![[Maple Math]](images/binomexp9.gif){kind=small}
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![[Maple Math]](images/binomexp10.gif){kind=small}
Using Tables and Computers
We will use tables generated by a computer to help us answer questions about binomial probabilities.
Example 1
If a fair coin is tossed 16 times, find the probability that there are at least 10 heads.
Solution .
To use the computer we need to know that n = 16 and that p = 1/2. To get the answer from the table that is generated we need to realize that the probability of at least 10 heads is obtained by
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Note that in the table below the values we want are in the column called prob(x).
(The other information in the table about the mean, variance, and standard deviation will be used in the next section.)
![[Maple OLE 2.0 Object]](images/binomexp11.gif){kind=small}
x prob(x) x*prob(x) ((x - mean)^2)*prob(x)
---------------------------------------------------------------
0 .15259E-4 0 .97661E-3
1. .24414E-3 .24414E-3 .11963E-1
2. .18311E-2 .36622E-2 .65921E-1
3. .85449E-2 .25635E-1 .21363
4. .27771E-1 .11108 .44436
5. .66650E-1 .33325 .59989
6. .12220 .73320 .48885
7. .17456 1.2219 .17459
8. .19639 1.5711 .19639E-8
9. .17455 1.5710 .17452
10. .12219 1.2219 .48871
11. .66650E-1 .73315 .59981
12. .27771E-1 .33325 .44431
13. .85449E-2 .11108 .21361
14. .18311E-2 .25635E-1 .65918E-1
15. .24415E-3 .36623E-2 .11963E-1
16. .15259E-4 .24414E-3 .97655E-3
SUM 1.0000 8.0001 4.0000
mean variance
2.0000
standard deviation
Note that E is used to indicate the exponent in scientific notation. Consequently P(10) = .12219, P(11) = .066650, P(12) = .027771, P(13) = .0085449, P(14) = .0018311, P(15) = .00024415, and P(16) = .000015259. Adding these values we obtain, .227
In some tables that you may use, you may see a zero listed for some values that are less than .0005
Example 2
Find the probability that number of heads is at least six and at most 8 when a coin is tossed 16 times.
Solution .
We need to find P(6) + P(7) + P(8). From the table generated for example one we get, .12220 + .17456 + .19639 = .49315
Example 3
A student who has not studied takes a 20 question mutiple choice history quiz. If the student guesses on each question ( and they all have 5 choices), find the probability that the student gets at least 70% correct.
Solution .
Since 70% of 20 is 14, we must find P(at least 14 correct) = P(14) + P(15) +...+P(20). Using n = 20 and p = 1/5, we obtain the following table.
x prob(x) x*prob(x) ((x - mean)^2)*prob(x)
---------------------------------------------------------------
0 .11529E-1 0 .18448
1. .57648E-1 .57648E-1 .51887
2. .13691 .27382 .54769
3. .20536 .61608 .20540
4. .21820 .87280 .21820E-8
5. .17456 .87280 .17453
6. .10910 .65460 .43636
7. .54550E-1 .38185 .49092
8. .22161E-1 .17729 .35455
9. .73870E-2 .66483E-1 .18467
10. .20313E-2 .20313E-1 .73125E-1
11. .46169E-3 .50786E-2 .22622E-1
12. .86564E-4 .10388E-2 .55399E-2
13. .13318E-4 .17313E-3 .10787E-2
14. .16647E-5 .23306E-4 .16647E-3
15. .16647E-6 .24971E-5 .20143E-4
16. .13006E-7 .20810E-6 .18729E-5
17. .76503E-9 .13006E-7 .12929E-6
18. .31876E-10 .57377E-9 .62477E-8
19. .83888E-12 .15939E-10 .18875E-9
20. .10486E-13 .20972E-12 .26844E-11
SUM 1.0000 4.0001 3.2000
mean variance
1.7889
standard deviation
Note that P(14) = .0000016647 and that the other probabilities we need are even smaller. There is not a very large probability that the student will get at least a "C" by just guessing.
We would expect that a student would get 4 correct by guessing.