The frequency distribution below summarize the home sale prices - 94862

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1. (5 points) The frequency distribution below summarize the home sale prices in a city for a specific month. Determine the width of each class. A. 20 B. 28 C. 30 D. 31 2. (5 points) The following frequency distribution analyzes the scores on a test. Find the class boundaries of scores interval 95-99. A. 95.5, 99.5 B. 95.5,100.5 C. 94.5,99.5 D. 94.5,100.5 3. (5 points) The histogram below represents the number of television sets per household for a sample of U.S. households. What is the maximum number of households having the same number of television sets? A. 20 B. 100 C. 5 D. 50 4. (5 points) The students in John’s math class took the Scholastic Aptitude Test. Their math scores are shown below. Find the mean score. 528 505 342 348 492 346 349 643 470 482 A. 450.5 B. 459.7 C. 460.5 D. 441.7 5. (5 points) Listed below are the amounts of time in months that the employees of a restaurant have been working at the restaurant. Find the median. 12 4 6 8.5 12 16 17 32 53 85 99 123 140 167 A. 24.5 months B. 58.7 months C. 17 months D. 32 months 6. (5 points) Listed below are the lengths in inches of each snake in the Atlanta Zoo’s reptile house. Find the mode. 9 15 78 13 16 101 19 10 14 17 102 A. 17 inches B. 13.9 inches C. no mode D. 78 inches 7. (5 points) The heights of a group of professional basketball players are summarized in the frequency distribution below. Find the mean height. . A. 78.2 in. B. 76.4 in. C. 74.4 in. D. 13.2 in. 8. (5 points) A survey of the 9854 vehicles on the campus of State University yielded the following pie chart. Find the number of hatchbacks. Round to the nearest whole number. A. 657 B. 36 C. 6307 D. 3547 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Express percents as decimals. Round dollar amounts to the nearest cent. 9. There are 31 participants in a special high-adventure camp. Following is a list of the age of the participants. 16, 18, 13, 24, 17, 17, 18, 14, 14, 16, 14, 20, 22, 21, 15 11, 13, 26, 27, 13, 16, 17, 17, 14, 19, 15, 17, 16, 19, 19, 28 a. (10 points) Prepare a frequency distribution of the participants' ages with a class width of 2, and another with class width of 5. (a) Class size = 2: Class Interval Frequency 11 - 12 1 13 - 14 7 15 - 16 6 17 - 18 7 19 - 20 4 21 - 22 2 23 - 24 1 25 - 26 1 27 - 28 2 Class size = 5: Class Interval Frequency 10-14 8 15-19 16 20-24 4 25-29 3 b. (10 points) Construct a histogram of the participants' age with a class width of 2 and another with a class width of 5. (a) Class size = 2: (b) Class size = 5: 10. I have a collection of 5 ancient gold coins. Their weights, in ounces, are 23.1, 18.6, 33.5, 12.4, and 27.1. a. (5 points) What is the mean weight of my ancient gold coins? Mean (M) = (23.1 + 18.6 + 33.5 + 12.4 + 27.1)/5 = 22.94 ounces b. (5 points) How do you consider this collection, a population or a sample? Why? This is a sample because there can be other ancient gold coins also in existence and ours is one small collection of 5 coins. c. (10 points) What is the variance and standard deviation in weight of my coin collection? Deviations from the mean(d = x – M) are respectively 0.16, -4.34, 10.56, -10.54 and 4.16 Squared deviations (d^2) are respectively 0.0256, 18.8356, 111.5136, 111.0916 and 17.3056 Variance = ?(d^2)/(N – 1) = 258.772/4 = 64.693 ounce^2 Standard deviation = ?variance = ?64.693 = 8.043 ounces 11. Below is a summary of test score in two sections. The questions and possible maximum scores are different in these two sections. We notice that Student A4 in Section A and Student B2 in Section B have the same numerical score. Test A Student Score Test B Student Score A1 70 B1 15 A2 42 B2 61 A3 53 B3 48 A4 61 B4 90 A5 22 B5 85 A6 87 B6 73 A7 59 B7 48 ----- ------ B8 39 a. (15 points) How do these two students stand relative to their own classes based on their z-scores? For Test A, mean is ? = 56.29 and SD is ? = 20.62 z = (x – ?)/? = (61 – 56.29)/20.62 = 0.2284 For Test B, mean is ? = 57.38 and SD is ? = 25.50 z = (x – ?)/? = (61 – 57.38)/25.50 = 0.142 Since both z values are positive, both students have scored higher than the average marks in their respective section b. (5 points) Which student performed better? Explain your answer based on z-score. Student A4 has performed better than student B2 since his z- score is higher.
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