E there are no outliers because all the observations

This preview shows page 13 - 17 out of 28 pages.

e. There are no outliers because all the observations are within 3 standard deviations of the mean. 34. a. x is 100 and s is 13.88 or approximately 14 b. If the distribution is bell shaped with a mean of 100 points, the percentage of NBA games in which the winning team scores more than 100 points is 50%. A score of 114 points is z = 1 standard deviation above the mean. Thus, the empirical rule suggests that 68% of the winning teams will score between 86 and 114 points. In other words, 32% of the winning teams will score less than 86 points or more than 114 points. Because a bell-shaped distribution is symmetric, approximately 16% of the winning teams will score more than 114 points. c. For the winning margin, x is 11.1 and s is 10.77. To see if there are any outliers, we will first compute the z -score for the winning margin that is farthest from the sample mean of 11.1, a winning margin of 32 points. 32 11.1 1.94 10.77 x x z s = = =
Image of page 13

Subscribe to view the full document.

Chapter 3 3 - 14 Thus, a winning margin of 32 points is not an outlier ( z = 1.94 < 3). Because a winning margin of 32 points is farthest from the mean, none of the other data values can have a z -score that is less than 3 or greater than 3 and hence we conclude that there are no outliers 35. a. x x n i = = = Σ 79 86 20 399 . . Median = 4.17 4.20 4.185 2 + = (average of 10th and 11th values) b. Q 1 = 4.00 (average of 5th and 6th values) Q 3 = 4.50 (average of 15th and 16th values) c. s x x n i = = = Σ ( ) . . 2 1 12 5080 19 08114 d. The distribution is significantly skewed to the left. e. Allison One: z = 412 399 08114 016 . . . . Omni Audio SA 12.3: z = ≈ − 2 32 399 08114 2 06 . . . . f. The lowest rating is for the Bose 501 Series. It’s z -score is: z = ≈ − 214 399 08114 2 28 . . . . This is not an outlier so there are no outliers. 36. 15, 20, 25, 25, 27, 28, 30, 34 Smallest = 15 i = = 25 100 8 2 ( ) Q 1 20 25 2 22 5 = + = . Median = + = 25 27 2 26 i = = 75 100 8 8 ( ) Q 3 28 30 2 29 = + = Largest = 34 37.
Image of page 14
Descriptive Statistics: Numerical Methods 3 - 15 15 20 25 30 35 38. 5, 6, 8, 10, 10, 12, 15, 16, 18 Smallest = 5 25 (9) 2.25 100 i = = Q 1 = 8 (3rd position) Median = 10 75 (9) 6.75 100 i = = Q 3 = 15 (7th position) Largest = 18 15 20 5 10 39. IQR = 50 - 42 = 8 Lower Limit: Q 1 - 1.5 IQR = 42 - 12 = 30 Upper Limit: Q 3 + 1.5 IQR = 50 + 12 = 62 65 is an outlier 40. a. Five number summary: 5 9.6 14.5 19.2 52.7 b. IQR = Q 3 - Q 1 = 19.2 - 9.6 = 9.6 Lower Limit: Q 1 - 1.5 (IQR) = 9.6 - 1.5(9.6) = -4.8 Upper Limit: Q 3 + 1.5(IQR) = 19.2 + 1.5(9.6) = 33.6 c. The data value 41.6 is an outlier (larger than the upper limit) and so is the data value 52.7. The financial analyst should first verify that these values are correct. Perhaps a typing error has caused 25.7 to be typed as 52.7 (or 14.6 to be typed as 41.6). If the outliers are correct, the analyst might consider these companies with an unusually large return on equity as good investment candidates. d.
Image of page 15

Subscribe to view the full document.

Chapter 3 3 - 16 20 35 -10 5 50 65 * * 41. a. Median (11th position) 4019 25 (21) 5.25 100 i = = Q 1 (6th position) = 1872 75 (21) 15.75 100 i = = Q 3 (16th position) = 8305 608, 1872, 4019, 8305, 14138 b. Limits: IQR = Q 3 - Q 1 = 8305 - 1872 = 6433 Lower Limit: Q 1 - 1.5 (IQR) = -7777 Upper Limit: Q 3 + 1.5 (IQR) = 17955 c. There are no outliers, all data are within the limits. d. Yes, if the first two digits in Johnson and Johnson's sales were transposed to 41,138, sales would have shown up as an outlier. A review of the data would have enabled the correction of the data.
Image of page 16
Image of page 17
You've reached the end of this preview.
  • Spring '08
  • Staff
  • Standard Deviation, Descriptive statistics, standard deviations, x=, Σxi

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern