#$&* course MTH152 Time: 3:04 PM Date: 8/2Note that this assignment is labeled MTH152-QA17 (1), because I want you to look at my method of working problem 5 specifically before I continue.
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Given Solution: 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. Self-critique: Good to go. Graphs and geometry is where my strong suit lies, I believe. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - Taking into account that 34% will lie 1 strd dev LOWER than the mean and 34% will lie 1 strd dev HIGHER than the mean, we see that the mean is 32, taken from n * p = avg. In this case, ‘n’ denotes the total number of events (here, 64) multiplied by the probability of success, defined as getting ‘heads’ (which is shown by 1/2. If this were a die and a specific number was denoted a success, this would be 1/6). - For my solution, I simply added and subtracted the mean by the strd dev to get 36, and 28. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. Self-critique: - I can understand and see the reasoning behind applying .68 * 200 = 136 flips will be heads total between 28 and 36….*** I think I just answered my own question. The 68% obviously is derived from the % of heads between 28 & 32 and 32 & 36, thus 136 heads will be, like I said before, the cumulative total between those two limits. However, please explain further the following excerpt from your Given Solution: 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. - ------------------------------------------------ Self-critique rating:
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Given Solution: 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. Self-critique: - Why wouldn’t one subtract 78.8 and 19.8? If one was making a generalization, wouldn’t the difference be used to show this, or is this another way of saying “the number of instances from one end of the spectrum below the mean to one end of the spectrum above the mean”? ------------------------------------------------ Self-critique rating: I think I got it.
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Given Solution: 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. Self-critique: Good to go. ------------------------------------------------ Self-critique rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: - My theory is that the first range, designated 150-170, has a 20 pt deviation. The strd dev = 20, so this is 1 strd dev . According to the chart, the proportion coinciding with 1 strd dev = 0.341, so 0.341 * 600 = 204.6 or 205 students. - The second range, 120-150, has a 30 pt. deviation, leading me to consider a 1.5 std deviation, which matches with 0.433 on the chart, so .433 * 600 = 259.8 or 260 students. - The third range 135-155 correlates a 15 - 5 point deviation, which I designate as .5 strd dev. Chart lines .5 with .191, so .191 * 600 = 114.6 or 115 students.
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Given Solution: We saw in the preceding problem that 170 lies one standard deviation above the mean. The proportion lying between the mean 150 and 170 is therefore 0.34 so that out of 600 outcomes, we would expect 0.34 * 600 = 204 to lie within the range 150 - 170. As seen in the previous problem, 120 corresponds to an outcome 1.5 standard deviations below the mean. The range 120-150 consists of all outcomes lying between the mean and 1.5 standard deviations below the mean. The proportion lying between these values is seen from the table to be 0.433, we expect that of 600 outcomes we will have 0.433 * 600 = 260 in the range 120-150. We have seen that 135 corresponds to an outcome 0.75 standard deviations below the mean, and that 155 corresponds to an outcome 0.25 standard deviations above the mean. Between 0.75 standard deviations below the mean and the mean we expect 0.273 of all outcomes, and between the mean and 0.25 standard deviations above the mean we expect another 0.099 of the outcomes. Since one range lies completely below and the other completely above the mean, there is no overlap and we expect that 0.273 + 0.099 = 0.372 of the 600 outcomes, or 0.372 * 600 = 223.2 of the outcomes will lie within the range 135-155. We have seen that 135 corresponds to an outcome 0.75 standard deviations below the mean, and that 120 corresponds to an outcome 1.5 standard deviations below the mean. Between 0.75 standard deviations below the mean and the mean we expect 0.273 of all outcomes, and between 1.5 standard deviations below the mean and the mean we expect 0..477 of the outcomes. Since both ranges lie completely below the mean, they overlap and we therefore expect that 0.477 - 0.273 = 0.204 of the 600 outcomes, or .204 * 600 = 122.4 of the outcomes will lie within the range 120-135. STUDENT COMMENT I understood most of these except for the very last one. INSTRUCTOR RESPONSE You should try to be more specific about what you do and do not understand, so I can focus my answer appropriately. However the following might be helpful: The solution basically says that 27.3% of outcomes lie between 135 and 150, while 47.7% lie between 120 and 135, so 47.7% - 27.3% = 20.4$ lie between 120 and 135. Self-critique: - My methods closely resemble yours. See my method of diverging a deviation relating to the difference of ranges taken from each score (e.g. 135-155 indicates a 15 pt and 5 pt deviation, which I labeled .5) ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. The table given in question `q005 can be used as a basis for answering the questions below. A student clicks a mouse repeatedly, as fast as possible. The times between consecutive clicks has a mean of .132 second with a standarad deviation of .016 second. Out of 100 clicks, how many would be expected to fall in each of the following ranges: .132 second to .148 second .148 second to .164 second .116 second to .148 second less than .116 second between .128 second and .140 second (Note: If you want to compare your performance to that of the student in this example, you may do so at http://vhcc2.vhcc.edu/dsmith/forms/ph1_timer_experiment.htm . This is intended only for your interest and you may ignore the part about submitting the document (though if you really want to you are welcome to submit all or part of it) ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. A runner runs a 100-meter race in an average of 11.52 seconds, with a standard deviation of .12 second. In how many out of 100 races would the runner expect a time of 11.40 seconds or better? In how many out of 100 races would the runner expect a time of 11.70 seconds or worse? In how many out of 100 races would the runner expect a time between 11.40 seconds and 11.70 seconds? What do you think is the fastest time the runner could expect in 100 races? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. What would you estimate to be the standard deviation of a distribution whose mean is 100, if 90% of the distribution lies between 75 and 125? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!