3 Example Homework Problems for Week 3 of Math 221

1. Week 3 Three Extra Homework
Examples (3x9)/W
MA 221
Statistics for Decision Making
Professor Brent Heard
Not to be copied or linked to
without my permission

2. (3x9)/W
• Number 11 Example
▫ On number 11 in the homework, they are just
trying to get you to calculate the number of
different combinations of a set of letters. There
will often be duplicates in these letters and that’s
where the problem is for some students.
▫ It’s really easy. I would recommend using either a
calculator or Excel

3. (3x9)/W
• Let’s say the question was “How many different
7 letter words (real or imaginary) can be formed
from the following letters?”
▫ A, B, C, D, A, B, E
▫ What you need to know….





We have a total of 7 letters
There are 2 A’s
There are 2 B’s
There is 1 C
There is 1 E
▫ Go on to the next page to see how easy it is

4. (3x9)/W
• We have the factorial of the total number of letters
in the numerator (7!)
• We have the factorials multiplied of the numbers of
the individual letters in the denominator
• So we have (7!)/(2!2!1!1!)
• 7! Is just 7x6x5x4x3x2x1 Your calculator should
have a factorial button, usually “x!” In other words
if you input a 7, then hit the x! button you will get
the answer
• 7! = 5040
• You can also use Excel by typing into an open
cell, =fact(7) and then hitting the enter key.

5. (3x9)/W
• So (7!)/(2!2!1!1!) = 5040/4 = 1260 (The answer)
▫ Remember (2!2!1!1!) = 2x1x2x1x1x1 = 4
• Let’s say we had 10 letters A,B,C, A, B, C, A, A, B, D (Or 4 A’s, 3 B’s, 2 C’s
and 1 D)
• The calculation would be
▫ (10!)/(4!3!2!1!) = 12600
 Or 3628800/(24x6x2x1) = 3628800/(288) = 12600

6. (3x9)/W
• Number 12 Example
▫ On number 12, they just want you to understand
about probabilities
▫ For example, let’s say a horse farm has 3 horses in
an 8 horse race. And they ask, “What is the
probability of those three horses finishing
first, second and third (regardless of order).
▫ This is easy, go to the next chart.

7. (3x9)/W
• The horse farm has three of the eight horses in
the race, thus there is a 3/8 chance of one
coming in first.
• There would be 2 of 7 horses left to come in
second
• There would be 1 of 6 horses left to come in third
• You just multiply these – see the next chart

8. (3x9)/W
▫ (3/8)x(2/7)x(1/6) = 0.017857 or 0.0179 rounded
to four decimal places
▫ Use your calculator or Excel
▫ In decimal form
 (3/8)x(2/7)x(1/6) = 0.375 x 0.285714 x 0.166667 =
0.017857 (I round a little there on a couple)
▫ It is important to note that this problem assumes
the horses are equal in ability, which is of course
not always the case.

9. (3x9)/W
• Number 15 Example
▫ This is a problem where they give you a
probability distribution. I use Excel on these, but
you could use a calculator, but there is not a
“magic formula” in Minitab for these type
problems.
▫ Example follows

10. (3x9)/W
• Let’s say we have students in a class take a quiz
with 8 questions. The number x of questions
answered correctly can be approximated by the
following probability distribution.
x
P(x)
0
0.03
1
0.02
2
0.06
3
0.06
4
0.09
5
0.22
6
0.27
7
0.18
8
0.07

11. (3x9)/W
• I start by putting these in Excel in vertical
columns, I can copy and paste the data easily
from the problems.
x
P(x)
0
1
2
3
4
5
6
7
8
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07

12. (3x9)/W
• I now put in another column where I multiply
these (x times P(x)). Unfortunately, I do not
have time to teach you how to do this in Excel,
but it is very easy. Notice that the new xP(x)
column just multiplies the previous two columns
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
0
0.02
0.12
0.18
0.36
1.1
1.62
1.26
0.56

13. (3x9)/W
• Now if I sum the xP(x) column, I get the mean of
the distribution. I did it a couple of cells below
and got 5.22 (That is my mean, rounded to one
decimal place it would be just 5.2)
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
0
0.02
0.12
0.18
0.36
1.1
1.62
1.26
0.56
5.22

14. (3x9)/W
• In calculating the variance, I need to add more
columns. Just to the right of xP(x), I calculated
each x value – the mean (I used the unrounded
5.22)
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
x - mean
0
-5.22
0.02
-4.22
0.12
-3.22
0.18
-2.22
0.36
-1.22
1.1
-0.22
1.62
0.78
1.26
1.78
0.56
2.78

15. (3x9)/W
• Just to the right of the “x-mean” column, I
squared each of those values… ^2 just means to
the second power or “squared”
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
x - mean (x - mean)^2
0
-5.22
27.2484
0.02
-4.22
17.8084
0.12
-3.22
10.3684
0.18
-2.22
4.9284
0.36
-1.22
1.4884
1.1
-0.22
0.0484
1.62
0.78
0.6084
1.26
1.78
3.1684
0.56
2.78
7.7284

16. (3x9)/W
• Just to the right of the “(x-mean)^2” column, I
multiplied each of those values by the respective
P(x) (For example, the first value of 0.817452 is
the result of 27.2484 times 0.03)
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
x - mean (x - mean)^2 (x - mean)^2 P(x)
0
-5.22
27.2484
0.817452
0.02
-4.22
17.8084
0.356168
0.12
-3.22
10.3684
0.622104
0.18
-2.22
4.9284
0.295704
0.36
-1.22
1.4884
0.133956
1.1
-0.22
0.0484
0.010648
1.62
0.78
0.6084
0.164268
1.26
1.78
3.1684
0.570312
0.56
2.78
7.7284
0.540988

17. (3x9)/W
• Now I just need to sum the (x-mean)^2 P(x)
column to get the variance. As you can see the
answer is 3.5116 or 3.5 rounded to one decimal.
x
P(x)
0
1
2
3
4
5
6
7
8
xP(x)
0.03
0.02
0.06
0.06
0.09
0.22
0.27
0.18
0.07
x - mean (x - mean)^2 (x - mean)^2 P(x)
0
-5.22
27.2484
0.817452
0.02
-4.22
17.8084
0.356168
0.12
-3.22
10.3684
0.622104
0.18
-2.22
4.9284
0.295704
0.36
-1.22
1.4884
0.133956
1.1
-0.22
0.0484
0.010648
1.62
0.78
0.6084
0.164268
1.26
1.78
3.1684
0.570312
0.56
2.78
7.7284
0.540988
5.22
3.5116

18. (3x9)/W
• Take the square root of the variance to get the
standard deviation. I did it right under the variance.
As you can see the answer is 1.87392 etc. or 1.9
rounded to the nearest tenth.
x
P(x)
0
1
2
3
4
5
6
7
8
(x (x - mean)^2
xP(x)
x - mean mean)^2 P(x)
0.03
0
-5.22
27.2484
0.817452
0.02
0.02
-4.22
17.8084
0.356168
0.06
0.12
-3.22
10.3684
0.622104
0.06
0.18
-2.22
4.9284
0.295704
0.09
0.36
-1.22
1.4884
0.133956
0.22
1.1
-0.22
0.0484
0.010648
0.27
1.62
0.78
0.6084
0.164268
0.18
1.26
1.78
3.1684
0.570312
0.07
0.56
2.78
7.7284
0.540988
5.22
3.5116
1.873926359

19. (3x9)/W
• To find the “expected value of the probability
distribution….” Well, you already have it – it’s
the mean (In our case 5.2)

20. (3x9)/W
•
•
•
•
Hope you enjoyed this…
More examples next week….
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