Stats Chapter 3 Homework

In June, an investor purchased 200 shares of Oracle (an information technology company) stock at $44 per share. In August, she purchased an additional 250 shares at $43 per share. In November, she purchased an additional 450 shares at $46 per share. What is the weighted mean price per share? (Round your answer to 2 decimal places.)

44.72 Multiply each value by its "weight" or number of times it happens to find the total. Then divide by the sum of the weights. In this case the weights are the number of shares purchased at each price. The weighted mean price per share is $44.72, found by 200($44) + 250($43) + 450($46)200 + 250 + 450 = $40,250900

The Cambridge Power and Light Company selected a random sample of 20 residential customers. Following are the amounts, to the nearest dollar, the customers were charged for electrical service last month: 54 76 68 54 26 45 75 40 66 26 45 75 40 66 75 26 45 75 40 66 Compute the arithmetic mean. (Round your answer to 2 decimal places.) Indicate whether it is a statistic or a parameter.

54.15 statistic a.The mean is the total divided by the number of values. Add the values and divide by 20. So, the mean equals $54.15, found by $1,083/20. b.This is a sample statistic because the power company likely serves more than 20 customers.

Compute the mean of the following population values: 5, 7, 5, 7, 6. (Round your answer to 1 decimal place.) Mean

6 The mean is the sum of all values divided by the number of values. Add the values and divide by five. So, the mean equals 30/5 = 6.0.

A sample of 25 undergraduates reported the following dollar amounts of entertainment expenses last year: 757 764 722 750 751 696 716 684 716 743 774 766 767 759 701 744 707 720 730 741 749 680 733 746 756 a. Find the mean, median, and mode of this information. b. What are the range and standard deviation? c. Use the Empirical Rule to establish an interval which includes about 95 percent of the observations. (Round your answers to 2 decimal places.)

a. Mean$734.88 Median$743.00 Mode$716 b. range 94 standard deviation 26.66 c. interval 681.56 up to 788.20 Explanation a.The mean is $734.88, found by $18,372/25. The median is $743 and there are two modes $716. b.The range is $94, found by $774 − 680, and the standard deviation is $26.66, found by the square root of 17,058.64/24. c.From $681.56 up to $788.20, found by $734.88 ± 2($26.66).

The annual report of Dennis Industries cited these primary earnings per common share for the past 5 years: $2.3, $1.33, $2.17, $4.25, and $4.2. If we assume these are population values. What is the arithmetic mean primary earnings per share of common stock? What is the variance?

a. 2.85 b. 1.37 Explanation a.The mean is the total divided by the number of values. So we have $2.85.b.A population variance is the mean of the squared deviation from the mean. In this case it isσ2=(2.30−2.850)2+...+(4.25−2.850)2+(4.20−2.850)25=1.37σ2=(2.30-2.850)2+...+(4.25-2.850)2+(4.20-2.850)25=1.37

The personnel files of all eight employees at the Pawnee location of Acme Carpet Cleaners Inc. revealed that during the last 6-month period they lost the following number of days due to illness: 3 1 0 0 2 11 5 3 All eight employees during the same period at the Chickpee location of Acme Carpets Cleaners revealed they lost the following number of days due to illness: 3 6 8 4 4 5 3 1 a Calculate the range and mean for the Pawnee location and the Chickpee location. (Round the "Mean" to 2 decimal places.) Range Mean b-1. Based on the sample data, which location has fewer lost days? Pawnee location Chickpee location b-2. Based on the sample data, which location has less variation? Chickpee location Pawnee location

a. Pawnee location Range 11 Mean 3.13 Chickpee location Range 7 Mean 4.25 b-1 Pawnee location b-2 Chickpee location Explanation a.The range is the difference between the largest and smallest values. The Pawnee range is 11, found by 11 - 0, and the Chickpee range is 7, found by 8 - 1.The mean is the total divided by the number of values. The Pawnee mean is 7, found by 25/8, and the Chickpee mean is 4.25, found by 34/8.

The ages of a sample of Canadian tourists flying from Toronto to Hong Kong were 34, 25, 57, 42, 50, 22, 56, 54, 41, and 38. a. Compute the range. b. Compute the standard deviation. (Round your answer to 2 decimal places.)

range 35 standard deviation 12.47 Explanation a.The range is the difference between the largest and smallest value or 35, found by 57 - 22.b.The sample mean is 41.9, found by 419/10.The differences between each value and the mean are: −7.9, −16.9, 15.1, .1, 8.1, −19.9, 14.1, 12.1, −.9 and −3.9, respectively. Their squares are 62.41, 285.61, 228.01, .01, 65.61, 396.01, 198.81, 146.41, .81, and 15.21. The sum of the squares is 1,398.9. So the sample variance is 155.433, found by 1,398.9/9 and the sample standard deviation is the square root of that number or 12.47.