500  .5 .
0.006 0.0015 
c) The outline says a. Ex  y   Ex   E y  and
Var x  y    x2   y2  2 xy  Var x   Var y   2Covx, y 
So E x1  x 2   0.200  0.100  0.300 and Varx1  x 2   0.006  .0015  2.0015   .0045 .
9
500y0711 12/26/06
d) The outline says If R  P1 R1  P2 R2 and P1  P2  1 , then ER   P1 ER1   P2 ER2  and
VarR  P12VarR1   P22VarR2   2P1 P2 CovR1 , R2   P12 0.006   1  P1 2 .0015   2P1 1  P1 0.0015




 P12 0.006  1  2P1  P12 .0015  2P1  2P12 0.0015
   .0030  .0030P1  .0015  .0105P12  .0060P1  .0015 . Using Calculus

this hits a minimum when 2.0105 P1  .0060 or P1  .2857 .
0.006  0.0015  .0030 P12
To find this by groping around, note the following.
If P1  0 , .0105P12  .0060P1  .0015  .0015 . Std deviation is .0387.
If P1  0.1 . .0105P12  .0060P1  .0015  .0105 0.01  .0060 0.1  .0015  .000105  .00060  .0015
= .001005. Std deviation is .0317.
If P1  0.2 . .0105P12  .0060P1  .0015  .0105 0.04   .0060 0.4  .0015  .00042  .0012  .0015
= .000720. Std deviation is .0268.
If P1  0.3 . .0105P12  .0060P1  .0015  .0105 0.09   .0060 0.3  .0015  .000945  .0018  .0015
=.000645. Std deviation is .0254.
If P1  0.4 . .0105P12  .0060P1  .0015  .0105 0.16   .0060 0.4  .0015  .00168  .0024  .0015 =.00078.
Std deviation is .0279.
If P1  0.5 . .0105P12  .0060P1  .0015 = .001125. Std deviation is .0335.
If P1  0.6 . .0105P12  .0060P1  .0015 = .001680. Std deviation is .0410.
If P1  0.7 . .0105P12  .0060P1  .0015 = .002445. Std deviation is .0494.
If P1  0.8 . .0105P12  .0060P1  .0015 = .003420. Std deviation is .0585.
If P1  0.9 . .0105P12  .0060P1  .0015 = .004605. Std deviation is .0679.
If P1  1.0 . .0105P12  .0060P1  .0015 = .00600. Std deviation is .0775.
We find a minimum between P1  0.2 and P1  0.3 . If you now try P1  0.20, P1  0.21 etc,, you should
find the lowest between .28 and .29, which is good enough.
[50]
10