week 6 - Exercise 88 Refer to the Baseball 2005 data, which...

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Exercise 88 Refer to the Baseball 2005 data, which reports information on the 30 major league teams for the 2005 baseball season. a. Select the variable team salary and find the mean, median, and the standard deviation. Mean = 73.06 Median = 66.20 Standard deviation = 34.23 b. Select the variable that refers to the age the stadium was built. (Hint: Subtract the year in which the stadium was built from the current year to find the stadium age and work with that variable.) Find the mean, median, and the standard deviation. Mean = 28.20 Median = 17.50 Standard deviation = 25.94 c. Select the variable that refers to the seating capacity of the stadium. Find the mean, median, and the standard deviation. Mean = 45,913 Median = 44,174 Standard deviation = 5,894 Exercise 56 Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study. a. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time? P(x=4) = 0.65610 b. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time? P(x=0) = 0.00010 c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time? P(x≥1) = 0.99990 n = 4, p = 0.9, q = 0.1 Binomial Distribution x p(x) 0 0.00010 1 0.00360 2 0.04860 3 0.29160 4 0.65610 Exercise 64 An internal study by the Technology Services department at Lahey Electronics
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revealed company employees receive an average of two emails per hour. Assume the arrival of these emails are approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday? P(x=1) = 0.270670566 b. What is the probability she received 5 or more email during the same period? P(x≥5) = 0.016563608 c. What is the probability she did not receive any email during the period? P(x=0) = 0.135335283 Exercise 50 Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles. Mean = 60,000, St dev = 2000, Normal Distribution a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more? P(x≥65200) = 0.004661188 = 0.4661% b. What percent of the trucks logged more than 57,060 but less than 58,280 miles? P(57,060 25) = 0.016947427 b. What is the likelihood the sample mean is greater than $22.50 but less than $25.00? P(22.5 < x-bar < 25) = 0.90440297 c. Within what limits will 90 percent of the sample means occur? Left Limit = 22.33691285 = $22.34
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week 6 - Exercise 88 Refer to the Baseball 2005 data, which...

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