Hillassig16

course Mth152

ñšÞÏdé—ÌЮâ®O¢Ï°€§êÜÞÔ声Ôõassignment #016

016. mean, std dev of freq dist (incl binomial)

Liberal Arts Mathematics II

10-27-2008

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10:34:59

`q001. Note that there are 8 questions in this assignment.

{}{}When rolling 2 dice a number of times, suppose you get a total of 5 on four different rolls, a total of 6 on seven rolls, a total of 7 on nine rolls, and total of 8 of six rolls a total of 9 on three rolls. What was your mean total per roll of the two dice?

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RESPONSE -->

4 x 5 = 20

7 x 6 = 42

9 x 7 = 63

6 x 8 = 48

3 x 9 = 27

20+42+63+48+27=200 is the total of all of the outcomes.

there are 29 outcomes(4+7+9+6+3)

to get the mean 200/29 = 6.9

confidence assessment: 3

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10:35:35

You obtained four 5's, which total 4 * 5 = 20.

You obtained seven 6's, which total 7 * 6 = 42.

You obtained nine 7's, which total 9 * 7 = 63.

You obtained six 8's, which total 6 * 8 = 48.

You obtained three 9's, which total 3 * 9 = 27.

The total of all the outcomes is therefore 20 + 42 + 63 + 48 + 27 = 200. Since there are 4 + 7 + 9 + 6 + 3 = 29 outcomes (i.e., four outcomes of 5 plus 7 outcomes of 6, etc.), the mean is therefore 200/29 = 6.7, approximately.

This series of calculations can be summarized in a table as follows:

Result Frequency Result * frequency

5 4 20

6 7 42

7 9 63

8 6 48

9 3 27

9 3 27

___ ____ ____

29 200

mean = 200 / 29 = 6.7

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RESPONSE -->

o.k.

I came up with a different answer for the mean

self critique assessment: 3

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10:54:28

`q002. The preceding problem could have been expressed in the following table:

Total Number of Occurrences

5 4

6 7

7 9

8 6

9 3

This table is called a frequency distribution. It expresses each possible result and the number of times each occurs.

You found the mean 6.7 of this frequency distribution in the preceding problem. Now find the standard deviation of the distribution.

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RESPONSE -->

6.9 - 5 = 1.9

6.9 - 6 = .9

6.9 - 7 = -0.1

6.9 - 8 = -1.1

6.9 - 9 = -2.1

1.9*1.9 = 3.61 3.61 x 4 = 14.44

.9 *.9 = 0.81 .81 x 7 = 5.67

-0.1 x -0.1 = 0.1 0.1 x 9 =0.9

-1.1 x -1.1x = 2.1 2.1 x 6 = 12.6

-2.1 x -2.1 = 4.41 4.41 x 3 =13.232

confidence assessment: 2

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10:55:17

We must calculate the square root of the 'average' of the squared deviation. We calculate the deviation of each result from the mean, then find the squared deviation. To find the total of the squared deviations we must add each squared deviation the number of times which is equal to the number of times the corresponding result occurs.

For example, the first result is 5 and it occurs four times. Since the deviation of 5 from the mean 6.7 is 1.7, the squared deviation is 1.7^2 = 2.89. Since 5 occurs four times, the squared deviation 2.89 occurs four times, contributing 4 * 2.89 = 11.6 to the total of the squared deviations.

Using a table in the manner of the preceding exercise we obtain

Result Freq Result * freq Dev Sq Dev Sq Dev * freq

5 4 20 1.7 2.89 11.6

6 7 42 .7 0.49 3.4

7 9 63 0.3 0.09 0.6

8 6 48 1.3 1.69 10.2

9 3 27 2.3 5.29 15.9

___ ____ ____ ___

29 200 41.7

mean = 200 / 29 = 6.7

'ave' squared deviation = 41.7 / (29 - 1) = 1.49

std dev = `sqrt(1.49) = 1.22

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RESPONSE -->

I drew a blank and could not finish this problem. Again some of my numbers don't match the numbers in the answer.

self critique assessment: 2

You just needed to add up the numbers in your last column to get the sum of the squared deviations, and applied by 29 to get the average squared deviation and finally take the square root to get the standard deviation. You did everything else; the only difference in the numbers is that you were using 6.9 for your mean and the given solution was using 6.7. There was nothing at all wrong with what you were doing.

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11:25:05

`q003. If four coins are flipped, the possible numbers of 'heads' are 0, 1, 2, 3, 4. Suppose that in an experiment we obtain the following frequency distribution:

# Heads Number of Occurrences

0 4

1 20

2 22

3 13

4 3

What is the mean number of 'heads' and what is the standard deviation of the number of heads from this mean?

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RESPONSE -->

result x frequency

0 4 = 0

1 20 = 20

2 22 = 44

3 13 = 39

4 3 = 12

62 115

115 / 62 = 1.85

1.85 - 0 = 1.85 1.85 x 1.85 = 3.42

1.85 - 1 = .85 .85 x .85 = .72

1.85 - 2 = .15 .15 x.15 = .02

1.85 - 3 = -1.15 -1.15 x -1.15 = 2.3

1.85 - 4 = -2.15 -2.15 x -2.15 = 4.62

4x3.42 = 13.68 , 20 x .72= 14.40 , 22 x .02 = .44

13 x 2.3 = 29.9, 4 x 4.62 = 18.48

76.9 / 62 = 1.24

square root of 1.24 = 1.113

confidence assessment: 2

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11:25:58

Using the table in the manner of the preceding problem we obtain the following:

Result Freq Result * freq Dev Sq Dev Sq Dev * freq

0 4 0 1.86 3.5 0

1 20 20 0.86 0.7 14

2 22 44 0.24 0.1 2

3 13 39 1.24 1.5 20

4 3 12 2.24 5.0 15

___ ____ ____ ___

62 115 51

mean = 115 / 62 = 1.86 approx. Note that the mean must be calculated before the Dev column is filled in.

'ave' squared deviation = 51 / 62 = .83

std dev = `sqrt(.83) = .91

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RESPONSE -->

o.k.

I don't understand what I did wrong but none of my calculations are the same as what is in the answer.

self critique assessment: 3

Your solution was correct. For some unaccountable reason the given solution failed to square the deviations.

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11:59:24

`q004. If we rolled 2 dice 36 times we would expect the following distribution of totals:

Total Number of Occurrences

2 1

3 2

4 3

5 4

6 5

7 6

8 5

9 4

10 3

11 2

12 1

What is the mean of this distribution and what is the standard deviation?

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RESPONSE -->

2x1=2

3x2=6

4x3=12

5x4=20

6x5=30

7x6=42

8x5 = 40

9x4 =36

10 x 3 = 30

11 x 2 =22

12 x 1 =12

2+6+12+20+30+42+40+36+30+22+12=252

1+2+3+4+5+6+5+4+3+2+1=36

252/36 = 7

(7-2 )+ (7-3 )+ (7-4) +( 7-5) +(7-6) + (7-7) + (7-8) +(7-9) + (7 - 10) + (7-11) + (7-12) = 5+4+3+2+1+0+1+2+3+4+5= 30

25+16+9+4+ 1+0+1+4+9+16+25=110

110 / 36 = 3.05

the square root of 3.05= 1.746

confidence assessment: 2

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12:00:24

Using the table in the manner of the preceding problem we obtain the following:

Result Freq Result * freq Dev Sq Dev Sq Dev * freq

2 1 2 5 25 25

3 2 6 4 16 32

4 3 12 3 9 37

5 4 20 2 4 16

6 5 30 1 1 5

7 6 42 0 0 0

8 5 40 1 1 5

9 4 36 2 4 16

10 3 30 3 9 27

11 2 22 4 16 32

12 1 12 5 25 25

___ ____ ____ ___

36 252 230

mean = 252 / 36 = 7. Note that the mean must be calculated before the Dev column is filled in.

'ave' squared deviation = 230 / 36 = 6.4 approx.

std dev = `sqrt(6.4) = 2.5 approx.

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RESPONSE -->

o.k.

am i missing a step or something? I am still not getting the correct answers

self critique assessment: 3

Something is wrong with the alignment of the columns in the given solution, and the numbers.

However your step

25+16+9+4+ 1+0+1+4+9+16+25=110

adds the square of the deviations, but does not consider the frequency of each. For example, 25 occurs once, but 16 occurs twice, 9 occurs 3 times, etc..

So the correct sum of square deviations is

1*25+2*16+3*9+4*4+ 5*1+6*0+5*1+4*4+3*9+2*16+25=260.

There are a total of 36 possibilities, so the mean of the squared deviations is 260 / 36 = 7.2, approx., and the standard deviation is the square root of the mean of the square deviations, which would b e sqrt(7.2) = 2.7, approx..

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12:17:15

`q005. If we flip n coins, there are C(n, r) ways in which we can get r 'heads' and 2^n possible outcomes. The probability of r 'heads' is therefore C(n, r) / 2^n. If we flip five coins, what is the probability of 0 'heads', of 1 'head', of 2 'heads', of 3 'heads', of 4 'heads', and of 5 'heads'?

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RESPONSE -->

there would be 2x2x2x2x2= 32 possible outcomes when flipping five coins to get 0 heads there would be only 1 way and that would be (t,t,t,t,t,) so 1 / 32 chance of getting 0 heads. To get 1 head there are five flips so five chances of getting 1 head so 5/32.

5 flips x 2 heads = possibility of 10

10/32 of 2 heads

5 flips x 3 heads = possible 15

15/32

5 flips x 4 heads = possible 20

20/32=5/16

5 flips x 5 heads = possible 25

25/32

confidence assessment: 3

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12:17:34

If we flip 5 coins, then n = 5.

To get 0 'heads' we find C(n, r) with n = 5 and r = 0, obtaining C(5,0) = 1 way to get 0 'heads' out of the 2^5 = 32 possibilities for a probability of 1 / 32.

To get 1 'heads' we find C(n, r) with n = 5 and r = 1, obtaining C(5,1) = 5 ways to get 1 'heads' out of the 2^5 = 32 possibilities for a probability of 5 / 32.

To get 2 'heads' we find C(n, r) with n = 5 and r = 2, obtaining C(5,2) = 10 ways to get 2 'heads' out of the 2^5 = 32 possibilities for a probability of 10 / 32.

To get 3 'heads' we find C(n, r) with n = 5 and r = 3, obtaining C(5,3) = 10 ways to get 3 'heads' out of the 2^5 = 32 possibilities for a probability of 10 / 32.

To get 4 'heads' we find C(n, r) with n = 5 and r = 4, obtaining C(5,4) = 5 ways to get 4 'heads' out of the 2^5 = 32 possibilities for a probability of 5 / 32.

To get 5 'heads' we find C(n, r) with n = 5 and r = 5, obtaining C(5,5) = 1 way to get 5 'heads' out of the 2^5 = 32 possibilities for a probability of 1 / 32.

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RESPONSE -->

o.k.

self critique assessment: 3

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12:58:21

`q006. The preceding problem yielded probabilities 1/32, 5/32, 10/32, 10/32, 5/32 and 1/32. On 5 flips, then, we the expected values of the different numbers of 'heads' would give us the following distribution: :

# Heads Number of Occurrences

0 1

1 5

2 10

3 10

4 5

5 1

Find the mean and standard deviation of this distribution.

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RESPONSE -->

1/32, 5/32, 10/32, 10/32, 5/32, 1/32

0x1=0

1x5=5

2x10=20

3x10=30

4x5=20

5x1=5

0+5+20+30+20+5 = 80

mean = 80 /32 = 2.5

2.5 + 1.5 + .5 + 1.5 + 1.5 +2.5 =

6.25 + 2.25 + .25 2.25 + 2.25 + 6.25 =

6.25 + 11.25 + 2.50 + 2.50 + 11.25 + 6.25 =40

40 / 32 = 1.25 square root of 1.25 = 1.118

confidence assessment: 3

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12:59:28

Using the table in the manner of the preceding problem we obtain the following:

Result Freq Result * freq Dev Sq Dev Sq Dev * freq

0 1 0 2.5 6.25 6.25

1 5 5 1.5 2.25 11.25

2 10 20 0.5 0.25 2.50

3 10 30 0.5 0.25 2.50

4 5 20 1.5 2.25 12.25

5 1 5 2.5 6.25 6.25

___ ____ ____ ___

32 80 32.00

mean = 80 / 32 = 2.5. Note that the mean must be calculated before the Dev column is filled in.

'ave' squared deviation = 40 / 32 = 1.25.

Thus std dev = `sqrt(1.25) = 1.12 approx.

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RESPONSE -->

That one was hard. I used the numbers from the problem # 6, not the numbers that I came up with in problem # 5.

self critique assessment: 3

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13:00:56

`q007. Suppose that p is the probability of success and q the probability of failure on a coin flip, on a roll of a die, or on any other action that must either succeed or fail. If a trial consists of that action repeated n times, then the average number of successes on a large number of trials is expected to be n * p. For large values of n, the standard deviation of the number of successes is expected to be very close to `sqrt( n * p * q ). For values of n which are small but not too small, the standard deviation will still be close to this number but not as close as for large n.

If the action is a coin flip and 'success' is defined as 'heads', then what is the value of p and what is the value of q?

For this interpretation in terms of coin flips, if n = 5 then what is n * p and what does it mean to say that the average number of successes will be n * p?

In terms of the same interpretation, what is the value of `sqrt(n * p * q) and what does it mean to say that the standard deviation of the number of successes will be `sqrt( n * p * q)?

How does this result compare with the result you obtained on the preceding problem?

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RESPONSE -->

I don't know what to do on this one

confidence assessment: 0

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13:01:23

We first identify the quantities p and q for a coin flip. Success is 'heads', which for a fair coin occurs with probability .5. Failure therefore has probability 1 - .5 = .5.

Now if n = 5, n * p = 5 * .5 = 2.5, which represents the mean number of 'heads' on 5 flips. The idea that the mean number of occurrences of some outcome with probability p in n repetitions is n * p should by now be familiar (e.g., from basic probability and from the idea of expected values).

For n = 5, we have `sqrt(n * p * q) = `sqrt(5 * .5 * .5) = `sqrt(1.25) = 1.12, approx..

In the preceding problem we found that the standard deviation expected on five flips of a coin should be exactly 1. This differs from the estimate `sqrt(n * p * q) by a little over 10%, which is a fairly small difference.

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RESPONSE -->

o.k.

will work on this

self critique assessment: 3

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13:09:02

`q008. Suppose that p is the probability of success and q the probability of failure on a coin flip, on a roll of a die, or on any other action that must either succeed or fail. If a trial consists of that action repeated n times, then the average number of successes on a large number of trials is expected to be n * p. The standard deviation of the number of successes is expected to be `sqrt( n * p * q ).

If the action is a roll of a single die and a success is defined as rolling a 6, then what is the probability of a success and what is the probability of a failure?

If n = 12 that means that we count the number of 6's rolled in 12 consecutive rolls of the die, or alternatively that we count the number of 6's when 12 dice are rolled. How many 6's do we expect to roll on an average roll of 12 dice? What do we expect is the standard deviation the number of 6's on rolls of 12 dice?

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RESPONSE -->

the probability of getting a 6 on a dice roll is 1 out of 6 (1/6) , so the probability of getting a 6 on 12 rolls would be 12 x 1/6 = 2

confidence assessment: 3

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13:09:16

We first identify the quantities p and q for a rolling a die. Success is defined in this problem as getting a 6, which for a fair die occurs with probability 1/6. Failure therefore has probability 1 - 1/6 = 5/6.

Now if n = 12, n * p = 12 * 1/6 = 2, which represents the mean number of 6's expected on 12 rolls. This is the result we would expect.

For n = 12, we have `sqrt(n * p * q) = `sqrt(12 * 1/6 * 5/6) = `sqrt(1.66) = 1.3, approx..

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RESPONSE -->

o.k.

self critique assessment: 3

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&#Your work looks good. See my notes. Let me know if you have any questions. &#