Assignment 22

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Algonquin College *

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QUA0003

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Statistics

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Apr 3, 2024

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Name: Ayah Alasttal Assignment #2 (6%) This assignment relates to the Course Learning Requirements: CLR2 Calculate Summary Measures Objective of this Assignment : Evaluate the calculation and interpretation of measures of central tendency (averages) and measures of dispersion. Instructions: Use Word or Excel to solve the following eight exercises, evaluated out of 37 points. Please show all your work. Exercises 1) Consider the following sample values: 1.3, 7.0, 3.6, 4.1, 5.0. (a)Compute the mean; (b) Calculate ∑ ( X − X ) . Round off to nearest 10 th . (4 points) (a) The mean = (1.3+7.0+ 3.6+ 4.1+ 5.0) / 5 = 21 / 5 = 4.2 (b) ∑ ( X − X ) = (1.3 – 4.2) + (7.0 – 4.2) + (3.6 – 4.2) + (4.1– 4.2) + (5.0 – 4.2) = -2.9 + 2.8 + -0.6 + -0.1 + 0.8 = 0 2) The Bookstall Inc. is a specialty bookstore concentrating on used books sold via the Internet. Paperbacks are $1 each, and hardcover books are $3.50. Of the 50 books sold last Tuesday morning, 40 were paperback, and the rest were hardcover. What was the weighted mean price of a book? Round to nearest 100-th. Round to nearest 100 th (2 points) The weighted mean = ($1 x 40) + ($3.50 x 10) / 50 = $1.5 3) Listed below is the number of hours that a sample of students spent travelling to college each week. Round off to nearest 100 th (4 points) 9.0 8.5 8.0 9.1 10.3 11.0 11.5 10.3 10.5 9.8 9.3 8.2 8.2 8.5 a) What is the median number of hours spent travelling per week? 8.0, 8.2, 8.2 , 8.5, 8.5, 9.0, 9.1, 9.3, 9.8, 10.3, 10.3 , 10.5, 11.0, 11.5 The median number = 9.1 + 9.3 / 2 = 9.2 b) What is the mode? The mode = 8.2, 8.5, 10.3

4) Gas prices in St. John’s, Newfoundland, have increased from 64.1 cents per litre for regular unleaded in 1990 to 129.6 cents per litre in 2013. What was the geometric mean annual increase for the period? (2 points) Starting (1990) = 64.1, Ending (2013) = 129.6, Number of years = 23 years The geometric mean = n √ (end value / start value) – 1 = 23 √ ( 129.6 / 64.1 ) – 1 = 0.0311. Percentage = 3.11% 5) The Business School at a local college offers eight sections of basic statistics. Following are the number of students enrolled in these sections: 34, 46, 52, 29, 41, 38, 36, and 28. Find (a) the range, (b) the arithmetic mean, (c) the mean deviation; (d) interpret the range and the mean deviation. (5 points) (a) The range = max – min = 52 – 28 = 24 (b) The arithmetic mean = sum / number of sections = 304 / 8 = 38 (c) The mean deviation = (|34 - 38| + |46 - 38| + |52 - 38| + |29 - 38| + |41 - 38| + | 38 - 38| + |36 - 38| + |28 - 38|) / 8 = 6.25 (d) Interpretation: the range of 24 means that there is a difference of 24 students between the most section with students and the fewest. For the mean deviation , it means the number of students enrolled in each section deviates from the arithmetic mean by around 6.25 students. 6) A consumer watchdog organization is concerned about credit card debt. A survey of 10 young adults with credit card debt of more than $2000 showed they paid an average of just over $100 per month against their balances. Listed below are the amounts ($) each young adult paid last month: 100, 126, 103, 93, 99, 113, 87, 101, 109, 100. Compute the variance and the standard deviation. (4 points) 87, 93, 99, 100, 100, 101, 103, 109, 113, 126 The mean = 1031 / 10 = 103.1 The variance = ( (100 - 103.1) 2 + (126- 103.1) 2 + (103- 103.1) 2 + (93- 103.1) 2 + (99- 103.1) 2 + (113- 103.1) 2 + (87- 103.1) 2 + (101- 103.1) 2 + (109- 103.1) 2 + (100 - 103.1) 2 ) / (10-1) The variance = 1058.9 / 9 = 117.66 The standard deviation = 10.85

7) Consider the salaries, in thousands of dollars, for a sample of 15 chief financial officers in the electronics industry. Round off to nearest 100 th (6 points) Salaries $516.0 548.0 566.0 534.0 586.0 529.0 546.0 523.0 538.0 523.0 551.0 552.0 486.0 558.0 574.0 486.0, 516.0, 523.0, 523.0, 529.0, 534.0, 538.0, 546.0, 548.0, 551.0, 552.0, 558.0, 566.0, 574.0, 586.0 a) Determine the mean, the median, and the standard deviation The mean = 8130 / 15 = 542 The median = 546 The standard deviation = √ ( (486- 542 ) 2 + (516- 542 ) 2 + (523- 542 ) 2 + (523- 542 ) 2 + (529- 542 ) 2 + (534- 542 ) 2 + (538- 542 ) 2 + (546- 542 ) 2 + (548- 542 ) 2 + (551- 542 ) 2 + (552- 542 ) 2 + (558- 542 ) 2 + (566- 542 ) 2 + (574- 542 ) 2 + (586- 542 ) 2 ) / (15-1) = √ 8808 / 14 = 25.08 b) Determine the coefficient of skewness. Coefficient of Skewness = (Mean - Median) / Standard Deviation = (542 - 546) / 25.08 = -4 / 25.08 = -0.156 8) The masses (in kilograms) of a sample of five boxes being sent by United Parcel Service (UPS) are: 12, 6, 7, 3, and 10. Round to nearest 100 th (10 points) 3, 6, 7, 10, 12 a) Obtain the range. The range = max – min = 12 – 3 = 9 b) Get the mean deviation. The mean = 38 / 5 = 7.6 Mean Deviation = (|12 - 7.6| + |6 - 7.6| + |7 - 7.6| + |3 - 7.6| + |10 - 7.6|) / 5 = (4.4 + 1.6 + 0.6 + 4.6 + 2.4) / 5 = 13.6 / 5 = 2.72 c) Find the standard deviation. The standard deviation = √ ( (3 - 7.6 ) 2 + (6 - 7.6 ) 2 + (7 - 7.6 ) 2 + (10 - 7.6 ) 2 + (12 - 7.6 ) 2 / (5-1) = √ 49.2 / 4 = 3.51 d) Using Chebyshev’s theorem, at least what percentage of the observations must be within 2.5 standard deviations of the mean? At least 84% e) Using the Empirical Rule, about 68 percent of the values would occur between what values?

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- 7.6 + 1 x (3.51) = 11.11 - 7.6 - 1 x (3.51) = 4.09 Therefore, it is between 11.11 and 4.09.

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