course Mth152 KټǭGxHۡ{assignment #016
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20:15:41 `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 --> four 5s= 4*5=20 seven 6s=7*6=42 nine 7s =9*7=63 six 8s =6*8=48 three9s = 3*9=27 20+42+63+48+27=200 4+7+9+6+3=29 the mean is 200/29 confidence assessment: 3
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20:18:47 `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 --> Calculate the Sq root of the avg. of the Sq. deviation Calculate the deviation of each result from the mean. Next solve the Sq. deviation confidence assessment: 2
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20:24:49 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 --> The mean is 115/62=1.86 avg Sq deviation = 51/62=.83 std dev= sqrt (.83)=.91 Results : 0, 1, 2 ,3 ,4 Freq: 4, 20,22,13,3 results* Freq: 0, 20 ,44,39,12 Dev.Sq: 1.86, 0.86,0.24,1.24,2.24 Dev.Sq: 3.5,0.7,0.1,1.5,5.0 Dev * freq: 0,14,2,20,15 self critique assessment: 2
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⛚} assignment #016 016. mean, std dev of freq dist (incl binomial) Liberal Arts Mathematics II 12-10-2008
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20:30:39 `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 --> four 5s seven 6s nine 7s six eights three 9s 4*5=20,7*6=42, 9*7=63, 6*8=48, 3*9= 27 20+42+63+48+27=200 4+7+9+6+3=29 The mean is 200/29 confidence assessment: 3
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20:33:29 `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 --> Calulate the Sq root of the avg of the sq deviation. Calculate the deviation of each result from the mean. Next solve the Squ. deviation. confidence assessment: 2
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20:40:22 `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=1.86 Avg Sq deviation =51/62= .83 std dev=sqrt(.83)= .91 Results: 0, 1, 2 ,3, 4, Freq: 4,20, 22, 13, 3 Results * Freq: 0, 20,44,39,12 Dev. Sq: 1.86, 0.86, 0.24, 1.24, 2.24 Dev. Sq: 3.5, 0.7, 0.1, 1.5, 5.0 Dev* freq: 0, 14,2, 20, 15 confidence assessment: 3
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20:46:01 `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 = 252/36= 7 avg Sq dev.= 230/36 =6.4 Results: 2,3,4,5,6,7,8,9,10,11,12 Freq: 1,2,3,4,5,6,5,4,3,2,1 Results*freq: 2,6,12,20,30,42,40,36,30,22,12 Dev.Sq: 5,4,3,2,1,0,1,2,3,4,5 Dev. Sq: 25,16,9, 4, 1 ,0, 1 4, 9,16,25 Dev*freq: 25, 32, 37,16,5, 0, 5, 16, 27, 32, 25 confidence assessment: 3
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20:51:20 `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 --> A coin flipped 5 times =n=5 0 heads is C(5,0)=1 2^5=32 1/32 1 heads isC(5,1) =5 2^5=32 5/32 2 heads is C(5,2)=10 2^5=32 10/32 3 heads is C(5,3)= 10 2^5=32 10/32 4 heads is C(5,4)=5 2^5=32 5/32 5heads is C(5,5)= 1 2^5=32 1/32 confidence assessment: 3
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20:52:54 `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 80/32=2.5 avg Sq dev.+40/32=1.25 std dev. sqrt(1.25) confidence assessment: 1
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