Ch 2 Test Remediation Work Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 60 F to 85 F. What is the probability that a randomly selected August day has a high temperature that exceeded 65 F? A) 0.04 B) 0.8 C) 0.4483 D) 0.2 2) High temperatures in a certain city for the month of August follow a uniform distribution over the interval 73 F to 103 F. Find the high temperature which 90% of the August days exceed. A) 100 F B) 103 F C) 83 F D) 76 F 3) A machine is set to pump cleanser into a process at the rate of 10 gallons per minute. Upon inspection, it is learned that the machine actually pumps cleanser at a rate described by the uniform distribution over the interval 9.5 to 12.5 gallons per minute. Find the probability that between 10.0 gallons and 11.0 gallons are pumped during a randomly selected minute. A) 0.33 B) 0.67 C) 1 D) 0 1) 2) 3) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 4) You are performing a study about the weight of preschoolers. A previous study found the weights to be normally distributed with a mean of 30 and a standard deviation of 4. You randomly sample 30 preschool children and find their weights to be as follows. 25 25 26 26.5 27 27 27.5 28 28 28.5 29 29 30 30 30.5 31 31 32 32.5 32.5 33 33 34 34.5 35 35 37 37 38 38 a) Draw a histogram to display the data. Is it reasonable to assume that the weights are normally distributed? Why? b) Find the mean and standard deviation of your sample. c) Is there a high probability that the mean and standard deviation of your sample are consistent with those found in previous studies? Explain your reasoning. 4) 1
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 5) The graph of a normal curve is given. Use the graph to identify the value of µ and. 5) 753 785 817 849 881 913 945 A) µ = 32, = 849 B) µ = 96, = 849 C) µ = 849, = 32 D) µ = 849, = 96 6) The normal density curve is symmetric about 6) A) Its mean B) A point located one standard deviation from the mean C) An inflection point D) The horizontal axis 7) The area under a standard normal density curve with mean of 0 and standard deviation of 1 is 7) A) 1 B) µ + 2(3 ) C) µ + 3 D) infinite 8) The highest point on the graph of the normal density curve is located at 8) A) µ + B) its mean C) an inflection point D) µ + 3 9) Approximately % of the area under the normal curve is between µ - and µ +. 9) A) 95 B) 68 C) 99.7 D) 50 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 10) The weight of 2-year old hyraxes is known to be normally distributed with mean µ = 2200 grams and standard deviation = 365 grams (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of hyraxes who weighed more than 2930 grams. (c) Suppose the area under the normal curve to the left of X = 2930 is 0.0228. Provide two interpretations of this result. 10) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 11) Find the area under the standard normal curve between z = 0 and z = 3. 11) A) 0.0010 B) 0.9987 C) 0.4641 D) 0.4987 12) Find the area under the standard normal curve between z = 1 and z = 2. 12) A) 0.5398 B) 0.8413 C) 0.1359 D) 0.2139 2
13) Find the area under the standard normal curve between z = -1.25 and z = 1.25. 13) A) 0.6412 B) 0.8817 C) 0.2112 D) 0.7888 14) For a standard normal curve, find the z-score that separates the bottom 30% from the top 70%. 14) A) -0.47 B) -0.53 C) -0.98 D) -0.12 15) Find the z-score that is less than the mean and for which 70% of the distribution's area lies to its right. A) -0.47 B) -0.98 C) -0.81 D) -0.53 15) 16) For a standard normal curve, find the z-score that separates the bottom 90% from the top 10%. 16) A) 1.28 B) 2.81 C) 1.52 D) 0.28 17) Scores on a standardized test are normally distributed with a mean of 96 and a standard deviation of 12. An individual's test score is found to be 128. Find the z-score corresponding to this value. A) -0.37 B) 0.38 C) -2.67 D) 2.67 17) 18) Use the standard normal distribution to find P(-2.25 < Z < 0). 18) A) 0.0122 B) 0.4878 C) 0.6831 D) 0.5122 19) Use the standard normal distribution to find P(Z < -2.33 or Z > 2.33). 19) A) 0.7888 B) 0.0198 C) 0.9809 D) 0.0606 Determine the area under the standard normal curve that lies a) above and b) below the given z-score. 20) z = -0.2 20) A) 0.4207, 0.4207 B) 0.5793, 0.5793 C) 0.5793, 0.4207 D) 0.4207, 0.5793 Determine the area under the standard normal curve that lies between: 21) z = 0.9 and z = 1.4 21) A) 0.1033 B) 0.9192 C) 0.1841 D) 0.8159 22) z = -2 and z = -0.2 22) A) 0.0228 B) 0.5793 C) 0.4207 D) 0.3979 Provide an appropriate response. 23) A physical fitness association is including the mile run in its secondary-school fitness test. The time of 470 seconds and a standard deviation of 50 seconds. Find the probability that a randomly selected boy in secondary school will take longer than 355 seconds to run the mile. A) 0.4893 B) 0.0107 C) 0.9893 D) 0.5107 24) A physical fitness association is including the mile run in its secondary-school fitness test. The time of 440 seconds and a standard deviation of 60 seconds. Find the probability that a randomly selected boy in secondary school can run the mile in less than 302 seconds. A) 0.4893 B) 0.5107 C) 0.9893 D) 0.0107 23) 24) 3
25) The tread life of a particular brand of tire is a random variable best described by a normal distribution with a mean of 60,000 miles and a standard deviation of 1400 miles. What is the probability a certain tire of this brand will last between 57,060 miles and 57,480 miles? A) 0.4920 B) 0.0180 C) 0.4649 D) 0.9813 26) A physical fitness association is including the mile run in its secondary-school fitness test. The time of 440 seconds and a standard deviation of 60 seconds. Between what times do we expect most (approximately 95%) of the boys to run the mile? A) between 345 and 535 sec B) between 341.3 and 538.736 sec C) between 0 and 538.736 sec D) between 322.4 and 557.6 sec 27) The length of time it takes college students to find a parking spot in the library parking lot follows a normal distribution with a mean of 4.5 minutes and a standard deviation of 1 minute. Find the cut-off time which 75.8% of the college students exceed when trying to find a parking spot in the library parking lot. A) 5.3 min B) 5.0 min C) 5.2 min D) 4.8 min 25) 26) 27) SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. 28) The board of examiners that administers the real estate broker's examination in a certain state found that the mean score on the test was 504 and the standard deviation was 72. If the board wants to set the passing score so that only the best 80% of all applicants pass, what is the passing score? Assume that the scores are normally distributed. 28) MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 29) A physical fitness association is including the mile run in its secondary-school fitness test. The time of 450 seconds and a standard deviation of 50 seconds. The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition. What time would the boys need to beat in order to earn a certificate of recognition from the fitness association? A) 386 sec B) 367.75 sec C) 532.25 sec D) 514 sec 29) 4
Answer Key Testname: CH 2 TEST REMEDIATION WORK 1) B 2) D 3) A 4) (a) It is not reasonable to assume that the heights are normally distributed since the histogram is skewed. (b) µ = 31, = 3.86 (c) Yes. The mean and standard deviation are close. 5) C 6) A 7) A 8) B 9) B 10) (a), (b) (c) The two interpretations are: (1) the proportion of hyraxes who weighed more than 2930 is 0.0228 and (2) the probability that a randomly selected hyrax weighs more than 2930 is 0.0228. 11) D 12) C 13) D 14) B 15) D 5
Answer Key Testname: CH 2 TEST REMEDIATION WORK 16) A 17) D 18) B 19) B 20) C 21) A 22) D 23) C 24) D 25) B 26) D 27) C 28) Let x be a score on this exam. Then x is a normally distributed random variable with µ = 504 and = 72. We want to find the value of x0, such that P(x > x0) = 0.80. The z-score for the value x = x0 is 29) A z = x 0 - µ = x 0-504. 72 P(x > x0) = P z > x 2-504 72 We find x 0-504 72-0.84. = 0.80 x0-504 = -0.84(72) x0 = 504-0.84(72) = 443.52 6