course Mth 152 ??????????????assignment #017017. normal-curve models
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23:06:44 `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 --> First: 6 bars with totals of 1, 5, 10, 10, 5, and 1 Second: 6 bars with totals of (1/32), (5/32), (10/32), (10/32), (5/32), and (1/32) confidence assessment: 3
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23:07:56 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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23:14:03 `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 --> 1 standard deviation above: 36 1 standard deviation below: 28 Between 28 and 36 heads: 136 Less than 28 heads: 32 confidence assessment: 3
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23:14:52 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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23:22:05 `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 --> Between mean and 1.25: .394 Normal between mean and 1.25: .394*200=78.6 Between mean and .25 st dev below: .099*200=19.8 Between 1.25 above and .25 below: 78.6+19.8=98.4 confidence assessment: 3
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23:22:44 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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23:25:14 `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 --> 170: 1 st dev above 120: 1.5 st dev below 135: 3/4 st dev below 155: 1/4 st dev above confidence assessment: 3
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23:25:32 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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???~??????????? assignment #017 017. `query 17 Liberal Arts Mathematics II 07-16-2007
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23:30:40 **** query problem 13.4.12 z score for KG's rebounds (.4 from bottom range 10-13)
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RESPONSE --> z = (489-538.2) / 38.8 = -1.3
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23:31:03 ** The z score for KG is his total number of rebounds minus the mean average number of rebounds for all the players and then divided by the standard deviation. In KG' s case: z = (489 - 538.2) / 38.8 = -1.3 **
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RESPONSE --> Ok.
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23:32:28 query problem 13.4.30 midquartile same as median? (Q1+Q3)/2
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RESPONSE --> No, it doesn't have to be the same as the median.
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23:32:53 ** If the median is the actual number in the middle, the it's not necessarily equal to the mean of the first and third quartile. There are different ways to see this. For example suppose that in a large set of numbers, the median number is at least 2 greater than the next smaller number and 2 smaller than the next greater number. Then if all the other numbers stay the same, but the median is increased or decreased by 1, it's still in the middle, so it's still the median. Since all the other numbers stay the same, the first and third quartiles are the same as before, so (Q1 + Q3) / 2 is still the same as before. However the median has changed. So if the median was equal to (Q1 + Q3) / 2, it isn't any more. And if it is now, it wasn't before. In either case we see that the median is not necessarliy equal to the midquartile. To be even more specific, the median of the set {1, 3, 5, 7, 9, 11, 13} is 7. The median of the set {1, 3, 5, 8, 9, 11, 13} is 8. The midquartile of both sets is the same, so for at least one of the two sets (namely the second, as you can verify for yourself) the median and the midquartile are different. **
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RESPONSE --> Ok.
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????{O?????L?assignment #018 018. `query 18 Liberal Arts Mathematics II 07-16-2007
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23:35:10 **** query problem 13.5.12 percent above 115 IQ, mean 100 std dev 15
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RESPONSE --> .159 above 115
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23:35:40 ** The z-score is measured relative to the mean. The mean is 100, and you need to measure the z score of 115. 115 is 15 units from the mean, which gives you a z-score of 15 / 15 = 1. The table tells you that .339 of the distribution lies between the mean and z = 1. You want the proportion beyond 115. Since half the distribution lies to the right of the mean, and .339 of the distribution lies between the mean and z = 1, we conclude that .5 - .339 = .159 of the distribution lies to the right of z = 1. It follows that .159, or 15.9% of the distribution exceeds an IQ of 115. **
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RESPONSE --> Ok.
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23:37:48 **** query problem 13.5.20 area between z=-1.74 and z=-1.14
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RESPONSE --> Between -1.14 and -1.74: .459 - .373 = .086
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23:38:21 ** According to the table the z score for -1.74 is .373 and the z score for -1.14 is .459, meaning that .373 of the distribution lies between the mean and z = -1.73 and .459 of the distribution lies between the mean and z = -1.14. Since -1.74 and -1.14 both lie on the same side of the mean, the region between the mean and -1.74 contains the region between the mean and -1.14. The region lying between z = -1.14 and z = -1.74 is therefore that part of the .459 that doesn't include the .373. The proportion between z = -1.14 and z = -1.74 is therefore .459 - .373 = .086, or 8.6%. **
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RESPONSE --> Ok.
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23:40:31 **** query problem 13.5.30 of 10K bulbs, mean lifetime 600 std dev 50, # between 490 and 720 **** How many bulbs would be expected to last between 490 and 720 hours? **** Describe in detail how you obtained your result.
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RESPONSE -->
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23:41:17 ** You first calculate the z value for each of the given lifetimes 490 hrs and 720 hrs. You should then sketch a graph of the distribution so you can see how the regions are located within the distribution. Then interpret what the table tells you about the proportion of the distribution within each region and apply the result to the given situation. The details: The displacement from the mean to 490 is 490 - 600 = - 110 (i.e., 490 lies 110 units to the left of the mean). The z value corresponding to 490 hours is therefore z = -110/50 = -2.2. The area of the region between the mean and z = -2.2 is found from the table to be .486. Similarly 720 lies at displacement 720- 600 = 120 from the mean, giving us z = 120/50 = 2.4. The area of the region between the mean and z = 2.4 is shown by the table to be .492. Since one region is on the negative side and the other on the positive side of the mean, the region lying between z = -2.2 and z = 2.4 contains .486 + .492 = .978 of the distribution. Out of 10,000 bulbs we therefore expect that .978 * 10,000 = 9780 of the bulbs will last between 490 and 720 hours. **
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RESPONSE --> Ok, I understand a little better.
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23:43:53 **** query problem 13.5.48 A's for > mean + 3/2 s What percent of the students receive A's, and how did you obtain your result?
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RESPONSE --> .500 - .433 = .067 of the area in which the A's are given.
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23:44:10 ** A's are given for z scores greater than 1.5. The area between mean and z = 1.5 is given by the table as .433. To the right of z = 1.5, corresponding to the A's, we have .500 - .433 = .067 or 6.7% of the total area. So we expect that 6.7% of the group will receive A's. **
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RESPONSE --> Ok.
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23:45:07 GENERAL ADVICE: To solve problems of the type covered in this section it is a good idea to follow a strategy something like the following: 1. Find the z-score(s) corresponding to the given values. 2. Look up the corresponding numbers on the table. 3. Sketch a graph of the normal distribution representing what the numbers in the table tell you. Be sure you understand that the table tells you the proportion of the distribution lying between the mean and the given z value. 4. Decide what region of the graph corresponds to the result you are trying to find. 5. Find the proportion of the total area lying within this region. 6. If necessary apply this proportion to the given numbers to get your final result. See how this procedure is applied in the given solutions. Then you should probably rework the section, being sure your answers agree with those given in the back of the text, and send me questions about anything you aren? sure you understand.
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RESPONSE --> Ok.
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