251y0213 10/07/02 ECO251 QBA1 Name __________________ FIRST HOUR EXAM SECTION MWF 10 11 TR 11 12:30 OCTOBER 1, 2002 Part I. (10 points)1. (Lind et. al.) A student takes a survey of heights of 450 college women and divides them into 10 classes.Her first two class midpoints are 62.5" and 65.5". There are 91 women in the first class and 110 women inthe second class. (5)a. What is the (width of the) class interval w ?b. What is the lower limit of the third class?c. What is the relative frequency of the second class?d. What is the cumulative relative frequency of the second class?e. If this distribution is skewed to the left, about what percent of the data is above the median?Solution: a) Since 65.5 - 62.5 = 3, the interval must be 3.b) The lower limit must be 67. If the class interval is 3, the class must extend by 1.5 on either side of the midpoint. The layout for the first three classes is given in the table below. 450nClass Midpoint fnfrelf FnFrelF 61 - 64 62.5 91 .2022 91 .202264 - 67 65.5 110 .2444 201 .446767 - 70 68.5 ? ? ? ?c) The relative frequency is .275 d) The cumulative relative frequency is .5025, which might be found as (about) the sum of .2022 and .2444.e) the median is defined as a point with 50% of the data above or below it.2. Indicate whether the following are: Nominal Data, Ordinal Data, Interval Data, Continuous Ratio Data or Discrete Ratio Data. (3) a. A recent cartoon suggested that the numbers on pro Football jerseys be replaced by their salaries. What kind of data would the numbers be before the change? Ans: Nominal.b. What kind of data would the numbers usually be considered after the change? Ans: Continuousratio.c. What kind of data is a team's score at half time? Ans: Discrete ratio.3. All my family doctor's patient files are coded as follows: FS (Adult females who currently smoke); FN (Adult females who do not currently smoke); MS (Adult males who currently smoke) and MN (Adult males who do not currently smoke). Are these categories mutually exclusive and collectively exhaustive? Ifyou say 'no' to either characteristic, explain. (2) Ans: The categories are mutually exclusive ( for instance no person who is coded FS will be in FN, MS or MN), but not collectively exhaustive because there is no reason to assume that all the family doctor's patients are adults.1251y0213 9/30/02Part II. Compute an appropriate answer, showing your work (15 Points maximum - if you do more than 15 points, only your right answers will be counted.):1) A sample of pipe outside diameters gives a mean of 16.0 inches and a standard deviation of 0.2 inches.a) If the median diameter is 15.9 inches and the mode is unknown(i) What is the maximum fraction of the pipes that could have a diameter above 16.4 inches? (1) Ans: Get a z-score. 22.0164.16sxxzk. We know from the Chebyshef rule that the fraction in the tail above k is less than 21k. Since,2kthis fraction is below %.2541(ii) Between what two diameters must at least 15/16 of the pipe diameters lie? (1) Ans: If only 21161k are outside the interval, we must have 162k or .4kThus the interval is 8.00.162.40.16 k or 15.2 to 16.8.(iii) Is this distribution skewed to the left or the right, or is it symmetrical? (1) Ans: Since the median is below the mean, it should be skewed to the right.b) If, instead, the median diameter is 16.0 inches and the mode is also 16.0 inches, between what two diameters must almost all the pipe diameters lie? (1) Ans: Since 6.0162.0316 is 3 standard deviations from the mean, the Empirical Rule (which applies because the mean, median and mode are equal) says that there will be almost none outside 15.4 to 16.6.c) What is the coefficient of variation for this sample of pipes? (1) Ans:025.0164.0xsC.2) The newest computer in the headquarters of a firm that you are liquidating is five months old and the oldest is 99 months old. a) If the (absolute) frequencies are to be presented in a line graph in seven classes, what intervals would you use? Explain your reasoning using an appropriate formula and use it to fill in the table below.(3)Ans: 43.137599so use 14. This is only a suggestion. Any number, like 15, somewhat above 13.57 will work, as long as you cover the range. Two possibilities are shown below. You should have shown only one.Class From to From toA 3 16.9 0 14.9B 17 30.9 15 29.9C 31 44.9 30 44.9D 45 58.9 45 59.9E 59 72.9 60 74.9F 73 86.9 75 89.9G 87 100.9 90 104.9b) What is the name of the type of graph that you are drawing (Is it a histogram?) and what would the x and y coordinates be of the last point on the line that you draw to represent the frequencies? (2) Ans: The graph is a frequency polygon and we must create an empty class to endit. For the first classification above, the interval is 14 and the midpoint of the last class on the table is 94, so the last point is 0,108 yx. For the second classification above, the interval is 15 and the midpoint of the last class on the table is 97.5, so the last point is 0,5.112 yx.23251y0213 9/30/023) For the numbers 70, 270, 470 and 670, compute the a) Geometric Mean b) Harmonic mean, c) Root-mean-square (2 each). Label each clearly. If you wish, d) Compute the geometric mean using natural or base 10 logarithms. (3 points if you need it here or two points if you need it in the next section - doing thisis insurance, you cannot get more than cannot get more than 15 points on part II or 25 points on part III unless you do the extra credit in part III. ) Solution: Note that 1480x. This is not used in any of the following calculations and there is no reason why you should have computed it!(a) The Geometric Mean. 415951610000595161000067047027070441321nnngxxxxxx 25.05951610000753.277. Do any of you really believe that 45951610000595161000041?(b) The Harmonic Mean. 0014925.00021277.00037037.00142857.04167014701270170141111xnxh 0216096.04100540240.0. So 103.18500540240.01111xnxh. As I explained several times, xn11 could not possibly be 1480141. (c) The Root-Mean-Square.
View Full Document
Unlocking...