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Name: ____________________ Period: _____ CHAPTER 6 REVIEW 1. A continuous random variable is a random variable that can assume: a. only countable values c. any value in one or more intervals b. only discrete frequencies d. no continuous random frequency 2. For a continuous random variable x, the probability that x assumes a value in an interval is: a. in the range zero to 1 b. greater than 1 c. less than zero d. greater than 2 3. For a continuous random variable x, the area under the probability distribution curve between any two points is always: a. greater than 1 b. less than zero c. equal to 1 d. in the range 0 and 1 4. For a continuous random variable x, the total probability of all (mutually exclusive) intervals within which x can assume a value is: a. less than 1 b. greater than 1 c. equal to 1 d. between 0 and 1 5. The total area under the probability distribution curve of a continuous random variable x is: a. less than 1 b. greater than 1 c. equal to 1 d. between 0 and 1 6. The probability that a continuous random variable x assumes a single value is always: a. less than 1 b. greater than zero c. equal to zero d. between zero and 1 7. Which of the following is not a characteristic of the normal distribution? a. The total area under the curve is 1.0 c. The curve is symmetric about the mean b. The value of the mean is 1.0 d. The two tails of the curve extend indefinitely 8. The total area under a normal distribution curve to the left of the mean is always: a. equal to 1 b. equal to zero c. equal to .5 d. greater than .5 9. For the standard normal distribution the mean is____ and the standard deviation is___. a. 1; zero b. .5;.5 c. zero; 1 d. 1; 1 10. For a normal distribution, the z value for an x value that is to the left of the mean is always: a. equal to zero b. negative c. less than 1 d. positive 11. P(0 < z < 1.53) = 12. P( z > -2.18) = 13. The random variable X is normally distributed with mean 80 and standard deviation 12. a) What is the probability that a value of X chosen at random will be between 65 and 95? b) What is the probability that a value of X chosen at random will be less than 74? 14. The scores on a placement test have a bell-shaped distribution with mean 400 and standard deviation 45. a) What percentage of people taking this exam will have scores of 310 or greater? b) What percentage of the people taking this test will have scores between 445 and 490? 15. Suppose X is the cost per gallon of gas at a pump anywhere in the U.S., and X is normally distributed with mean 2.25 and standard deviation 0.2. If you fill up at a random gas pump, what is the probability that the gas is less than $2.36? 16. In the same situation as the previous example, let X = price per gallon of gas at a random pump. Mean = 2.25, SD = 0.20. Find the gas price where 58.71% of gas prices fall below the value. 17. The _____________________________ is the distribution of all values of the statistic when all possible samples of size n are taken from the population. It is usually displayed as a table, probability histogram, or formula. 18. The ______________________is the distribution of all values of the sample proportions when all possible samples of size n are taken from the population. 19. The ________________________is the distribution of all values of the sample mean when all possible samples of size n are taken from the population. 20. Which Estimators Target the Population Parameter? 21. Central Limit Theorem (using samples):Let X represent the height of a U.S. resident. X is normally distributed, with a population mean of 71 inches and standard deviation of 5 inches. Suppose we take a sample of size n = 64. Find the probability that the sample mean will be larger than 72. 22. When throwing three darts, the average score is 45 with a standard deviation of 11.5. What is the mean and standard deviation for the sampling distribution of the mean for the average score when each of the 40 students in class throws three darts? 23. What is the probability that the average score of the forty students in class will be less than 42 when throwing three darts? Mean: 42 S.D=11.5 24. 6.6- Normal Approximation to Binomial Distribution: The probability of an airline flight arriving on time is 90%. Use the normal approximation to find the probability that at least 300 of a random sample of 350 flights willarrive on time. 25. It is estimated that 10% of the vehicles entering Canada from the United States carry undeclared goods. Use the normal approximation to calculate the probability that a search of 500 randomly selected vehicles will find more than 60 with undeclared goods. CHAPTER 6 TEST 1. C 2. A ANSWER KEY STATISTICS 3. D 4. C 5. C 6. B 7. C 8. C 9. C 10. B Answers 11) 1.53 P(0 < z < 1.53) = Normalcdf(0,1.53)= .4370 12) -2.18 P( z > -2.18) Normalcdf(-2.18, 1e^99)= .9854 13) a) Find the z-scores for 65 and 95, and use those z-scores as your lower and upper limit when computing the probability using normal cdf( lower limit (find z score of 65), upper limit (find z-score of 95))=.7888 b) Same as above accept when computing the probability, use normalCDF(1e^99,(find z-score of 74))=.3085 14) a) NormalCDF((find z-score of 310), 1 e^99)= .9772 97.72 % of the people taking this test will have scores of 310 or greater. b) NormalCDF((find z-score of 445), (find z-score of 490) = .1359 13.59 % of the people taking this test will have scores between 445 and 490. 15. We want P(X < 2.36), but we only have a table for the standard normal random variable Z. Thus we have to compute the z-score: (x-u)/standard dev.= .55; then we compute normal(-1 e^99, .55) <~ this is because the graph is shaded to the left (“less than”) = .7088 16. To find an x value, you have to use the following: InvNorm(Prob to the left, mean, standard deviation); InvNorm(.5871, 2.25, .2)= $2.11 17) sampling distribution of a statistic 18)sampling distribution of the proportion 19)sampling distribution of the mean 20) 1. The sample mean targets the population mean. 2. The sample variance targets the population variance. 1. The sample median does not target the population median. 2. The sample standard deviation does not target the population standard deviation. 3. The sample range does not target the population range. 21) 6.5: Central Limit Theorem- Where the z-score is found by: (x-u)/(standard deviation/(square root(n)))= 1.6 Normalcdf(1.6, 1e^99)=.0548 22) Again, this is a Central Limit Theorem question ,where you are finding values from a sample. You must note that the mean will still reflect the population mean, however it is the standard deviation that changes. The standard deviation for a sample is: original population standard deviation/square root (n). So in this case the mean remains 45, whereas the standard deviation is (11.5/sq.rt(40))=1.8183 23) z score= (42-45)/(11.5/sq.rt((40))= -1.65 NormalCDF(-1e^99, -1.65)=.0496 24) Here, n = 350, p = 0.90, and q = 0.10. u= 315 Standard Deviation= (Square root(350*.90*.10)= 5.6 Continuity correction tells us to use 299.5 instead of 300 bc it states at least (refer to chart in your notes, section 6.6), so our x-value we will use to find its correlating z-score is 299.5. The Z-Score= 299.5-315/(5.6)= -2.77 To find the probability shaded to the right (note: at least), compute the following: normalCDF(-2.77, 1e^99)= .9971 25) n = 500, p = 0.10, and q = 0.90. Mean= 500*.1= 50 Standard Deviation= (Square root(500*.90*.10)= 45 Continuity correction tells us to use 60.5 instead of 60 bc it states more than (refer to chart in your notes, section 6.6), so our x-value we will use to find its correlating z-score is 60.5. The Z-Score= 60.5-50/(45)= .23 To find the probability shaded to the right (note: at least), compute the following: normalCDF(.23, 1e^99)= .409