course Mth 152 I understood about half of this... I really need to study this chapter.One thing I've come to understand... though this has no real bearing on anything... is that I hate the term z-score I know that sounds stupid but it sounds like some bad ""hip street name"" from the early 1980's... I think of that kid with the flip up sunglasses on that bad sitcom spin off (of the Cosby Show I think).... Anyways, my thoughts on that.
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10:44:28 **** query problem 13.4.12 z score for KG's rebounds (.4 from bottom range 10-13)
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RESPONSE --> Okay, I had to study this set of problems in the book a few times because I thought they wanted something different than they actually did... I was thinking they just wanted me to add all of his rebounds up and compare the mean.... after figuring it out here is what I got: I took KG's total amount of rebounds (921) and subtracted a mean of all of the other player's rebounds (892.8) and got a total of 28.2 Then I found the standard deviation, which came out to 38.8, and I divided 38.8/28.2 for an answer of 1.375
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10:44:44 ** 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 --> Okay, I got something different on that one
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10:47:38 query problem 13.4.30 midquartile same as median? (Q1+Q3)/2
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RESPONSE --> Yes... the second quartile, or Q2 is defined as being ""just the median, the middle item when the number of items is odd, or the mean of the two middle items when the number of items is even""
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10:48:28 ** 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 --> Oh, okay... I see, I guess the mid quartile is not the same after all.
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{HdwЇ assignment #018 {~ήǪSi`患 Liberal Arts Mathematics II 07-17-2006
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10:56:15 **** query problem 13.5.12 percent above 115 IQ, mean 100 std dev 15
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RESPONSE --> The percent of people who's IQ is above 115 is roughly 16% (take from .159). I found this by doing the following.... taking the mean of 100 and comparing the z score of 115... thus 15 units for this are from the mean. or z score= 15/15=1 As .339 of the distribution (by looking at the table) lies between the mean and z we look at the next step... the remaining proportion of students who had an IQ higher than the 115 rest of the populus.
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10:56:54 ** 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 --> More or less what I got, yes. This problem was a bit hard for me so I'm surprised I got it right... I followed the answers to similiar problems in the book and compared them to examples in the book.
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10:57:38 ** 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 --> This one had me confused so I will monitor how you worked this
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11:01:59 **** 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 --> First, I think I have this correct... lets find the displacement from the 490 hour bulbs.... 490-600= -110 -110 divided by the std dev of 50=-2.2 Consulting the chart for the area of region between the mean and z=-2.2 we find the figure of .486 Okay, lets look at the 720 bulbs.... 720-600= 120 120/50= 2.4 the area of region is .492 Comparing one side on the negative and the other on the positive we add the .486 and .492 area of regions together.... for a total of .978 Now, since we are looking at 10,000 bulbs I think we are to multiply the .978 times the 10,000.... for a total of 9780 that will last in the given area.
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11:02:10 ** 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 --> Okay, whew! I got this one correct!
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11:04:27 **** 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 --> Okay, the A grades are given to the students who have a z score that is above a mark of 1.5 As such, I had to find the area between the mean and z= 1.5 on the table, which was .433 Glancing to the right of z=1.5, which is for those who had the A grades (thus above the 1.5 mark), I find .500 Now I took .500 -.433 to find that .067 or 6.7%
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11:04:32 ** 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 --> Okay, good
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11:04:38 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 arent sure you understand.
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RESPONSE --> Thanks for the info.
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