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6416 ECO - Dr. Richard Hofler - Exam 1 PRACTICE VERSION
NOTE: NEED STANDARD NORMAL TABLE FOR THIS EXAM.
REMINDER: A few students have been using electronic devices to cheat during exams. This is very unfair to the vast majority of honest students. In order to keep those few from gaining an unfair advantage, the following rule must be obeyed: NO ONE MAY USE ANY ELECTRONIC DEVICE DURING AN EXAM FOR ANY PURPOSE AT ALL. This includes, but is not limited to, cell phones, PDA's and programmable watches. The only exception to this rule is that a non-programmable calculator is allowed if needed. Any person caught violating this rule will automatically receive a zero for the relevant exam plus other penalties. TURN OFF YOUR CELL PHONE.
IMPORTANT NOTE ABOUT EXAM 1 Exam 1 will test your knowledge of material that you should have known before this class started. I will include questions on both i) material we covered in class and ii) material in the notes that we did not cover in class. Unless I told you otherwise about specific topics, you must know all of the material in the notes from the first two classes.
Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. ____ 1. In a cumulative frequency distribution, the last class will always have a cumulative frequency equal to a. one. b. 100%. c. the total number of elements in the data set. d. None of these answers is correct. Exhibit 2-1 The numbers of hours worked (per week) by 400 statistics students are shown below. Number of hours 0- 9 10 - 19 20 - 29 30 - 39 ____ Frequency 20 80 200 100
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2. Refer to Exhibit 2-1. The relative frequency of students working 9 hours or less a. is 20. b. is 100. c. is 0.95. d. 0.05. e. cannot be determined from the information given. 3. The sample variance a. is always smaller than the true value of the population variance b. is always larger than the true value of the population variance c. could be smaller, equal to, or larger than the true value of the population variance d. can never be zero e. both c and d are correct answers
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4. The variance of a sample of 81 observations equals 64. The standard deviation of the sample equals a. 0 b. 4096 c. 8 d. 6,561 e. None of these answers is correct. 5. The coefficient of correlation a. is the same as the covariance b. can be larger than 1 c. cannot be larger than 1 d. cannot be negative e. None of these answers is correct. 6. The range of probability is a. any value larger than zero b. any value between minus infinity to plus infinity c. zero to one d. any value between -1 to 1 e. None of these answers is correct. 7. An experiment consists of making 80 telephone calls in order to sell a particular insurance policy. The random variable in this experiment is the number of sales made. This random variable is a a. discrete random variable b. continuous random variable c. complex random variable d. None of these answers is correct. 8. A description of the distribution of the values of a random variable and their associated probabilities is called a a. probability distribution b. random variance c. random variable d. expected value e. None of these answers is correct. 9. Variance is a. a measure of the average, or central value of a random variable b. a measure of the dispersion of a random variable c. the square root of the standard deviation d. the sum of the deviation of data elements from the mean e. None of these answers is correct. Exhibit 5-4 A local bottling company has determined the number of machine breakdowns per month and their respective probabilities as shown below. Number of Breakdowns 0 1 2 3 4
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Probability 0.12 0.38 0.25 0.18 0.07
____ 10. Refer to Exhibit 5-4. The probability of at least 3 breakdowns in a month is a. 0.5 b. 0.10 c. 0.30 d. 0.90 e. None of these answers is correct. ____ 11. If you are conducting an experiment where the probability of a success is .02 and you are interested in the probability of 4 successes in 15 trials, the correct probability function to use is the a. standard normal probability density function b. normal probability density function c. Poisson probability function d. binomial probability function e. None of these answers is correct. ____ 12. In the textile industry, a manufacturer is interested in the number of blemishes or flaws occurring in 100 each feet of material. The probability distribution that has the greatest chance of applying to this situation is the a. normal distribution b. binomial distribution c. Poisson distribution d. uniform distribution e. None of these answers is correct. ____ 13. Which of the following is not a characteristic of the normal probability distribution? a. symmetry b. The total area under the curve is always equal to 1. c. 99.72% of the time the random variable assumes a value within plus or minus 1 standard deviation of its mean d. The mean is equal to the median, which is also equal to the mode. e. None of these answers is correct. Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. ____ 14. Refer to Exhibit 6-3. What percent of players weigh between 180 and 220 pounds? a. 34.13% b. .6826% c. 0.3413% d. 68.24% e. None of these answers is correct. Exhibit 6-5 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. ____ 15. Refer to Exhibit 6-5. What is the probability that a randomly selected item will weigh more than 10 ounces? a. 0.3413 b. 0.8413 c. 0.1587 d. 0.5000 e. None of these answers is correct.
Practice Exam 1 PROBLEMS
POINT VALUES: 2 PER QUESTION
Write your answer in the space provided on this page. 1. A manufacturing company has 5 identical machines that produce nails. The probability that a machine will break down on any given day is 0.1. Define a random variable x to be the number of machines that will break down in a day. Compute the probability that 3 machines will break down.
2. The average number of calls received by a switchboard in a 1-minute period is 7. What is the probability that between 10:00 and 10:01 the switchboard will receive exactly 5 calls?
3.
The length of time patients must wait to see a doctor in a local clinic is uniformly distributed between 15 minutes and 1 hours. Determine the expected waiting time.
4. The contents of soft drink containers are normally distributed with a mean of 16 ounces and a standard deviation of one ounce. What is the probability that a randomly selected container will contain more than 15 ounces of soft drink?
Practice Exam 1 PROBLEMS
POINT VALUES: 2 PER QUESTION ANSWERS 1. f ( x) =
0.0081. Use binomial probability function with n=5, p=0.1 & x = 3.
5! n! 0.13 (1 0.1)(5 3) p x (1 p ) ( n x ) = f (3) = 3!(5 3)! x!(n x)!
= (10) (.001) (.81) = 0.0081 (using calculator) 2. f ( x) =
0.128. Use the formula for Poisson probability function where = 7 and x = 5.
= f (5) =
75 e 7 (16,807)(.000912) = = 0.1277 (NOTE: if use calculator and store the result of 5! 120 x! exp(-7), will get 0.1277. However, = 0.126 if use only 4 digits displayed by calculator)
x e
3. 4.
45 minutes; E(x) = (75 +15)/2 = 45 .8413. 15 is < the mean, so the answer will be 50% plus the area from 16 to 15. 15 is 1 std. deviation below the mean. Using the table, that corresponds to an area of .1587 below z= -1, which means that there is .5000 - .1587 = 0.3413 from z = -1 to the mean. So, the total area is .5000 + .3413 = .8413.
ECO 6416 - Dr. Richard Hofler - Exam 1 PRACTICE VERSION Answer Section
MULTIPLE CHOICE 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. C D C C C C A A B E D C C E C
ECO 6416 - Dr. Richard Hofler - Exam 1 PRACTICE VERSION
Answers to Selected Multiple Choice Questions
14. .5762 or 57.62% of the players will weigh between 180 and 220 pounds. 180 is < the mean and 220 is > mean. Each value is the same distance away from the mean, so do the calculation for one of the two values and double your result to answer the question. So, (220-200)/25 = .8. The table displays a value of .2119 for s = .8. This means that .2119 of the area to the right of the mean is above s = .8. It also means that .5000 .2119 = .2881 of the area is between the mean and 220. So, double this to find that 2 x .2881 = .5762 or 57.62% of the players will weigh between 180 and 220 pounds. 15. The prob. that an item will weigh more than 10 oz. is .1587 or 15.87%. 10 is > mean. So, (10-8)/2 = 1.0. The table shows a value of .1587 for s = 1.0. This means that .1587 of the area to the right of the mean is above s = 1 (or a weight of 10 oz.) So, the prob. that an item will weigh more than 10 oz. is .1587 or 15.87%.