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ASSIGNMENT/PROJECT VII MATH 566 1. (Assessing normality). Many times in statistics it is necessary to see if a distribution of data values is approximately normal. One technique is to draw a histogram for the data and see if it is approximately bell-shaped. The following table lists the number of branches of the 50 top libraries: 67, 84, 80, 77, 97, 59, 62, 37, 33, 42, 36, 54, 18, 12, 19, 33, 49, 24, 25, 22, 24, 29, 9, 21, 21, 24, 31, 17, 15, 21, 13, 19, 19, 22, 22, 30, 41, 18, 20, 22, 26, 33, 14, 14, 16, 22, 26, 10, 16, 25 (i) Construct a frequency distribution for the data using class width of 10. (ii) Draw a histogram from the distribution given in (i). 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 1 14 17 7 3 2 2 1 2 1 (iii) Describe the shape of the distribution. The shape is approximately normal; it is skewed up and to the right. (iv) Based on your answer to (iii), do you think that the distribution is approximately normal? Yes, it is approximately normal. 1 In addition to the histogram, distributions that are normal have about 68% of the values fall within 1 standard deviation of the mean, about 95% of the values fall within 2 standard deviation of the mean, and almost 100% of the values fall within 3 standard deviation of the mean (see Figure 6-5, page 268). (v) Find the mean and standard deviation for the data. (vi) What percent of data values fall within 1 standard deviation of the mean? 82% is one standard deviation of the mean (vii) What percent of data values fall within 2 standard deviation of the mean? 92% is within two standard deviation of the mean (viii) What percent of data values fall within 3 standard deviation of the mean? 98% is within three standard deviation of the mean (ix) How do your answers to (vi), (vii), and (viii) compare to 68%, 95%, and 100%, respectively? Comparing 68% to 82% there is a significant gap, 95% to 92 the gap has minimized and same with comparing 100% to 98% 2. (Smart people). Assume that you are thinking about starting a Mensa chapter in your home town, which has a population of 10,000 people. You need to know how many people would qualify for Mensa, which requires an IQ score of at least 130. You realize that IQ scores is normally distributed with a mean of 100 and a standard deviation of 15. Complete the following: 2 (i) Find the approximate number of people in your home town that are eligible for Mensa. Get the z score: z = (130-100)/15 = 2 Use the z table to get probability (z > 2) = 0.02275 Multiply by the 10000 people: 227.5 about 228 people (ii) Is it reasonable to continue your quest for a Mensa chapter in your home town? Yes, it is reasonable, since there are enough people in the home town to start a chapter (iii) How would you proceed to find out about how many of the eligible people would actually join the new chapter? Be specific about your methods of gathering data. You can put up flyers at the library, bus stops, local colleges, schools, and other populated areas. You can encourage people to get IQ tested. You can ask your friends to spread the word. You can send mass e-mails about the project to family and friends (iv) What would be the minimum IQ score needed if you wanted to start an Ultra-Mensa club that include only the top 1% of IQ scores? The z score for the top 1% is (from a table): 2.3263 Use the formula: x = sigma x z + mu x = 2.3263 x 15 + 100 x = 134.8945 3 3. p. 314 Using the standard normal distribution find each Probability. a. P(0 < z < 2.07) = 0.4808 b. P(-1.83 < z < 0) = 0..4664 c. P(-1.59 < z < 2.01) = 0.9219 d. P(1.33 < z < 1.88) = 0.0617 e. P(-2.56 < z < 0.37) = .06391 f. P(z > 1.66) = 0.0485 g. P(z < 2.03) = 0.9788 h. P(z > 1.19) = 0.1170 i. P(z < 1.93) = 0.9732 j. P(z > 1.77) = 0.0383 -x has a normal distribution with a mean of 30 and a standard deviation of 4 -z has a normal distribution with a mean of 0 and a standard deviation of 1 4 4. p. 316 The time it takes for a certain pain reliever to begin to reduce symptoms is 30 minutes, with a standard deviation of 4 minutes. Assuming the variable is normally distributed, find the probability that it will take the medication a. Between 34 and 35 minutes to begin to work. a. P(34<x<35) = P(34-30)/4<Z<35-30)/4) = P(1<Z<1.25) = P(Z<1.25)P(Z<1) = 0.894 – 0.841 = 0.053 b. More than 35 minutes to begin to work. b. P(x>) =P(Z>1.25)=1-P(Z<1.25) = 1-0.894 = 0.106 c. Less than 25 minutes to begin to work. c. P(x<25) = P(z<1.25) = 0.106 d. Between 35 and 40 minutes to begin to work. d. P(35<x<40) = P(1.25<Z<2.5) = P(Z<2.5)-P(Z<1.25)=0.994-0.894 = 0.1 5 5. P.317 The average number of years a person takes to complete a graduate degree program is 3. The standard deviation is 4 months. Assume the variable is normally distributed. If an individual enrolls in the program, find the probability that it will take: 4 months = 0.333333 years a. More than 4 years to complete the program. Z= (4-3)/0.333333 = 3 P(z > 3) from a table = 0.00135 b. Less than 3 years to complete the program. Z = 0, since 3 years in the mean P(z < 0) = 0.5 , because half the data is below the mean c. Between 3.8 and 4.5 years to complete the program. Z (3.8) = (3.8-3)/0.3333 = 2.4 Z (4.5) = (4.5-3)/0.3333 = 4.5 P(24 < z < 4.5) from a table = 0.008194 d. Between 2.5 and 3.1 years to complete the program. Z (2.5) = (2.5-3)/0.33333 = -1.5 Z (3.1) = (3.1-3)/.33333 = 0.3 P(-1.5 < z < 0.3) from a table = 0.5511 6 6. Data Project 1 p. 319. Select a variable (interval or ratio) and collect 30 data values. a. Construct a frequency distribution for the variable. (a) Collect a data of size n=30 or more and construct a frequency distribution table. b. Use the procedure described in the Critical Thinking Challenge on page 317 to graph the distribution on normal probability paper. (b) Find the mean and standard deviation of the data. c. Can you conclude that the data are approximately normally distributed? Explain your answer. (c) Find the approximate probability for each class interval given in part (a). (d) refer to the critical thinking challenges problem for drawing the graph on normal probability paper. use the cdf (cumulative distribution function) 7 1. Make a table, as shown. Boundaries Frequency Frequency Cumulative Frequency 89.5–104.5 104.5–119.5 119.5–134.5 134.5–149.5 149.5–164.5 164.5–179.5 Low Cumulative Percent 24 62 72 26 12 _4__ 200 Freq 89.5-104.5 104.5-119.5 Center Point 97 112 Cumm Freq Cumm Freq % 24 62 24 86 12% 43% 2,328 6,944 Probability of Rages 9% 28% 119.5-134.5 127 72 158 79% 9,144 37% 134.5-149.5 142 26 184 92% 3,692 15% 149.5-164.5 164.5-179.5 157 172 12 4 196 200 98% 100% Total Wtd Avg 1,884 688 24,680 4,113 8% 3% Mean Std Dev of Freq 108.36 27.47 Range Cumm Normal Distribution 34% 55% 75% 89% 96% 99% 8 Wtd Avg Gap to reach Mean Percentage of 2nd range Offset into 2nd Range Wtd Mean 1,785 25.7% 3.86 108.36