Week 3 Three Extra Homework Examples (3x9)/WMA 221 Statistics for Decision Making

Professor Brent HeardNot to be copied or linked to without my permission 3 x 9

WS4DM

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•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

(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

(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.

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•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

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•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.

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•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

(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.

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•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

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•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 0 1 2 3 4 5 6 7 8P(x) 0.03 0.02 0.06 0.06 0.09 0.22 0.27 0.18 0.07

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•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 0.031 0.022 0.063 0.064 0.095 0.226 0.277 0.188 0.07

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•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 columnsx P(x) xP(x)

0 0.03 01 0.02 0.022 0.06 0.123 0.06 0.184 0.09 0.365 0.22 1.16 0.27 1.627 0.18 1.268 0.07 0.56

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•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) xP(x)

0 0.03 01 0.02 0.022 0.06 0.123 0.06 0.184 0.09 0.365 0.22 1.16 0.27 1.627 0.18 1.268 0.07 0.56

5.22

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•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)xP(x)

x - mean

0 0.03 0 -5.221 0.02 0.02 -4.222 0.06 0.12 -3.223 0.06 0.18 -2.224 0.09 0.36 -1.225 0.22 1.1 -0.226 0.27 1.62 0.787 0.18 1.26 1.788 0.07 0.56 2.78

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•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)xP(x)

x - mean

(x - mean)^2

0 0.03 0 -5.22 27.24841 0.02 0.02 -4.22 17.80842 0.06 0.12 -3.22 10.36843 0.06 0.18 -2.22 4.92844 0.09 0.36 -1.22 1.48845 0.22 1.1 -0.22 0.04846 0.27 1.62 0.78 0.60847 0.18 1.26 1.78 3.16848 0.07 0.56 2.78 7.7284

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•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)

xP(x)x - mean

(x - mean)^2

(x - mean)^2 P(x)

0 0.03 0 -5.22 27.2484 0.8174521 0.02 0.02 -4.22 17.8084 0.3561682 0.06 0.12 -3.22 10.3684 0.6221043 0.06 0.18 -2.22 4.9284 0.2957044 0.09 0.36 -1.22 1.4884 0.1339565 0.22 1.1 -0.22 0.0484 0.0106486 0.27 1.62 0.78 0.6084 0.1642687 0.18 1.26 1.78 3.1684 0.5703128 0.07 0.56 2.78 7.7284 0.540988

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•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)xP(x)

x - mean

(x - mean)^2

(x - mean)^2 P(x)

0 0.03 0 -5.22 27.2484 0.8174521 0.02 0.02 -4.22 17.8084 0.3561682 0.06 0.12 -3.22 10.3684 0.6221043 0.06 0.18 -2.22 4.9284 0.2957044 0.09 0.36 -1.22 1.4884 0.1339565 0.22 1.1 -0.22 0.0484 0.0106486 0.27 1.62 0.78 0.6084 0.1642687 0.18 1.26 1.78 3.1684 0.5703128 0.07 0.56 2.78 7.7284 0.540988

5.22 3.5116

(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)xP(x)

x - mean

(x - mean)^2

(x - mean)^2 P(x)

0 0.03 0 -5.22 27.2484 0.8174521 0.02 0.02 -4.22 17.8084 0.3561682 0.06 0.12 -3.22 10.3684 0.6221043 0.06 0.18 -2.22 4.9284 0.2957044 0.09 0.36 -1.22 1.4884 0.1339565 0.22 1.1 -0.22 0.0484 0.0106486 0.27 1.62 0.78 0.6084 0.1642687 0.18 1.26 1.78 3.1684 0.5703128 0.07 0.56 2.78 7.7284 0.540988

5.22 3.5116

1.873926359

(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)

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•Hope you enjoyed this…•More examples next week….•Visit me at www.facebook.com/statcave

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