Assignment 16

course Mth 152

10/31 3:00 Pm I've been fairly sick with the flu for the past week so that is why my assignments aren't up to date. I will get them caught up this weekend.

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution.

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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

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Question: `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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Your solution: 5*4=20, 6*7=42, 7*9=63, 8*6=48, 9*3=27

20+42+63+48+27=200

There are 29 rolls total so you divide 200 by 29 for 200/29=6.9 for the mean total per roll of the dice.

confidence rating:

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Given Solution: 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

Self-critique: I understood this well.

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Self-critique rating:3

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Question: `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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Your solution: 5 deviates from 6.7 by 1.7 so you must square 1.7 for 1.7^2=2.89*4 since there are four occurrences for 11.6

6 deviates by .7 so you follow the same example and .7^2=.49*7 =3.4

7 deviates by .3^2=.09*9=.81

8 deviates by 1.3^2=1.69*6=10.2

9 deviates by 2.3^2=5.29*3=15.9

These total up to 41.7 so you take the 41.7/ (29-1) for the deviation for 1.49 and take the square root of that for

1.22

confidence rating:

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Given Solution: 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

Self-critique:

I understood this ok.

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Self-critique rating:2

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Question: `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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Your solution: 0*4=0 frequency 1.86 deviation from mean 3.5 squared 3.5

1*20=20 frequency .86 deviation .74 squared

2*22=44 frequency .14 deviation .02 squared

3*13=39 frequency 1.14 deviation 1.3 squared

4*3=12 frequency 2.14 deviation 4.6 squared

62 total frequency with 115 frequencies 115/62=1.86

61/62=.98 for average square deviance

Standard is .99

confidence rating:

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Given Solution: 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 14

1 20 20 0.86 0.7 14

2 22 44 0.14 0.1 2

3 13 39 1.14 1.3 17

4 3 12 2.14 4.6 14

___ ____ ____ ___

62 115 61

mean = 115 / 62 = 1.86 approx.

Note that the mean must be calculated before the Dev column is filled in.

'ave' squared deviation = 61 / 62 = .98

std dev = `sqrt(.98) = .99

Self-critique: I understood this ok but could you explain how the last column comes together so it’s a little clearer for me?

I'm not sure what you need to have cleared up. The numbers in the given solution seem to match those in your solution.

If you can clarify exactly what you mean, I can try to clarify further.

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Self-critique rating:2

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

Total Number of Occurrences Deviation SQ. Dev *Frequency

2 1 2 5 25 25

3 2 6 4 16 32

4 3 12 3 9 27

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

Total= 36 252 210

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

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Your solution: 252/36=7 mean of the distribution 210/36=5.8 square root of 5.8=2.4

confidence rating:

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Given Solution: 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 27

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

___ ____ ____ ___

36 252 210

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

'ave' squared deviation = 210 / 36 = 5.8 approx.

std dev = `sqrt(5.8) = 2.4 approx.

Self-critique: Ok after working this one I think I understand how to calculate these numbers better.

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Self-critique rating:3

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Question: `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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Your solution: n=5 because they equal the coins there are 32 ways possible 2^5. 0 heads = C(5,0) 1 way to get 0 heads for 1/32

1 head is C(5, 1) or 5 ways to get 1 head for 5/32

2 heads is C(5,2) or 10 ways to get 2 heads for 10/32

3 heads is C(5,3) or 10 ways to get 3 heads for 10/32

4 heads is C(5,4) or 5 ways to get 4 heads for 5/32

And 5 heads is C(5,5) or 1 way to get 5 heads for 1/32

confidence rating:

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Given Solution: 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.

Self-critique: I understood this fairly well.

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Self-critique rating:3

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Question: `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 Frequency Deviation Sq. Deviation *Frequency

0 1 0 2.5 6.25

1 5 5 1.5 11.25

2 10 20 .5 2.5

3 10 30 .5 2.5

4 5 20 1.5 12.25

5 1 5 2.5 6.25

Total= 32 80 32.00

Find the mean and standard deviation of this distribution.

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Your solution:80/32=2.5

40/32=1.25 Square root =1.12

confidence rating:

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Given Solution: 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 40.00

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.

Self-critique:Ok I understood this up until the average squared deviation. In your answer, where exactly does the 40 come from?

The squared deviations * frequency numbers are 6.25, 11.25, 2.50, 2.50, 12.25 and 6.25. These add up to 40.

I agree that this wasn't clear in the given table, mainly because the table was simply incorrect. I corrected the table above.

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Self-critique rating:2

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Question: `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?

Heads has .5 value because it is success, and taking that from 1 we have the value of a failure, also .5.

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?

if n=5 then 5*.5=2.5

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)?

You take the square root of the number (5*.5*.5)or 1.25 for 1.12 approx.

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

It is slightly different.

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Your solution:

Solution above—could you explain the q?

p is the probability of success

q is the probability of failure

p + q = 1 so q = 1 - p.

confidence rating:

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Given Solution: 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.

Self-critique:

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Self-critique rating:

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Question: `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?

Success is 1/6 for rolling a 6 and failure is 5/6 for rolling anything other than a 6

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?

For 12 rolls you would expect 1/6 of them to be 6’s=12*1/6=2

Taking the deviation you expect there to be the square root of(12*1/6*5/6) or the square root of 1.66 or 1.3 approximately

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Your solution: Ok I think I understand q as in it representing the ‘failures’, right?

right; q represents the probability of failure where p represents probability of success

confidence rating:

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Given Solution: 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..

Self-critique: I think I understood this a bit better for this problem.

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Self-critique rating:3

&#Good responses. See my notes and let me know if you have questions. &#