course Mth 152 ?????????????assignment #017017. normal-curve models
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14:48:58 `q001. Note that there are 5 questions in this assignment. {}{}Sketch a histogram, or bar graph showing the distribution of the number of ways to get 0, 1, 2, 3, 4 and 5 'heads' on a flip of 5 coins. Your histogram should show a bar for 0, 1, 2, 3, 4 and 5 'heads', and the height of a bar should represent the number of ways of getting that number of 'heads'. Sketch also a histogram showing the probabilities of the different outcomes. Describe both of your histograms.
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RESPONSE --> ? confidence assessment: 0
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14:49:04 Your first histogram should show 6 bars, one for each of the possible outcomes 0, 1, 2, 3, 4, 5. These bars should sit on top of a horizontal axis, like the x axis, and each should be labeled just below that axis with the outcome (0, 1, 2, 3, 4 and 5). The heights of the bars will be 1, 5, 10, 10, 5 and 1, representing the numbers of possible ways for the six different outcomes to occur. Your second histogram should have the same description as the first, except that the heights of the bars will be 1/32 = .0325, 5/32 = .1625, 10/32 = .325, 10/32 = .325, 5/32 = .1625 and 1/32 = .0325. The vertical scales of the two histograms may of course be different, and both histograms may even look identical except for the labeling of the vertical axis. Note that the bar representing, say, 2 will extend along the x axis from x = 1.5 to x = 2.5. Note also that the distribution is symmetric about the central x value x = 2.5, which occurs on the boundary between the x = 2 and x = 3 bars of the graph. That is, the distribution to the left of x = 2.5 is a mirror image of the distribution to the right of x = 2.5.
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RESPONSE --> ok self critique assessment: 3
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14:50:33 `q002. If we toss 64 coins, then the mean number of 'heads' is mean = n * p = 64 * 1/2 = 32 and the standard deviation of the number of 'heads' is very close to std dev = `sqrt( n * p * q ) = `sqrt( 64 * 1/2 * 1/2) = 4. If we toss 64 coins a large number of times we expect that about 34% of the tosses will lie between 1 standard deviation lower than the mean and the mean, and that 34% of the tosses will lie between the mean and 1 stardard deviation higher than the mean. What number is 1 standard deviation lower than the mean and what number is one standard deviation higher than the mean? Out of 200 flips of 64 coins, how many would we expect to give us between 28 and 36 'heads' (use the percents given above, don't use combinations)? {}How many would we expect to give less than 28 'heads' (again base your answer on the percents given above)?
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RESPONSE --> ? confidence assessment: 0
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14:51:01 The mean is 32 and the standard deviation is 4. An outcome one standard deviation lower than the mean is 32-4 = 28. An outcome one standard deviation higher than the mean will be 32 + 4 = 36. Note that we therefore expect that 34 percent of our outcomes will lie between 28 and 32, while another 34 percent lie between 32 and 36. Out of 200 repetitions of the 64-flip experiment, we would therefore expect that 34% will lie between 28 and 32 while another 34% lie between 32 and 36. Thus a total of 68% lie between 28 and 36. Since 68% of 200 is .68 * 200 = 136, our expectation is that 136 of the 200 outcomes will lie between 28 and 32. Since we expect that half of the outcomes, or 50%, will be less than the mean 32, then since 34% lie between 28 and 32, this leaves 16% of the outcomes falling below 28. Since 16% of 200 is 32, we expect that on the average 32 of 200 outcomes will lie below 28. Note that we are 'fudging' a bit on this solution. If we had a histogram of this distribution, the bar representing 28 would actually extend from 27.5 to 28.5 on the x axis. Similarly the bar representing 32 would extend from 31.5 to 32.5, and the bar representing 36 would extend from 35.5 to 36.5. This needn't concern you much if the idea hasn't already occurred to you that there are nine, not eight outcomes from 28 through 36. The problem is resolved as follows To represent the outcomes from 28 to 36 we would have to go from the middle of the bar representing 28 to the middle of the bar representing 36. The number of outcomes calculated here would therefore include only half of the outcomes 28 and 36, plus the remaining seven bars representing 29 - 35.
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RESPONSE --> ok self critique assessment: 3
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14:52:07 `q003. A more detailed breakdown of proportions of 'normal' distributions (i.e., distributions based on the probabilities associated with large numbers of coin flips) which fall into various ranges is as follows: std dev prop 0.25 0.099 0.50 0.191 0.75 0.273 1.00 0.341 1.25 0.394 1.50 0.433 1.75 0.460 2.00 0.477 The column 'std dev' stands for the number of standard deviations from the mean, and 'prop' stands for the proportion of all occurrences lying between the mean and the given number of standard deviations above the mean. What proportion of a normal distribution is expected to lie between the mean and 1.25 standard deviations from the mean? If a certain quantity is normally distributed, then given a sample of 200 instances how many would lie between the mean and 1.25 standard deviations above the mean? Given a sample of 200 instances how many would lie between the mean and 0.25 standard deviations below the mean? Given a sample of 200 instances how many would lie between the 1.25 standard deviations above the mean and 0.25 standard deviations below the mean?
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RESPONSE --> ? confidence assessment: 0
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14:52:35 According to the table, the proportion 0.394 of the distribution will lie between the mean and 1.25 standard deviations above the mean. This is consistent with the information given in the preceding problem, that 34% or 0.34 of the distribution should lie between the mean and one standard deviation above the mean. Given 200 instances, we would therefore expect 0.394 * 200 = 78.6 of the outcomes to lie between the mean and one standard deviation above the mean. The number of instances lying between the mean and 0.25 standard deviations below the mean should, because of the symmetry of the distribution, be the same as the number of instances between the mean and 0.25 standard deviations above the mean. According to the information given here, the portion of the distribution should account for 0.099 of the entire distribution. If there are 200 total occurrences, then 0.099 * 200 = 19.8 of the occurrences should lie between the mean and 0.25 standard deviations below the mean. As we have seen here 78.6 of the outcomes should lie between the mean and 1.25 standard deviations above the mean, while 19.8 should lie between the mean and 0.25 standard deviations below the mean. There is no overlap between these regions, since one lies entirely below of the mean while the other lies entirely above the mean. So the total number lying between the two given extremes must be 78.6 + 19.8 = 98.4. Note that this corresponds to 0.394 + 0.099 of the distribution.
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RESPONSE --> ok self critique assessment: 3
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14:53:32 `q004. If scores on a certain test are normally distributed with average 150 points standard deviation of 20 points, then how many standard deviations above or below the distribution is each of the following scores: 170, 120, 135, 155?
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RESPONSE --> 1, 2, 1, 1 confidence assessment: 1
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14:53:48 Since 170 is 20 units above the mean of 150, and the standard deviation is 20, we see that 170 lies exactly one standard deviation above the mean. We see that 120 lies 30 units below the mean of 150, which is 30/20 = 1.5 times the standard deviation 20. Thus 120 lies 1.5 standard deviations below the mean. 135 lies 15 units below the mean, or 15/20 = 0.75 of a standard deviation below the mean. 155 lies 5 units above the mean, or 5/20 = 0.25 of a standard deviation above the mean. Note that we could use the proportion given in the preceding problem to determine what proportion of a distribution lie between the mean and each given value.
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RESPONSE --> ok self critique assessment: 3
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14:55:41 `q005. The table from the previous problem is given again here: std dev prop 0.25 0.099 0.50 0.191 0.75 0.273 1.00 0.341 1.25 0.394 1.50 0.433 1.75 0.460 2.00 0.477 If scores on a certain test are normally distributed with average 150 points standard deviation of 20 points, then out of 600 people taking the test how many are expected to score in each of the following ranges: 150 - 170 120 - 150 135 - 155 120-135?
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RESPONSE --> ? confidence assessment: 0"