IT 403 Practice Problems (1-2) Answers

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1 IT 403 Practice Problems (1-2) Answers #1. Using Tukey's Hinges method ('Inclusionary'), what is Q3 for this dataset? a. 7 b. 11 c. 12 d. 15 c (12) #2. How do quartiles and percentiles relate? Q0 (minimum) = 0 percentile, Q1 = 25 percentile, Q2 = 50 percentile, Q3 = 75 percentile, and Q4 (maximum) is 100 percentile. #3. What is the Y-axis of histograms? Frequency, count, or proportion, percentage. #4. Draw the histogram in each case. Note: [a,b) denotes an interval that is closed on the left (includes a) and open on the right (does not include b). Bin Count [0,1) 1 [1,2) 3 [2,3) 5 [3,4] 1 Bin Count [0,1) 3 [1,2) 5 [2,4] 2 Bin Count [0,1) 2 [1,2) 4 [2,2.5) 3 [2.5,3]

2 #5. Why are outliers important? Outliers could represent erroneous data, in which case they must be corrected or omitted. If correct, they might be the most important data points in the dataset. In business they can significantly affect the bottom line; in science, they might be the key to a scientific breakthrough. #6. Mean and median differ (sometimes significantly) if the data had skewed distribution. The mean is pulled towards: a. the tail (either side) b. the dense area a (the tail) #7. When (for what kind of data, for what purpose) is median a better (or more appropriate) measure of center? Since median is robust (less sensitive) to extreme values such as outliers, it is more appropriate if the data has a skewed, non-symmetric distribution. #8. What happens to mean and Q2 (median) for a dataset that contains n observations, a. if every observation is increased by 7? Both mean and median are increased by 7. b. if the largest observation is increased by 1000? The mean is increased by 1000 / n, the median is unchanged if n 3. #9. What happens to SD for a dataset a. if every observation is increased by 7? -- the SD is unchanged because the spread is unchanged. b. if every observation is multiplied by 3? -- the SD is multiplied by 3 because the spread is multiplied by 3. #10. [Exercise 1.93, p. 59; condensed] For a test score dataset with mean = 288 and stdev = 38, assuming the distribution is approximately normally distributed (i.e., N(288, 38)), what is the range of scores that includes 95% of them? According to the rule, the 95% range is +/- 2 sigma away from the mean. So for the distribution N(288, 38), the range is between: (2 * 38) = 212 and (2 * 38) = 364. #11. [Exercise 1.95&96, p. 60] For the distribution in the previous question, give the z-score for the scores 350 and 240.

3 For XX = 350, ZZ = = = For XX = 240, ZZ = = = #12. [Exercise 1.97, p. 64] For the same dataset, find the proportion of students who received a score < 350. What about the proportion for those who scored >= 350? The proportion of students who scored < 350, i.e. X < 350, can be obtained by looking up the Normal Table, for Z = 1.63 as in the previous question. The proportion is Then the proportion of students who scored >= 350, i.e., X >= 350, is the complement of X < 350. So we get = #13. [Exercise 1.98, p. 64] Give the proportion of students who scored between 300 and 350. The proportion for 300 <= X < 350 can be obtained by subtracting the proportion of X < 300 from X < 350. From the previous question, we know for X < 350 the proportion is For XX = 300, ZZ = proportion for X < 300 is So = = 12 = , and the 38 #14. [Exercise 1.99, p. 66] For the same dataset, what score is needed to be in the top 20%? The question is essentially asking for the X value which separates the distribution between lower 80% and upper/top 20%. From the Normal table, the closest Z score that has 0.8 (for the lower proportion) is So the score X = (0.84 * 38) = =

4 #15. [Exercise 1.100, p. 66] Find the score that 75% of students will exceed (i.e., score above). The question is essentially asking for the X value which separates the distribution between lower 25% and upper 75%. From the Normal table, the closest Z score that has 0.25 (for the lower proportion) is So the score X = (-0.67 * 38) = = #16. For a population of giraffes, the heights are measured. The sample mean is 16.5 ft; the sample standard deviation is 3.8. What percentage of giraffes are taller than 20 ft. The distribution is N(16.5, 3.8). For XX = 20, ZZ = = = 0.921, and the proportion of X < 20 is according to the Normal table. So the proportion for X >= 20 is = , and the percentage is 17.88%. #17. What is the 93rd percentile of height for the giraffes in the previous problem? According to the Normal table, the Z score for which the proportion is 0.93 is So we can obtain the X value by (1.48 * 3.8) = = (ft). #18. For a standard normal curve, where are Q1 and Q3 located? In percentile, Q1 is 25% and Q3 is 75%. According to the Normal table, the Z scores for 0.25 and 0.75 are and #19. The heights of a population of 100,000 men are normally distributed with a mean of 68 inches with SD = 3. About how many men have heights a. below 65 inches?

5 b. above 74 inches? c. between 62 and 74 inches? The distribution is N(68, 3). For X = 65, Z = By the rule, we know the proportion below -1 sigma is 16%. That means 16% of the men have heights below 65 inches. So out of 100,000 men, ( * 0.16 =) 16,000 men are shorter than 65 inches. For X = 74, Z = 2.0. By the rule, we know the proportion above 2 sigma is 2.5%. Similar to the same calculation as the previous question, we get ( * =) 2,500 men are taller than 74 inches. The area between 62 and 74 inches are +/- 2 sigma from the mean, and we know 95% of the data fall in this range. So we get ( * 0.95 =) 95,000 men are between 62 and 74 inches tall. #20. IQ scores are normally distributed with mean 100 and standard deviation 15. Find the IQ scores that correspond to these percentiles: a. 50 b. 85 c. 95 d a. The 50 percentile is the mean. So the IQ score is 100. b. The 85 percentile is at Z score of So the IQ score is (1.04 * 15) = = c. The 95 percentile is at Z = So the IQ score is (1.64 * 15) = = d. The 99.9 percentile is (or starts at) Z = So the IQ score is (3.08 * 15) = = #21. IQ scores are normally distributed with mean = 100 and SD = 15. You took the test and the score was 135. i. What is your percentile? The distribution is N(100, 15). For XX = 135, ZZ = = 35 = By the standard normal table, the proportion of z < 2.33 is So the percentile is 99.01%.

6 ii. How many persons out of one billion have an IQ score greater than 175? Use the Extreme Values of the Normal Distribution table to obtain the proportion. For XX = 175, ZZ = = 75 = 5. Since z = 5.0 is not in the standard normal table, we use 15 the Extreme Values of the Normal Distribution table. According to the table, the proportion for z > 5.0 is 2.867e -07, which is * So in one billion people, there are 1,000,000,000 * (2.867 * 10-7 ) = people who have IQ scores > 175. #22. Construct the normal plots by hand of this dataset: In the class, SPSS was used to create a normal plot automatically (to save time). In SPSS, normal plots are called a Q-Q plot, and it has the Z-score (as Expected Normal ) on the Y-axis instead of X-axis shown in the textbook. Raw Theoretical Data X Z X mean stdev To produce a normal plot by hand, you follow the steps described in the textbook p Arrange the observed data values from smallest to largest. Record what percentile of the data each value occupies. For example, the smallest observation in a set of 20 is at the 5% point, the second smallest is at the 10% point, and so on. 2. Do Normal distribution calculations to find the values of z corresponding to these same percentiles. For example, z = is the 5% point of the standard Normal distribution, and z = is the 10% point. We call these values of Z Normal scores. 3. Plot each data point x against the corresponding Normal score. If the data distribution is close to any Normal distribution, the plotted points will lie close to a straight line. This figure (below, next page) shows what Excel gives you as an X-Y scatter plot where X-axis is the Z-scores and the Y-axis is the X values.

7 Normal Quantile Plot