MATH 221 FINAL EXAM REVIEW(Version 2) QUESTIONS WITH ANSWERS: DEVRY UNIVERSITY

MATH 221 FINAL EXAM REVIEW (Version 01) : DEVRY UNIVERSITYFinal ExamMATH 221 FINAL EXAM REVIEW(Version 2) QUESTIONS WITH ANSWERS: DEVRY UNIVERSITY Review Questions You should work each of the following on your own, then review the solutions guide. DO NOT look at the solutions guide first. 1. Determine whether the following are nominal, ordinal, interval, and ratio. a. Daily temperatures in Ripon, WI b. Test scores in statistics class Solution: (a) would be interval as there is no zero while (b) would be ratio as there is a zero. 2. The following numbers represent the weights in pounds of six 7-year old children in Mrs. Jones' 2nd grade class. {25, 60, 51, 47, 49, 45} Find the mean; median; mode; variance; standard deviation. Solution: This would be a sample from the class mean = 46.166 (=AVERAGE) median = 48 (=MEDIAN) mode does not exist (looking at the data) variance = 134.5667 (=VARIANCE.S) standard deviation =11.60029 (=STDEV.S) 3. If the variance is 846, what is the standard deviation? Solution: standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data: 34, 38, 22, 21, 29, 37, 40, 41, 22, 20, 49, 47, 20, 31, 34, 66. Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 | 2 1 9 2 0 0 3 | 4 8 7 1 4 4 | 0 1 9 7 5 | 6 | 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. Find the regression equation of the following data. This study source was downloaded by from CourseH on :52:19 GMT -05:00 MATH 225 FINAL EXAM REVIEW(VERSION 1) DEVRY UNIVERSITY X 6 5 7 6 5 6 8 9 4 y Solution: y-hat = 1.32x + 24.32, found by using Data and then Data Analysis menu option. 6. To predict the annual rice yield in pounds we use the equation yˆ = 859 + 5.76x1 + 3.82x2 where x1 represents the number of acres planted (in thousands) and where 2x represents the number of acres harvested (in thousands) and where r2 = 0.94. a. Predict the annual yield when 3200 cares are planted and 3000 are harvested b. Interpret the results of r2 value. Solution: (a) yˆ= 859 + 5.76*3200 + 3.82*3000 = 859 + 18432 + 11460 = 30751 which is 30,751,000 pounds of rice (b) 94% of the variation in the annual rice yield can be explained by the number of acres planted and harvested. The remaining 6% is unexplained and is due to other factors or to chance. 7. The Student Services office did a survey of 500 students with the following results: Transfer Non-transfer Total Part-time Full-time Total a. Find the probabilities that a student is a transfer student. b. Find the probability that a students is part-time. c. Find the probability that a student is a transfer student and a part-time student d. Find the probability that a student is a transfer student, given that the student is part-time, P(transfer|part-time). Solution: (a) The total number of transfer students is 270. The total number of students in the survey is 500. P(Transfer) = 270/500 = .54 (b) The total number of part time students is 210. The total number of students in the survey is 500. P(Part Time) = 210/500 = .42 (c) From the table we see that there are 100 students which are both transfer and part time. This is out of 500 students in the sample. So, 100/500 = .20 (d) This is conditional probability and so we must change the denominator to the total of what has already happened. There are 100 students which are This study source was downloaded by from CourseH on :52:19 GMT -05:00 both transfer and part time. There are 210 part time students. P(transfer | part time) = 100/210 ≈ .4762. 8. A shipment of 40 television sets contains 3 defective units. How many ways can a vending company can buy five of these units and receive no defective units? Solution: There are 37 sets which are not defective. There are 37C5 ways to get 5 sets with none defective. 37C5 = 435, 897. In Excel, it would be =COMBIN(37,5). Thus, there are 435,897 ways to get 5 sets with nondefective. 9. A company claims that 61% of consumers know about their product. If the company asks 18 consumers, what is the probability that (a) exactly 10 will say yes, that (b) between 10 and 12 will say yes. Solution: For part A, it would be =BINOM.DIST(10,18,0.61,FALSE) to get 0.167 or 16.7%. For part B, it would be =BINOM.DIST(12,18,0.61,TRUE)- BINOM.DIST(10,18,0.61,TRUE) to get 0.3634 or about 36%. 10.The mean number of cars per minute going through the Eisenhower turnpike automatic toll is about 7. Find the probability that exactly 3 will go through in a given minute. Solution: =(3,7,FALSE) to get 0.0521 or about 5%. 11.On a dry surface, the braking distance (in meters) of a certain car is a normal distribution with mu = μ = 45.1 m and sigma = σ = 0.5 a. Find the braking distance that represents the 91st percentile. b. Find the probability that the braking distance is less than or equal to 45 m c. Find the probability that the braking distance is greater than 46.8 m d. Find the probability that the braking distance is between 45 m and 46.8 m. Solution: (a) =NORM.INV(0.91, 45.1, 0.5) to get 45.77m. (b) =NORM.DIST(45, 45.1 0.5, TRUE) to get 0.4207. (c) find the probability of less than 46.8 first and then subtract that probability from 1. So it would be =NORM.DIST(46.8, 45.1, 0.5, TRUE) which is 0.99966. More than 46.8 would be 1 = 0.99966 or 0.00034. (d) find the probability of each and subtract the larger from the smaller. In one cell, =NORM.DIST(46.8, 45.1, 0.5, TRUE) – NORM.DIST(45, 45.1, 0.5, TRUE) to get 0.579.