course Mth 152 I am not sure why but my answers are off on the rounding part. æý–¨{ª~ŧiŽƒŠ…ßwõ¼çôåŠÅõ¼assignment #016
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21:45:03 `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 --> 200/29=6.896 = 6.9 rounded confidence assessment: 2
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21:47:22 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 --> I got the 200/29. When I do the operation on my calculator I got 6.89 which I rounded to 6.9. self critique assessment: 2
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21:59:02 `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 --> Using the standard deviation of 6.9 which the answer I got from the first problem the deviations would be -1.9 four times, -.9 seven times, .1 nine times, 1.1 six times, 2.1 three times. The standard deviation always shows 0. confidence assessment: 2
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22:00:50 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 see that I needed to find the square root to get the standard deviation. self critique assessment: 2
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22:21:42 `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 --> The mean is 115/62= 1.85 = 1.9 -1.9 3.6 14.4 -.9 .81 16 .01 .0 0 1.1 1.2 15.6 2.1 4.4 13.2 59.2 59.2 / (62-1) = .97 = .98 confidence assessment: 1
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22:24:23 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 --> I see where I went wrong. I rounded 1.85 up to 1.9 instead of 1.86 which threw the rest of the information off. I have the basic understanding of how it works I just need to work on the rounding. self critique assessment: 2
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22:33:09 `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 --> The mean is 252/36=7 The standard deviation is 729/35 = 20.8 square root is 4.56 confidence assessment: 2
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22:39:50 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 --> I went back and looked at my chart and numbers . I multiplied by the wrong number half way down but I went back and corrected it. Divided by 35 which is 36-1 and got 6 then found the square root which gave me 2.45. self critique assessment: 2
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22:48:48 `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 --> 0 heads - C(5,0) / 2^5=1/32 1 head - C(5,1) / 2^5 = 5/32 2 heads - C(5,2)/ 2^5 = 10/32 3 heads - C(5,3) / 2^5 = 10/32 4 heads -C(5,4)/2^5 = 5/32 5 heads - C(5,5,)/2^5= 1/32 confidence assessment: 2
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22:49:21 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 --> correct self critique assessment: 3
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22:57:17 `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 --> The mean is 81/32 = 2.5 The standard deviation 40/31= 1.29 then the square root would be 1.135 rounded to 1.14 confidence assessment: 1
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22:58:58 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 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 --> When we do the standard deviation are we not supossed to take the number and subtract one from it then get the square root. self critique assessment: 2
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23:08:20 `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 --> For sucess p will be 1 and q would be 0. For 5 flips we would have 5*1/2. =2.5 sqrt for (5*1/2*1/2) =sqrt 1.25 =1.12 confidence assessment: 1
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23:10:10 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 --> I see that I should have used 1/2 on the first part of the problem. But I did however get the second part of the question correct. self critique assessment: 2
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23:18:00 `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 --> Success - 1/6 Failure - 5/6 n*p - 12*1/6 sq rt (12*1/6*5/6)= sq rt 2 = 1.4 confidence assessment: 1
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23:18:56 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 --> Correct just a little diffeence in the rounding. self critique assessment: 3
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