#$&* course MTH 152 I'm beginning to worry if I'll have enough time to complete the remainder of the coursework. Please provide insight if this will be a problem. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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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: My average resulted in 6.896, which I rounded to 6.9 ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - To find the standard deviation (hereafter abbreviated as std dev) one must calculate the mean, the average deviation (hereafter abbreviated as avg dev) from the mean, the squared deviation, then take the square root of the preceding result. - (20, 42, 63, 48, 27) = avg of 40 - (20, 2, 23, 8, 13) = 13.2 avg deviation - (400, 4, 529, 64, 169) = 1166 / 4 = 291.5****** - Sqrt` (291.5) = 17.073 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: It seems significantly easier to make a frequency distribution table rather than going through the previous process ------------------------------------------------ Self-critique rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - I made sure to copy down the definition of a standard deviation a few times to get it in my head. “The standard deviation is the square root of the average of the squared deviations” - So that said I’ll first have to gather the data in question, the average to find the deviation, compare the deviation with the set, square the values, and taking from the rule of when dealing with a set of numbers where n < 30 then (avg sqrdev)/(# of sqrdev - 1) - The mean number of heads (Result * Frequency) was about 115/5 = 23 - Standard deviation (std dev) = 74 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: - Why wouldn’t you use the /(n-1) rule with binomial-normal distribution? I thought that ‘n’ denoted the total number of results?
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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: - Only thing I need to work on is knowing which column to work with which. - For instance, - Deviation = avg (R*F) - (R*F)_n - Where (R*F)_n denotes each element within the R*F column - Square dev = straightforward, squared deviation values - Sq Dev Freq = (square dev)_n * (Freq)_n ------------------------------------------------ Self-critique 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: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. The preceding problem yielded probabilities 1/32, 5/32, 10/32, 10/32, 5/32 and 1/32. On 5 flips, then, the expected values of the different numbers of 'heads' obtained on 32 flips 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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: ------------------------------------------------ Self-critique rating: ********************************************* 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? 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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. STUDENT QUESTION Solution above—could you explain the q? INSTRUCTOR RESPONSE p is the probability of success q is the probability of failure p + q = 1 so q = 1 - p. Self-critique: ------------------------------------------------ Self-critique rating: ********************************************* 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? 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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: ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. The mean of a set of numbers in which 1.05 occurs 4 times, 1.03 occurs 3 times, 1.06 occurs 10 times and 1.08 occurs 3 times is 105.2. On the average by how much do the numbers in this set deviate from their mean? (You would have answered this question in the preceding qa). What is the standard deviation of this set of numbers? How does the standard deviation compare with the average deviation from the mean? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. On four flips of a coin there are 2^4 = 16 possible outcomes. In 16 trials of four flips each we would expect 'Heads' to come up 0 times on C(4, 0) trials, 1 time on C(4, 1) trials, 2 times on C(4, 2) trials, 3 times on C(4, 3) trials and 4 times on C(4, 4) trials. Make a table representing these results and use it to calculate the mean and standard deviation of the number of heads, as expected when doing a large number of four-flip trials. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!