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31-2 A probability function is a rule of correspondence

31-3 What is a synonym or example of a variable?

38-3 Define the following term

45-1 "Which of the following ""probability mass functions"" "

56-1 Find P(T <= 0)

63-1 The random numbers generator

119-2 Approximately how many A's were there

120-1 A standard normal distribution has

121-2 What proportion of samples would have 4800 cells

131-2 find K so that the probability that a sample value is

135-1 variance = 0.000052. What is the probability

136-3 "normal variable, then the area to the left "

137-1 What area under the standard normal curve falls outside

137-2 "and a variance of 9, what percent of the population"

138-1 then P(Z > -0.38) is

138-2 then P(XBAR > 15) is

139-1 Find the 33rd percentile of the distribution of lifetimes

141-1 A Z-Score is a

141-3 Z-scores provide information about the location of

142-1 standard deviation = 10. Her Z-score is

142-2 Charlie's raw score on the test is

143-1 below approximately what percent of the students taking the test

144-1 better than what percent of the persons taking the test

144-2 will have a standard deviation equal to

145-2 A score of 85 from this population has a Z-score

146-2 The Z-score corresponding to the 52nd percentile is

147-1 number of students scoring between 70 and 82 is

149-1 Pr(-.25 < Z) is

149-2 what score will be associated with a standard Z score of 1.5

150-1 Pr(Z <= 1.65 or Z > 3.0) is

150-2 Pr(Z > +1.96 or Z < -1.65) is

155-2 then P(69.5 <= X <= 75)

166-1 of cases are likely to be between 86 and 93 in a normal

166-2 percentile rank does a score of 42 fall

167-1 What score has a percentile rank of 33%

167-2 scores for the middle 50% of the data

168-1 What percentage of the scores are above 78?

168-2 has an X value equal to

169-2 In a frequency distribution with a median of 50

170-1 21% of the observations lie below it?

171-1 what is the percentage of scores likely to fall below 550?

172-2 and variance 4 will fall between -9 and -4?

173-1 The probability that a value between 7 and 9 is obtained is

176-1 The standard normal score Z is:

176-2 what percentage of his students will earn an A?

177-1 Rods are too long to be useable

177-2 "are too short, what is the cut off length between ""too"

185-1 how much time should we give them?

189-1 "600 yard run-walk represent a normal distribution, how"

190-1 the monthly food expenditures of families

193-2 guarantees the battery to last 30 months.

200-1 The average weekly food expenditure was $70.00

202-1 a floor manager of a large department store

206-2 a floor manager of a large department store is studying

207-1 A floor manager of a large department store is studying the buying

213-1 If 91% of the bike-commuters take longer to reach campus than you

213-2 P(-1 < X < 5) = _______________

529-2 Which of the following random variables are continuous

530-3 What is the principal distinction between a discrete

531-2 The number of individuals in a family is a continuous variable.

531-3 Variables in which measurement is always approximate because

532-2 A continuous variable:

743-1 The breaking strength of a cable is a discrete variable.

1776-4 "If each score is raised by 7 points, what percentage"

1859-3 "If the mean depth of a river is 2 feet, it would be safe "

1860-1 The mean number of children per family in LA is quoted as 2.2.

2035-1 What percent of the area of a distribution lies between the first and

2054-1 The 70th percentile of the distribution of a random variable X

2054-2 Not more than 10% of a set of measurements can be above the 95th

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Q: A probability function is a rule of correspondence or equation that: a) Finds the mean value of the random variable. b) Assigns values of x to the events of a probability experiment. c) Assigns probablities to the various values of x. d) Defines the variability in the experiment. e) None of the above is correct.

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Q: What is a synonym or example of a variable? a) constant b) characteristic which takes on different values c) number of ears on humans d) parameter

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Q: Define the following term and give an example of its use. Your example should not be one given in class or in a handout. Constant

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Q: A sample space consists of the following elementary events: - the traffic-light is red - the traffic-light is orange - the traffic-light is green Upon this a random variable X is defined as follows: red = 1 orange = 2 green = 3 Which of the following "probability mass functions" of the discrete random variable X is absolutely wrong? a. 5/6 | b. 5/6 | | | 4/6 | 4/6 | * * p(X) | p(X) | * * 3/6 | 3/6 | * * | | * * 2/6 | * * * 2/6 | * * | * * * | * * 1/6 | * * * 1/6 | * * * | * * * | * * * ---+--+--+------> X ---+--+--+------> X 1 2 3 1 2 3 c. 5/7 | d. 5/7 | * | | * 4/7 | 4/7 | * p(X) | p(X) | * 3/7 | * * 3/7 | * | * * | * 2/7 | * * 2/7 | * | * * | * 1/7 | * * * 1/7 | * * * | * * * | * * * ---+--+--+------> X ---+--+--+------> X 1 2 3 1 2 3

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Q: Suppose that the random variable T has the following probability distribution: t | 0 1 2 ---------------------- P(T = t) | .5 .3 .2 a. Find P(T <= 0) b. Find P(T >= 0 and T < 2) c. Compute E(T), the mean of the random variable T.

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Q: The random numbers generator of a computer produces values that are uniformly distributed from zero to one. A programmer doesn't want his program to print the same message everytime that a user reaches a certain point in the program. He wants the program to print: Hooray] 20% of the time, Yep. 40% of the time, Bullseye] 5% of the time, and Fantastic] 35% of the time. He can do this by including in the program an instruction that tells the program to do different things depending on the random number generated. a. Sketch the distribution of random numbers and indicate areas and boundary values so that the 4 comments will appear as desired.

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Q: One hundred students took a test on which the mean score was 73 with a variance of 64. A grade of A was given to all who scored 85 or better. Approximately how many A's were there, assuming scores were normally distributed? (Choose the closest.) 1. 42 2. 7 3. 58 4. 5 5. 22

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Q: A standard normal distribution has: a. the mean equal to the variance b. mean equal 1 and variance equal 1 c. mean equal 0 and variance equal 1 d. mean equal 0 and standard deviation equal 0 e. none of these

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Q: Use the model that the number of cells in a sample of kidney tissue is normally distributed with a mean of 4200 and a standard deviation of 300 to answer the following questions: (a) What proportion of samples would have 4800 cells or more? (b) What proportion of samples would have from 3700 to 4400 cells?

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Q: If a normal distribution has mean 200 and standard deviation 20, find K so that the probability that a sample value is less than K is .975. a. 239 b. 204 c. 210 d. 215 e. 220 E. 220 F. 230 G. 239 H. 250

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Q: The thickness of the individual cards produced by a certain playing card manufacturer is normally distributed with mean = 0.01 inches and variance = 0.000052. What is the probability that a deck of 52 cards is more than 0.65 inches in thickness? A. .001 B. .006 C. .023 D. .036 E. .067 F. .087 G. .159 H. .184

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Q: If Z is a standard normal variable, then the area to the left of Z = 0.65 is: a. 0.35 d. 0.2578 b. 0.2242 e. 0.7422 c. 0.65

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Q: What area under the standard normal curve falls outside the Z values -2.5 and 2.5? a. .0062 b. .9876 c. .0124 d. .4938 e. .5062

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Q: If the life of wild pheasants follows a normal distribution with a mean of 9 months and a variance of 9, what percent of the population will be less than 11 months of age? (Note that MU = 9 and SIGMA(X)**2 = 9.) (a) 34.13 (c) 74.86 (b) 84.13 (d) 62.93

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Q: If Z is the standard normal random variable, then P(Z > -0.38) is (a) .1480 (c) .3520 (b) .2960 (d) .6480

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Q: If XBAR is the mean of a sample from a normal distribution with MU = 10, SIGMA(X)**2 = 25 and n = 9, then P(XBAR > 15) is: (a) .001350 (c) .98778 (b) .998650 (d) .15866

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Q: The distribution of lifetimes for a certain type of light bulb is normally distributed with a mean of 1000 hours and a standard deviation of 100 hours. Find the 33rd percentile of the distribution of lifetimes. a. 560 b. 330 c. 1044 d. 1440 e. none of these

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Q: A Z-Score is a a. raw score with a mean of zero; b. raw score with a mean of 50; c. standard score with a mean of zero; d. standard score with a mean of 50.

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Q: Z-scores provide information about the location of raw scores a. below the mean in units of the range of the distribution; b. above the mean in units of the standard deviation of the distribution; c. above and below the mean in units of the range of the distribution; d. above and below the mean in standard deviation units from the mean.

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Q: Mary has a raw score of 40 in a distribution of scores with mean = 30, range = 60, and standard deviation = 10. Her Z-score is: a) -1.00 d) +1.00 b) -0.67 e) +10.00 c) +0.67

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Q: Charlie's Z-score is 1.15 on a classroom examination. The mean score for the class is 50, the range is 25, and the standard de- viation is 10. Charlie's raw score on the test is: a) 11.15 c) 61.50 b) 51.15 d) 77.75

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Q: A person with a Z-Score of -2.00 has performed below approximately what percent of the students taking the test? a) 2 percent d) 84 percent b) 15 percent e) 97 percent c) 50 percent

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Q: Suppose you were told that scores on an examination were converted to standard scores with a mean = 500, range of 800, and a standard deviation of 100. A person with a score of 600 has performed bet- ter than what percent of the persons taking the test? a) 20 percent d) 84 percent b) 50 percent e) 97.5 percent c) 57 percent

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Q: If each of a set of raw scores is transformed into a Z-score, the new distribution will have a standard deviation equal to a. zero. b. one. c. the mean of the original distribution. d. the standard deviation of the original distribution. e. a variable, depending upon the shape and spread of the original distribution.

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Q: In a population there are 60 scores; the distribution has a mean of 45 and a standard deviation of 25. A score of 85 from this population: a. has a Z-score of 1.60. b. has a Z-score of 1.00. c. has a Z-score of -1.00. d. it is impossible to compute Z without additional information e. none of the above

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Q: The Z-score corresponding to the 52nd percentile is: a. 2.06 b. 2.05 c. 1.99 d. 0.48 e. 0.05

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Q: Assume that the test scores of 600 students are normally distri- buted with a mean of 76 and a standard deviation of 8. The number of students scoring between 70 and 82 is: a. 328 b. 164 c. 260 d. 136 e. 272

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Q: Pr(-.25 < Z) is a. greater than .5000 b. less than .5000 c. equal to .5000 d. not possible to determine without more information.

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Q: Given that a distribution has a mean of 32 and a standard deviation of 4, what score will be associated with a standard Z score of 1.5? a) 26 b) 32 c) 38 d) 40

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Q: Pr(Z <= 1.65 or Z > 3.0) is 1) 0.0508 2) 0.9518 3) 0.9482 4) 0.0482 5) None of the above

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Q: Pr(Z > +1.96 or Z < -1.65) is 1) 0.025 2) 0.05 3) 0.0745 4) 0.0495 5) None of the above

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Q: The height of male college freshmen has a normal distribution with mean 71 inches and standard deviation 3 inches. If X is the height of a male college freshman selected at random, then P(69.5 <= X <= 75) = a. .5997 b. .6915 c. .2167 d. .9082 e. none of these

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Q: What percent of cases are likely to be between 86 and 93 in a normal distribution with mean 87 and variance 4? a. 30.85% d. 69.02% b. 30.72% e. none of these c. 49.87%

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Q: In a normal distribution with mean 30 and variance 25, at what percentile rank does a score of 42 fall? a. .82% b. 49.18% c. 50.82% d. 99.18% e. none of these

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Q: Suppose a set of data has a normal distribution with mean 43 and variance 9. What score has a percentile rank of 33%? a. 44.32 d. 41.68 b. 39.04 e. none of these c. 41.47

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Q: In a normal distribution with mean 3 and variance 49, what are the upper and lower limit scores for the middle 50% of the data? a. -29.83 and 35.83 b. -1.31 and 7.69 c. -1.69 and 7.69 d. 3.00 and 24.00 e. none of these

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Q: Consider a normal distribution with a mean of 86 and a standard deviation of 16. What percentage of the scores are above 78? a. 69.15% b. 2.28% c. 97.72% d. 77.34% e. none of these

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Q: The 67th percentile of a normal distribution with mean 6 and variance 9 has an X value equal to: a. 12.66 d. 8.22 b. 6.75 e. none of these c. 8.24

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Q: In a frequency distribution with a median of 50 and a standard deviation of 4, what score corresponds to a standard score of 1.0? a. 12.5 d. cannot be determined without additional information b. 54 e. none of the above is true c. 46

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Q: A normal distribution has mean 10 and variance 100. What is the number such that 21% of the observations lie below it? a. 11.9 d. 9.193 b. 1.9 e. none of these c. 4.4

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Q: If certain scores are distributed normally with mean 500 and variance 625, what is the percentage of scores likely to fall below 550? (1) 97.72% (4) 65.54% (2) 84.13% (5) none of these (3) 47.72%

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Q: What proportion of cases in a normal distribution with mean -7 and variance 4 will fall between -9 and -4? a. .9759 d. .7745 b. .4649 e. none of these c. .5919

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Q: A normally distributed variable has a mean of 10 and a standard devia- tion of 2. The probability that a value between 7 and 9 is obtained is: a. .6247 d. .0668 b. .3085 e. none of these c. .2417

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Q: The standard normal score Z is: a) Normally distributed with a mean of zero and with a standard deviation of one. b) Calculated by the formula Z = [X - MU]/[SIGMA]. c) Used to find the probabilities associated with any normal distribution. d) All of the above are correct. e) None of the above are correct.

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Q: A physical education instructor told his class that they could earn an A for the triple-jump if they could jump further than 24 feet. If the distances jumped by students are normally distributed with a mean of 22 feet and a standard deviation of 3 feet, what percentage of his students will earn an A? a) 0.0228 b) 0.2486 c) 0.2514 d) 0.4772 e) None of the above are correct.

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Q: Rods produced by G&R Company are normally distributed with a mean of 66 cm. and a standard deviation of 2 cm. Rods are too long to be useable if they are longer than 68.5 cm. What percentage of these rods are too long? a) 0.1056 b) 0.1151 c) 0.3849 d) 0.3944 e) None of the above are correct.

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Q: Rods produced by G&R Company are normally distributed with a mean of 66 cm. and a standard deviation of 2 cm. If the shortest 4 percent are too short, what is the cut off length between "too short" and "acceptable length"? a) 62.5 b) 63.36 c) 64.25 d) 65.96 e) None of the above are correct.

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Q: The average time students need to finish a particular test is 70 minutes with a standard deviation of 12 minutes. (Assume that these times are normally distributed.) If we want 90% of the students to have suffi- cient time to finish the test, how much time should we give them? a. 54.64 minutes b. 85.36 minutes c. 136.48 minutes d. 254.32 minutes

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Q: If the times recorded for a group of 150 high school students measured on the 600 yard run-walk represent a normal distribution, how would you answer the following: Given: XBAR = 2 minutes SIGMA = 12 seconds n = 150 a. time for the best student? b. time for the worst student? c. number of students worse than 2 minutes, 15 seconds? d. number of students better than 1 minute, 42 seconds? e. what is the mode? f. what is the median?

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Q: Assuming that the monthly food expenditures of families of a certain size in one economic group are approximately normally distributed with a mean of $130 and a standard deviation of $20: a. What proportion of the expenditures are less than $90? b. What percentage of the expenditures are between $100 and $120? c. What percentage of the expenditures are either less than $120 or more than $150? d. Above what value does the top 14 percent of the expenditures lie?

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Q: Suppose the length of life of certain kinds of batteries is normally distributed with MU = 36 months, SIGMA = 4 months. The company guar- antees the battery to last 30 months. What proportion of the batter- ies will they have to make an adjustment on?

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Q: The U.S. Department of Commerce has just completed a sample survey of weekly food expenditures. A simple random sample of 100 families was taken. The average weekly food expenditure was $70.00 per week, with a standard deviation of $8.00. You may assume expenditures in the population to be normally distributed. a. What proportion of the families spent $85.00 or more per week on food? Be sure to diagram your problem solution] b. Using the information above, find the expenditure value above which 80% of the families lie.

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Q: Suppose a floor manager of a large department store is studying buying habits of their customers. a) If he is willing to assume that monthly income of these customers is distributed normally, what proportion of the income should he expect to fall in the interval determined by MU +/- 1.2(SIGMA)? b) What proportion of the income should he expect to be greater than MU + SIGMA? c) Still assuming normality, what is the probability that a customer selected at random will have an income exceeding the population mean by 3*SIGMA?

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Q: Suppose a floor manager of a large department store is studying the buying habits of the store's customers. a) If he is willing to assume that monthly income of these customers is distributed normally and SIGMA = $500, find the proportion of customers exceeding the population mean by $375. b) Find the proportion of customers within $125 of the population mean.

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Q: A floor manager of a large department store is studying the buying habits of the store's customers. Suppose the manager has someone tell him that monthly income of these customers is distributed nor- mally with a population mean of $600 and standard deviation of $500. a) What proportion of the customers should he expect to have incomes less than $600? b) What proportion should he expect to have incomes less than $725?

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Q: Assume that commuting time via bicycle to the campus is a normal random variable with mean MU = 8 minutes and standard deviation SIGMA = 2 minutes. If 91% of the bike-commuters take longer to reach campus than you do, then your commuting time is approximately: (a) 5.32 (d) 10.68 (b) 5.71 (e) 4.88 (c) 7.54

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Q: If X is a random variable from a normal distribution with mean = 2.0 and variance = 4.0 then P(-1 < X < 5) = _______________.

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Q: Which of the following random variables are continuous and which are discrete? a. IQ b. number of kittens in a litter c. number of responses made by a rat in a bar-pressing situation d. the rate of bar pressing (responses/time) 1. a, b continuous; c, d discrete 2. a, c, d continuous; b discrete 3. d continuous; a, b, c discrete 4. a continuous; b, c, d discrete 5. none of these.

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Q: What is the principal distinction between a discrete and continuous random variable? Give an example of each.

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Q: True or False? If False, correct it. The number of individuals in a family is a continuous variable.

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Q: Variables in which measurement is always approximate because they permit an unlimited number of intermediate values are: a. nominal. b. discrete. c. ordinal. d. continuous. e. interval.

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Q: A continuous variable: a) may take on only integer values (e.g. 0, 1, 2, ...). b) may take on only a finite number of different values. c) may take on an infinite number of values. d) must be any nonnegative real number.

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Q: True or False? If False, correct it. The breaking strength of a cable is a discrete variable.

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Q: Consider a normal distribution with MU = 67 and SIGMA**2 = 144. If each score is raised by 7 points, what percentage of the new scores is less than 74? a. 72% b. 88% c. 50% d. 52% e. none of these

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Q: True or False? If the mean depth of a river is 2 feet, it would be safe for a non- swimmer to wade across.

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Q: True or False? The mean number of children per family in LA is quoted as 2.2. Since very few families have 2/10 of a child, this figure must be wrong.

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Q: What percent of the area of a distribution lies between the first and third quartiles? a. 25 b. 50 c. 68 d. 75 e. The question can't be answered without knowledge of the specific distribution.

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Q: True or False? If False, correct it. The 70th percentile of the distribution of a random variable X is an x-value which is exceeded by 70% of the population of X.

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Q: True or False? If False, correct it. Not more than 10% of a set of measurements can be above the 95th percentile.

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A: c) Assigns probabilities to the various values of x.

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A: b) characteristic which takes on different values

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A: Definition: A characteristic or measurement that only takes one unchanging value. Example: Suppose that we accurately record the amount of change carried by each person attending a class on a particular day. The mean for that population is a constant, say, 53 cents. It has one value which will not change. (On the other hand, if we consider drawing a random sample of 5 from the class and calculating a sample mean, we are dealing with a random variable. If we draw one sample it will have one mean, say, 82 cents. If we draw another sample, it almost always will have another mean, say, 45 cents, etc.)

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A: b. Because here the sum of the probabilities is greater than 1.

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A: a. P(T <= 0) = .5 b. P(T >= 0 and T < 2) = P(T = 0 or T = 1) = P(T = 0) + P(T = 1) = .5 + .3 = .8 c. E(T) = SUM(t * P(T = t)) = (0*.5) + (1*.3) + (2*.2) = 0.7

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A: This question can have many correct answers. One of them follows. --------------------- | | | || | REL. | | | || | FREQ. | | | || | | | | || | --------------------------------- 0 .2 .6 .65 1 VALUES OF RANDOM NUMBER If the random number is between 0 and .2 print Hooray] .2+ and .6 print Yep. .6+ and .65 print Bullseye] .65+ and 1 print Fantastic]

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A: 2. 7

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A: c. mean equal 0 and variance equal 1.

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A: (a) .0228 Z = (4800 - 4200)/300 = 2.00 Area for Z = 2.00 is .4772. Therefore, the desired area is .5000 - .4772 = .0228. (b) .7011 Z(1) = (3700 - 4200)/300 = -1.67 Z(2) = (4400 - 4200)/300 = 0.67 Prob(-1.67 < Z <= 0) = 0.4525 Prob(0 < Z < 0.67) = 0.2486 Therefore, the desired area is 0.4525 + 0.2486 = .7011.

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A: a. 239 (K - 200)/20 = 1.96 Therefore, K = 239

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A: B. .006 MU (deck) = 52 * .01 = .52 Var(deck) = 52 * .000052 = .002704 Z = (.65 - .52)/SQRT(.002704) = .13/.052 = 2.5 P(Z > 2.5) = .0062

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A: e. 0.7422

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A: c. .0124

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A: (c) 74.86 Z = (11 - 9)/3 = .67 P(Z < .67) = (.2486) + (.5000) = .7486 = 74.86%

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A: (d) .6480 From the table of cumulative normal distribution, the area under the curve to the right of Z = -0.38 is .6480.

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A: (a) .001350 P(XBAR > 15) = P(Z > (15 - 10)/(5/3)) = P(Z > 3) = .001350

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A: e. none of these P(z<=?) = .33 z = -.44 -.44 = (x-1000)/(100) x = -44 + 1000 = 956

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A: c. standard score with a mean of zero.

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A: d. above and below the mean in standard deviation units from the mean.

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A: d) +1.00 Z = (X - MU)/Standard Deviation = (40 - 30)/10 = 1

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A: c) 61.50 Z = (X - MU)/Standard Deviation 1.15 = (X - 50)/10 X = 61.50

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A: e) 97 percent Z = -2 Area to left of Z = .0228 Total area = 1 Percentage of people above that person = 1 - .0288 = .9712*100 = 97.12 percent

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A: d) 84 percent Z = (X - MU)/Standard Deviation Z = (600 - 500)/100 = 1 Area between Z and Mean = .3413 Area to Left of Z = .5 + .3413 = .8413 Percentile Rank = 100*.8413 = 84.13 Therefore, the person has performed better than 84 percent of the people.

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A: b. one. Z-score = (X - MU)/SIGMA VAR(Z) = 1.0

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A: a. has a Z-score of 1.60. Z-score = (X - MU)/SIGMA = (85 - 45)/25 = 1.60

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A: (e)

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A: a. 328 Z = (X - MU)/SIGMA Z-Score for 70 = (70-76)/8 = -.75 Z-Score for 82 = (82-76)/8 = .75 The area between -.75 and .75 = .2734 + .2734 = .5468 Number of students = 600*.5468 = 328

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A: a. greater than .5000. Pr(-.25 < Z) = .5000 + .0987 = .5987.

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A: c) 38 Z = (X - MU)/SIGMA 1.5 = (X - 32)/4 X = 32 + 1.5*4 = 32 + 6 = 38

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A: (2) 0.9518 P(Z <= 1.65 or Z > 3.0) = P(Z <= 1.65) + P(Z > 3.0) = .9505 + .0013 = .9518

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A: (3) 0.0745 Pr(Z > +1.96 or Z < -1.65) = P(Z > 1.96) + P(Z < -1.65) = .0251 + .0494 = .0745

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A: a. .5997 Z(1) = (69.5 - 71)/3 = -.5 Z(2) = (75 - 71)/3 = 1.33 P(69.5 <= X <= 75) = P(-.5 <= Z <= 1.33) = .1915 + .4082 = .5997

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A: d. 69.02% Mean = 87 Variance = 4 SIGMA = 2 Z = (X - MU)/SIGMA Z to the left of mean = (86 - 87)/2 = -1/2 = -.5 Z to the right of mean = (93 - 87)/2 = 6/2 = +3 Area between Z values of -.5 to +3 = .1915 + .4987 = .6902. Percentage of cases between 86 and 93 = 69.02%

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A: d. 99.18% Mean = 30 Variance = 25 SIGMA = 5 Z = (X - MU)/SIGMA = (42 - 30)/5 = 2.4 Area to the right of Z = 2.4 is .0082. Percentile rank = 100 - (.0082*100) = 99.18%

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A: d. 41.68 Z value corresponding to .33 = -.44 Z = (X - MU)/SIGMA = (X - 43)/3 (-.44) * 3 = X - 43 X = 43 - 1.32 = 41.68

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A: c. -1.69 and 7.69 Mean = 3, Variance = 49, SIGMA = 7 We want the limits for the middle 50% of the data, i.e. from the 25th percentile to the 75th percentile. Z value on left of mean corresponding to 25th percentile = -.67 Z value on right of mean corresponding to 25th percentile = .67 Z = (X - MU)/SIGMA -.67 = (X - 3)/7 -4.69 = X - 3 X(lower limit) = -1.69 Similarly, X(upper limit) = 3 + 4.69 = 7.69

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A: a. 69.15% Mean = 86 SIGMA = 16 X = 78 Z = (X - MU)/SIGMA = (78 - 86)/16 = -8/16 = -.5 Area to the right of Z value of -.5 = .1915 + .5 = .6915 Percentage of scores above 78 = 69.15%.

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A: e. none of these Area for 67th percentile = .17 (between Z and mean) Z value corresponding to .17 = .44 Z = (X - MU)/SIGMA = (X - 6)/3 X = (.44 * 3) + 6 = 7.32

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A: d. cannot be determined without additional information

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A: b. 1.9 Z = (X - MU)/SIGMA An area corresponding to the 21st percentile is the same as an area = .29 between X and the mean. Z value corresponding to this area = -.81 -.81 = (X - MU)/SIGMA = (X - 10)/10 -8.1 = X - 10 X = 1.9

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A: (1) 97.72% Mean = 500 Variance = 625 SIGMA = 25 Z = (X - MU)/SIGMA = (550 - 500)/25 = 2 Area beyond Z value of 2 = .0228 or 2.28% 2.28% of scores fall above 550 Percentage of scores falling below 550 = 100% - 2.28% = 97.72%

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A: d. .7745 Z value for an observation of -9 = (X - MU)/SIGMA = (-9 - (-7))/2 = -1 Z value for an observation of -4 = (-4 - (-7))/2 = 1.5 Area between Z values of -1 to 1.5 = .3413 + .4332 = .7745

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A: c. .2417 Z value associated with 9 = (X - MU)/SIGMA = (9 - 10)/2 = -.5 Z value associated with 7 = (7 - 10)/2 = -1.5 Area between Z values of -1.5 and -.5 = .4332-.1915 = .2417

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A: d) All of the above are correct.

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A: c) 0.2514 Z = [X - MU]/[SIGMA] = [24 - 22]/[3] = .666 Prob.(Z>.666) = .2514

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A: a) 0.1056 Z = [X - MU]/[SIGMA] = [68.5 - 66]/[2] = 1.25 Prob.(Z>1.25) = .1056

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A: a) 62.5 Prob.(Z= 46) = p(Z >= [46-40]/[8]) = p(Z >= .75) = .2266 Expected frequency E(f) = N * p = (10,000) * (.2266) = 2266

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A: b. 85.36 minutes The upper 10% of the students (needing the most time) correspond to a Z-score in a standard normal distribution of 1.28 1.28 = [X-70]/[12] implies X = 85.36

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A: a. Using the empirical rule, the expected Z score for the fastest runner would be near -3, which converts to an approximate time of 84 seconds. ((120 - 3*12) = 120 - 36 = 84) b. Again using the empirical rule, the expected Z score for the slowest runner would be near +3, which converts to an approximate time of 156 seconds. ((120 + 12*3) = 156) c. Z = 135 - 120/12 = 1.25 Area to the right of Z = 1.25 is .1056 .1056 * 150 = 16 students d. (102 - 120)/12 = -1.5 Area to the left of Z = -1.5 is .0668 .0668 * 150 = 10 students e. two minutes f. two minutes

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A: a. Z = (90 - 130)/20 = -2 percentage = .5 - .4772 = 2.28 percent b. Z = (100 - 130)/20 = -1.5 Z = (120 - 130)/20 = - .5 percentage = .4332 - .1915 = 24.17 percent c. Z(120) = -.5 percentage = .5 - .1915 = 30.85 percent Z(150) = 1.0 percentage = .5 - .3413 = 15.87 percent total percentage = 46.72 percent d. .5 - .14 = .36 Z = 1.08 1.08 = (value - 130)/20 value = $151.60

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A: P(X < 30) = P(Z < (30 - 36)/4) = P(Z < -1.5) = .0668 or 6.68%

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A: If available, consult file of graphs and diagrams that could not be computerized. a) Z = (X - MU)/SIGMA = (85 - 70)/8 = 1.875 Area beyond this Z value is .0301, so 3.01% of the families spent 85 dollars or more per week. b) A cumulative Z value such that 80% lies above it or 20% lies below it is -.84. Z = (X - MU)/SIGMA -.84 = (X - 70)/8 X = 63.28 Therefore, 80% of the families lie above the expenditure value of $63.28/week.

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A: a) P(MU <= X <= 1.2SIGMA) = .3849 P(X is in interval MU +/- 1.2SIGMA) = 2(.3849) = .7698 b) P(X > (MU + SIGMA)) = (.5 - .3413) = .1587 c) P(X > MU + (3*SIGMA)) = (.5 - .4987) = .0013

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A: a) Z = 375/500 = .75 P(Z > .75) = (0.5 - .2734) = .2266 b) Z = 125/500 = .25 P(-.25 <= Z <= .25) = 2(.0987) = .1974

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A: a) .5 b) Z = (725 - 600)/500 = .25 P(Z < .25) = .5 + .0987 = .5987

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A: (a) 5.32 Area beyond Z = .09 Z = -1.34 Z = (X - MU)/SIGMA -1.34 = (X - 8)/2 -2.68 = X - 8 5.32 = X

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A: P(-1 < X < 5) = .8664 P(-1 < X < 5) = P[ -1 - 2 < X - 2 < 5 - 2] = P[(-1 - 2)/2 < (X - 2)/2 < (5 - 2)/2] = P[-1.5 < Z < 1.5] = .8664

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A: 5. none of these

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A: The principal distinction is that a discrete random variable can assume a countable number of values, while a continuous random variable can assume an uncountably infinite number of values. Examples of a discrete random variable: a. The number of heads obtained when a coin is flipped three times. b. The number that turns up when a die is rolled. c. The number of people waiting in line at a movie theater Examples of a continuous random variable: a. The height of a human b. The amount of rainfall c. Time required to run a mile

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A: False - The number of individuals in a family is a discrete variable, since the values it can assume are only whole numbers.

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A: d. continuous.

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A: c) may take on an infinite number of values.

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A: False, the breaking strength of a cable is a continuous variable.

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A: c. 50% Since each score has been raised by 7 points, the mean will also be raised by 7 points. The new mean is 74. 50% of the scores will be less than the mean of a normal distribution.

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A: False. Actual depth may exceed the mean depth in some places.

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A: False

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A: b. 50 Between the twenty-fifth percentile, and the seventy-fifth percen- tile, the percentage of area is 75 - 25 = 50%.

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A: False, the 70th percentile value is exceeded by 30% of the population.

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A: False, only 5 percent of the measurements are above the 95th per- centile.

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