Stats Finals Chapter 6-8
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(−2.13 ≤ z ≤ −0.36) =
.3428
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≤ 1.24) =
.8925
Consider the probability distribution of a random variable x. Is the expected value of the distribution necessarily one of the possible values of x?
No. The expected value can be a value different from the exact value of x.
What does the random variable for a binomial experiment of n trials measure?
The random variable measures the number of successes out of n trials
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 40; 𝜎 = 15 P(50≤x≤70) =.2286
Find z such that 89.8% of the standard normal curve lies to the left of z
z= 1.27
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the right of z = −2.12 is .983
Are your finances, buying habits, medical records, and phone calls really private? A real concern for many adults is that computers and the Internet are reducing privacy. A survey conducted by Peter D. Hart Research Associates for the Shell Poll was reported in USA Today. According to the survey, 54% of adults are concerned that employers are monitoring phone calls. Use the binomial distribution formula to calculate the probability of the following. (a) Out of five adults, none is concerned that employers are monitoring phone calls. (b) Out of five adults, all are concerned that employers are monitoring phone calls. (c) Out of five adults, exactly three are concerned that employers are monitoring phone calls.
A- .021 B- .046 C- .333
A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 91 months and a standard deviation of 3.6 months. When this computer-relay microchip malfunctions, the entire satellite is useless. A large London insurance company is going to insure the satellite for 50 million dollars. Assume that the only part of the satellite in question is the microchip. All other components will work indefinitely. (a) If the satellite is insured for 84 months, what is the probability that it will malfunction before the insurance coverage ends? (b) If the satellite is insured for 84 months, what is the expected loss to the insurance company? (c) If the insurance company charges $3 million for 84 months of insurance, how much profit does the company expect to make? (d) For how many months should the satellite be insured to be 97% confident that it will last beyond the insurance date
A- .0262 B- $1310000 C- $1690000 D- 84 months
Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean 𝜇 = 6900 and estimated standard deviation 𝜎 = 2700. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection. (a) What is the probability that, on a single test, x is less than 3500? (b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? What is the probability of x < 3500? (c) Repeat part (b) for n = 3 tests taken a week apart. (d) Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased? If a person had x < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse?
A- .1039 B- The probability distribution of x is approximately normal with 𝜇x = 6900 and 𝜎x = 1909.19. .0375 C- .0146 D- The probabilities decreased as n increased It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.
Before 1918, approximately 55% of the wolves in a region were male, and 45% were female. However, cattle ranchers in this area have made a determined effort to exterminate wolves. From 1918 to the present, approximately 70% of wolves in the region are male, and 30% are female. Biologists suspect that male wolves are more likely than females to return to an area where the population has been greatly reduced. (a) Before 1918, in a random sample of 12 wolves spotted in the region, what is the probability that 9 or more were male?. (b) What is the probability that 9 or more were female? (c) What is the probability that fewer than 6 were female? (d) For the period from 1918 to the present, in a random sample of 12 wolves spotted in the region, what is the probability that 9 or more were male? (e) What is the probability that 9 or more were female?.002 (f) What is the probability that fewer than 6 were female?.882
A- .134 B- .036 C- .527 D- .493 E- .002 F- .882
Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 81 tons of coal into each car. The actual weights of coal loaded into each car are normally distributed, with mean 𝜇 = 81 tons and standard deviation 𝜎 = 0.6 ton. (a) What is the probability that one car chosen at random will have less than 80.5 tons of coal? What is the probability that 36 cars chosen at random will have a mean load weight x of less than 80.5 tons of coal? (b) Suppose the weight of coal in one car was less than 80.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? (c) Suppose the weight of coal in 36 cars selected at random had an average x of less than 80.5 tons. Would that fact make you suspect that the loader had slipped out of adjustment? Why?
A- .2023 0 B- No C- Yes, the probability that this deviation is random is very small.
Aldrich Ames is a convicted traitor who leaked American secrets to a foreign power. Yet Ames took routine lie detector tests and each time passed them. How can this be done? Recognizing control questions, employing unusual breathing patterns, biting one's tongue at the right time, pressing one's toes hard to the floor, and counting backwards by 7 are countermeasures that are difficult to detect but can change the results of a polygraph examination†. In fact, it is reported in Professor Ford's book that after only 20 minutes of instruction by "Buzz" Fay (a prison inmate), 85% of those trained were able to pass the polygraph examination even when guilty of a crime. Suppose that a random sample of five students (in a psychology laboratory) are told a "secret" and then given instructions on how to pass the polygraph examination without revealing their knowledge of the secret. What are the following probabilities? (a) all the students are able to pass the polygraph examination. (b) more than half the students are able to pass the polygraph examination. (c) no more than half of the students are able to pass the polygraph examination. (d) all the students fail the polygraph examination
A- .444 B- .973 C- .027 D- 0
Trevor is interested in purchasing the local hardware/sporting goods store in the small town of Dove Creek, Montana. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 55% of the business days it is open. Estimate the probability that the store will gross over $850 for the following. (Round your answers to three decimal places.) (a) at least 3 out of 5 business days (b) at least 6 out of 10 business days (c) fewer than 5 out of 10 business days (d) fewer than 6 out of the next 20 business days If the outcome described in part (d) actually occurred, might it shake your confidence in the statement p = 0.55? Might it make you suspect that p is less than 0.55? Explain. (e) more than 17 out of the next 20 business days If the outcome described in part (e) actually occurred, might you suspect that p is greater than 0.55? Explain.
A- .593 B- .504 C- .262 D- .006 Yes. This is unlikely to happen if the true value of p is 0.55. E- .001 Yes. This is unlikely to happen if the true value of p is 0.55
Police response time to an emergency call is the difference between the time the call is first received by the dispatcher and the time a patrol car radios that it has arrived at the scene. Over a long period of time, it has been determined that the police response time has a normal distribution with a mean of 9.2 minutes and a standard deviation of 2.1 minutes. For a randomly received emergency call, find the following probabilities. (a) the response time is between 5 and 11 minutes. (b) the response time is less than 5 minutes. (c) the response time is more than 11 minutes
A- .7824 B- .0228 C- .1949
A person's blood glucose level and diabetes are closely related. Let x be a random variable measured in milligrams of glucose per deciliter (1/10 of a liter) of blood. Suppose that after a 12-hour fast, the random variable x will have a distribution that is approximately normal with mean 𝜇 = 86 and standard deviation 𝜎 = 23. Note: After 50 years of age, both the mean and standard deviation tend to increase. For an adult (under 50) after a 12-hour fast, find the following probabilities. (a) x is more than 60. (b) x is less than 110. (c) x is between 60 and 110 (d) x is greater than 125 (borderline diabetes starts at 125)
A- .8708 B- .8508 C- .7216 D- .0446
A research team conducted a study showing that approximately 15% of all businessmen who wear ties wear them so tightly that they actually reduce blood flow to the brain, diminishing cerebral functions. At a board meeting of 20 businessmen, all of whom wear ties, what are the following probabilities? (a) at least one tie is too tight (b) more than two ties are too tight (c) no tie is too tight. (d) at least 18 ties are not too tight.
A- .961 B- .595 C- .039 D- .405
Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.7 millimeters (mm) and a standard deviation of 1.0 mm. For a randomly found shard, find the following probabilities. (a) the thickness is less than 3.0 mm. (b) the thickness is more than 7.0 mm. (c) the thickness is between 3.0 mm and 7.0 mm
A- 0.446 B- .0107 C- .9447
The resting heart rate for an adult horse should average about 𝜇 = 45 beats per minute with a (95% of data) range from 20 to 70 beats per minute. Let x be a random variable that represents the resting heart rate for an adult horse. Assume that x has a distribution that is approximately normal. (a) Use this "rule of thumb" to estimate the standard deviation of x distribution. (b) What is the probability that the heart rate is less than 25 beats per minute? (c) What is the probability that the heart rate is greater than 60 beats per minute? (d) What is the probability that the heart rate is between 25 and 60 beats per minute? (e) A horse whose resting heart rate is in the upper 16% of the probability distribution of heart rates may have a secondary infection or illness that needs to be treated. What is the heart rate corresponding to the upper 16% cutoff point of the probability distribution?
A- 12.5 beats B- .0548 C- .1151 D- .8301 E- 57 beats per minute
A consumer products magazine indicated that the average life of a refrigerator before replacement is 𝜇 = 14 years with a (95% of data) range from 8 to 20 years. Let x = age at which a refrigerator is replaced. Assume that x has a distribution that is approximately normal. (a) Use this "rule of thumb" to approximate the standard deviation of x values. (b) What is the probability that someone will keep a refrigerator fewer than 11 years before replacement? (c) What is the probability that someone will keep a refrigerator more than 18 years before replacement? (d) Assume that the average life of a refrigerator is 14 years, with the standard deviation given in part (a) before it breaks. Suppose that a company guarantees refrigerators and will replace a refrigerator that breaks while under guarantee with a new one. However, the company does not want to replace more than 5% of the refrigerators under guarantee. For how long should the guarantee be made
A- 3 years B- 0.1587 C- .0918 D- 9.1 Years
Quick Start Company makes 12-volt car batteries. After many years of product testing, the company knows that the average life of a Quick Start battery is normally distributed, with a mean of 45.6 months and a standard deviation of 6.7 months. (a) If Quick Start guarantees a full refund on any battery that fails within the 36-month period after purchase, what percentage of its batteries will the company expect to replace? (b) If Quick Start does not want to make refunds for more than 13% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries
A- 7.6 % B- 38 months
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 14 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with 𝜎 = 0.22 gram. (a) Find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.10 for the mean weights of the hummingbirds.
A- Lower limit 3.07 upper limit 3.23 margin of error .08 B- 𝜎 is known normal distribution of weights C- We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval. D- 8 hummingbirds
Suppose x has a distribution with 𝜇 = 56 and 𝜎 = 8 (a) If random samples of size n = 16 are selected, can we say anything about the x distribution of sample means? (b) If the original x distribution is normal, can we say anything about the x distribution of random samples of size 16? (c) Find P(52 ≤ x ≤ 57).
A- No, the sample size is too small. B- Yes, the x distribution is normal with mean 𝜇 x = 56 and 𝜎 x = 2. C- .6687
Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 16 Allen's hummingbirds has been under study in Arizona. The average weight for these birds is x = 3.15 grams. Based on previous studies, we can assume that the weights of Allen's hummingbirds have a normal distribution, with 𝜎 = 0.22 gram. (a) When finding an 80% confidence interval, what is the critical value for confidence level? (b) margin of error? (c) what conditions are necessary for your calculations? (d) Interpret your results in the context of this problem. (e) Which equation is used to find the sample size n for estimating 𝜇 when 𝜎 is known? (f) Find the sample size necessary for an 80% confidence level with a maximal margin of error E = 0.16 for the mean weights of the hummingbirds.
A- Zc= 1.28 B- Margin of error= .08 lower limit 3.08 uppef limit 3.22 C- 𝜎 is known normal distribution of weights D- We are 80% confident that the true average weight of Allen's hummingbirds falls within this interval. E- n = (z𝜎 𝜎/E)^2 F- 4 hummingbirds
(a) If we have a distribution of x values that is more or less mound-shaped and somewhat symmetric, what is the sample size n needed to claim that the distribution of sample means x from random samples of that size is approximately normal? (b) If the original distribution of x values is known to be normal, do we need to make any restriction about sample size in order to claim that the distribution of sample means x taken from random samples of a given size is normal?
A- n ≥ 30 B- no
(a) The video presents various estimated statistics regarding health. Which of the following is NOT one of the statistics discussed in the video? (b) The video presents various statistics regarding health. Which of the following describes an estimate of a population mean (𝜇)? (c) The video says that researchers at the University of Michigan found that having close community ties can reduce heart attack risk for those over 50 by 22%. Which of the following describes a 22% decrease in risk?
A- the percentage of men who walk a dog on a daily basis B- the average number of hours of sleep a teenager gets per night C- risk of a heart attack of those with "close community ties": 0.08 risk of a heart attack of those without "close community ties": 0.10
uppose x has a distribution with a mean of 20 and a standard deviation of 3. Random samples of size n = 36 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. (b) Find the z value corresponding to x = 20.5. (c) Find P(x < 20.5) (d) Would it be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 20.5? Explain.
A- x has an approximately normal distribution with mean 𝜇x = 20 and standard deviation 𝜎x = .5 B- Z= 1 C- P(x < 20.5) = .8413 D- No, it would not be unusual because more than 5% of all such samples have means less than 20.5.
Suppose x has a distribution with 𝜇 = 10 and 𝜎 = 7. (a) If a random sample of size n = 42 is drawn, find 𝜇x, 𝜎 x and P(10 ≤ x ≤ 12). (b) If a random sample of size n = 68 is drawn, find 𝜇x, 𝜎 x and P(10 ≤ x ≤ 12). (c) Why should you expect the probability of part (b) to be higher than that of part (a)? (Hint: Consider the standard deviations in parts (a) and (b).)
A- 𝜇x = 10 𝜎 x = 1.0801 P(10 ≤ x ≤ 12) = .4678 B- 𝜇x = 10 𝜎 x = .8489 P(10 ≤ x ≤ 12) = .4909 C- The standard deviation of part (b) is smaller than part (a) because of the larger sample size. Therefore, the distribution about 𝜇x is narrower
Raul received a score of 76 on a history test for which the class mean was 70 with a standard deviation of 3. He received a score of 79 on a biology test for which the class mean was 70 with standard deviation 7. On which test did he do better relative to the rest of the class?
History Test
Consider two normal curves. If the first one has a larger mean than the second one, must it have a larger standard deviation as well? Explain your answer.
No. The values of 𝜇 and 𝜎 are independent.
Does a raw score less than the mean correspond to a positive or negative standard score? What about a raw score greater than the mean?
Raw scores less than the mean will have negative standard scores; raw scores above the mean will have positive standard scores.
What does it mean to say that the trials of an experiment are independent?
The outcome of one trial does not affect the probability of success on any other trial.
What is the standard error of a sampling distribution?
The standard deviation
What is the standard deviation of a sampling distribution called?
The standard error
Consider the following scores. (i) a score of 40 from a distribution with mean 50 and standard deviation 10(ii) a score of 45 from a distribution with mean 50 and standard deviation 5 How do the two scores compare relative to their respective distributions?
They are both 1 standard deviation below their respective means.
True or false? If the sample mean x of a random sample from an x distribution is relatively small, when the confidence level c is reduced, the confidence interval for 𝜇 becomes shorter.
True. As the level of confidence decreases, the maximal error of estimate decreases.
In a carnival game, there are six identical boxes, one of which contains a prize. A contestant wins the prize by selecting the box containing it. Before each game, the old prize is removed and another prize is placed at random in one of the six boxes. Is it appropriate to use the binomial probability distribution to find the probability that a contestant who plays the game five times wins exactly twice? Check each of the requirements of a binomial experiment and give the values of n, r,and p.
Yes. The five trials are independent, have only two outcomes, and have the same P(success); n = 5, r = 2, p = 1/6
Sam computed a 90% confidence interval for 𝜇 from a specific random sample of size n. He claims that at the 90% confidence level, his confidence interval contains 𝜇. Is this claim correct? Explain.
Yes. The proportion of all confidence intervals based on random samples of size n that contain 𝜇 is 0.90
Find z such that 84% of the standard normal curve lies to the right of z.
Z= -0.99
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 100; 𝜎 = 13 P(x≥120) =.0618
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 114; 𝜎 = 10 P(x≥90) =.9918
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 14.2; 𝜎 = 3.6 P(8≤x≤12) =.2282
Find z such that 5.1% of the standard normal curve lies to the left of z.
-1.64
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≥ 2.09) =
.0183
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≤ −2.04)=
.0207
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≥ 1.44) =
.0749
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≤ −0.12)=
.4522
Consider a normal distribution with mean 26 and standard deviation 7. What is the probability a value selected at random from this distribution is greater than 26?
.5
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≤ 0) =
.5
Suppose an x-distribution has mean 𝜇 = 4. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81. (a) What is the value of the mean of each of the two x distributions? (b) For which x distribution is P(x > 5.00) smaller? Explain your answer. (c) For which x distribution is P(3.00 < x < 5.00) greater? Explain your answer.
A- For n = 49, 𝜇x = 4 For n = 81, 𝜇x = 4 B- The distribution with n = 81 because the standard deviation will be smaller. C- The distribution with n = 81 because the standard deviation will be smaller.
Suppose an x-distribution has mean 𝜇 = 6. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81. (a) What is the value of the mean of each of the two x distributions? (b) For which x distribution is P(x > 7.50) smaller? Explain your answer. (c) For which x distribution is P(4.50 < x < 7.50) greater? Explain your answer.
A- For n = 49, 𝜇x = 6 For n = 81, 𝜇x = 6 B- The distribution with n = 81 because the standard deviation will be smaller. C- The distribution with n = 81 because the standard deviation will be smaller.
Suppose an x-distribution has mean 𝜇 = 7. Consider two corresponding x distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81. (a)What is the value of the mean of each of the two x distributions? (b) For which x distribution is P(x > 8.75) smaller? Explain your answer. (c) For which x distribution is P(5.25 < x < 8.75) greater? Explain your answer.
A- Recall that the x-distribution in this case has a mean of 𝜇 = 7. The mean of the x distribution does not depend on n. The mean 𝜇x of the distribution with sample size n = 49 is 𝜇x = 𝜇 = 7 . The mean 𝜇x of the distribution with sample size n = 81 is 𝜇x = 𝜇 = 7 . B- Given distributions with sample sizes n = 49 and n = 81, the distribution with a smaller value of P(x > 8.75) is the distribution with n = 81 because its standard deviation is smaller C- Given distributions with sample sizes n = 49 and n = 81, the distribution with the greater value of P(5.25 < x < 8.75), which contains the mean 𝜇 = 7, is the distribution with n = 81 because its standard deviation is smaller
A- Consider two x distributions corresponding to the same x distribution. The first x distribution is based on samples of size n = 100 and the second is based on samples of size n = 225. Which x distribution has the smaller standard error? B- Explain
A- The distribution with n = 225 will have a smaller standard error. B-Since 𝜎x = 𝜎/√n, dividing by the square root of 225 will result in a small standard error regardless of the value of 𝜎.
At wind speeds above 1000 centimeters per second (cm/sec), significant sand-moving events begin to occur. Wind speeds below 1000 cm/sec deposit sand and wind speeds above 1000 cm/sec move sand to new locations. The cyclic nature of wind and moving sand determines the shape and location of large dunes. At a test site, the prevailing direction of the wind did not change noticeably. However, the velocity did change. Sixty-five wind speed readings gave an average velocity of x = 1075 cm/sec. Based on long-term experience, 𝜎 can be assumed to be 245 cm/sec. Find a 95% confidence interval for the population mean wind speed at this site. (b)Does the confidence interval indicate that the population mean wind speed is such that the sand is always moving at this site? Explain.
A- lower limit 1015 cm/sec upper limit 1135 cm/sec B- Yes. This interval indicates that the population mean wind speed is such that the sand is always moving at this site.
Let x represent the dollar amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. Based on a newspaper article, the mean of the x distribution is about $22 and the estimated standard deviation is about $6. (a) Consider a random sample of n = 40 customers, each of whom has 10 minutes of unplanned shopping time in a supermarket. From the central limit theorem, what can you say about the probability distribution of x, the average amount spent by these customers due to impulse buying? What are the mean and standard deviation of the x distribution? (b) Is it necessary to make any assumption about the x distribution? Explain your answer. (c) What is the probability that x is between $20 and $24? (d) Let us assume that x has a distribution that is approximately normal. What is the probability that x is between $20 and $24? (e) In part (b), we used x, the average amount spent, computed for 40 customers. In part (c), we used x, the amount spent by only one customer. The answers to parts (b) and (c) are very different. Why would this happen? (f) In this example, x is a much more predictable or reliable statistic than x. Consider that almost all marketing strategies and sales pitches are designed for the average customer and not the individual customer. How does the central limit theorem tell us that the average customer is much more predictable than the individual customer?
A- The sampling distribution of x is approximately normal with mean 𝜇x = 22 and standard error 𝜎x = $0.95 B- It is not necessary to make any assumption about the x distribution because n is large C- .9652 D- .2611 E- The standard deviation is smaller for the x distribution than it is for the x distribution F- The central limit theorem tells us that the standard deviation of the sample mean is much smaller than the population standard deviation. Thus, the average customer is more predictable than the individual customer
Thirty small communities in Connecticut (population near 10,000 each) gave an average of x = 137.5 reported cases of larceny per year. Assume that 𝜎 is known to be 40.7 cases per year. (a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (d) Compare the margins of error for parts (a) through (c). As the confidence levels increase, do the margins of error increase? (e) Compare the lengths of the confidence intervals for parts (a) through (c). As the confidence levels increase, do the confidence intervals increase in length?
A- lower limit 125.28 upper limit 149.72 margin of error 12.22 B- lower limit 122.94 upper limit 152.06 margin of error 14.56 C- lower limit 118.33 upper limit 156.67 margin of error 19.17 D- As the confidence level increases, the margin of error increases. E- As the confidence level increases, the confidence interval increases in length.
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 42 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that 𝜎 = 7.70 ml/kg for the distribution of blood plasma. (a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (b) Interpret your results in the context of this problem. (c) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.10 for the mean plasma volume in male firefighters.
A- lower limit 34.43 upper limit 40.57 margin of error 3.07 B- We are 99% confident that the true average blood plasma volume in male firefighters falls within this interval. C- 90 male firefighters
Total plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 43 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that 𝜎 = 7.60 ml/kg for the distribution of blood plasma. (a) Find a 99% confidence interval for the population mean blood plasma volume in male firefighters. What is the margin of error? (B) Interpret your results in the context of this problem. (C) Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.10 for the mean plasma volume in male firefighters.
A- lower limit 34.51 upper limit 40.49 margin of error 2.99 What conditions are necessary for your calculations? 𝜎 is known n is large B- We are 99% confident that the true average blood plasma volume in male firefighters falls within this interval. C- 88 male firefighters
Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma.† Over a period of months, an adult male patient has taken seven blood tests for uric acid. The mean concentration was x = 5.37 mg/dl. The distribution of uric acid in healthy adult males can be assumed to be normal, with 𝜎 = 1.81 mg/dl. (a) Find a 95% confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Find the sample size necessary for a 95% confidence level with maximal margin of error E = 1.02 for the mean concentration of uric acid in this patient's blood
A- lower limit 4.03 upper limit 6.71 margin of error 1.34 B- 𝜎 is known normal distribution of uric acid C- We are 95% confident that the true uric acid level for this patient falls within this interval D- 13 blood tests
Consider an x distribution with standard deviation 𝜎 = 27. (a) If specifications for a research project require the standard error of the corresponding x distribution to be 3, how large does the sample size need to be? (b) If specifications for a research project require the standard error of the corresponding x distribution to be 1, how large does the sample size need to be?
A- n= 81 B- n= 729
Describe how the variability of the distribution changes as the sample size increases.
As the sample size increases, the variability decreases.
True or false? A larger sample size produces a longer confidence interval for 𝜇.
False. As the sample size increases, the maximal error decreases, resulting in a shorter confidence interval
True or false? Every random sample of the same size from a given population will produce exactly the same confidence interval for 𝜇.
False. Different random samples may produce different x values, resulting in different confidence intervals.
True or false? If the sample mean of a random sample from an x distribution is relatively small, then the confidence interval for 𝜇 will be relatively short.
False. The maximal error of estimate controls the length of the confidence interval regardless of the value of x.
True or false? If the original x distribution has a relatively small standard deviation, the confidence interval for 𝜇 will be relatively short.
True. As 𝜎 decreases, E decreases, resulting in a shorter confidence interval
True or false? The value zc is a value from the standard normal distribution such that P(-zc < z < zc) = c.
True. By definition, critical values zc are such that 100c% of the area under the standard normal curve falls between -zc and zc.
True or false? Consider a random sample of size n from an x distribution. For such a sample, the margin of error for estimating 𝜇 is the magnitude of the difference between x and 𝜇.
True. By definition, the margin of error is the magnitude of the difference between x and 𝜇.
True or false? The point estimate for the population mean 𝜇 of an x distribution is x, computed from a random sample of the x distribution.
True. The mean of the x distribution equals the mean of the x distribution and the standard error of the x distribution decreases as n increases.
Find z such that 60% of the standard normal curve lies to the left of z
Z= .26
Find the z value such that 85% of the standard normal curve lies between −z and z.
Z= 1.44
Find z such that 4.6% of the standard normal curve lies to the right of z.
Z= 1.69
What does a standard score measure?
the number of standard deviations a measurement is from the mean
For a binomial experiment, how many outcomes are possible for each trial? What are the possible outcomes?
two; success or failure
Find z such that 8% of the standard normal curve lies to the right of z.
z= 1.4
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 104; 𝜎 = 11 P(x≥120) =.0735
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 16.0; 𝜎 = 4.5 P(10 ≤ x ≤ 26) = .895
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 38; 𝜎 = 16 P(50≤x≤70) =0.2038
In a certain article, laser therapy was discussed as a useful alternative to drugs in pain management of chronically ill patients. To measure pain threshold, a machine was used that delivered low-voltage direct current to different parts of the body (wrist, neck, and back). The machine measured current in milliamperes (mA). The pretreatment experimental group in the study had an average threshold of pain (pain was first detectable) at 𝜇 = 3.11 mA with standard deviation 𝜎 = 1.13 mA. Assume that the distribution of threshold pain, measured in milliamperes, is symmetric and more or less mound-shaped.
(a) Use the empirical rule to estimate a range of milliamperes centered about the mean in which about 68% of the experimental group will have a threshold of painfrom 1.98 mA to 4.24 mA (b) Use the empirical rule to estimate a range of milliamperes centered about the mean in which about 95% of the experimental group will have a threshold of painfrom .85 mA to 5.37 mA
At Burnt Mesa Pueblo, archaeological studies have used the method of tree-ring dating in an effort to determine when prehistoric people lived in the pueblo. Wood from several excavations gave a mean of (year) 1260 with a standard deviation of 28 years. The distribution of dates was more or less mound-shaped and symmetric about the mean. Use the empirical rule to estimate the following.
(a) a range of years centered about the mean in which about 68% of the data (tree-ring dates) will be foundbetween 1232 and 1288 A.D. (b) a range of years centered about the mean in which about 95% of the data (tree-ring dates) will be foundbetween 1204 and 1316 A.D. (c) a range of years centered about the mean in which almost all the data (tree-ring dates) will be foundbetween 1176 and 1344 A.D.
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(−2.02 ≤ z ≤ 1.08)=
.8382
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(−1.10 ≤ z ≤ 2.64)=
.8602
Let z be a random variable with a standard normal distribution. Find the indicated probability. P(z ≥ −1.27) =
.8979
A normal distribution has 𝜇 = 24 and 𝜎 = 5. (a) Find the z score corresponding to x = 19. (b) Find the z score corresponding to x = 36. (c) Find the raw score corresponding to z = −3. (d) Find the raw score corresponding to z = 1.3.
A- -1 B- 2.4 C- 9 D- 30.5
A normal distribution has 𝜇 = 24 and 𝜎 = 5. (a) Find the z score corresponding to x = 19. (b) Find the z score corresponding to x = 37 (c) Find the raw score corresponding to z = −3 (d) Find the raw score corresponding to z = 1.
A- -1 B- 2.6 C- 9 D- 30.5
Richard has just been given a 4-question multiple-choice quiz in his history class. Each question has five answers, of which only one is correct. Since Richard has not attended class recently, he doesn't know any of the answers. Assuming that Richard guesses on all four questions, find the indicated probabilities. (a)What is the probability that he will answer all questions correctly? (b) What is the probability that he will answer all questions incorrectly? (c) What is the probability that he will answer at least one of the questions correctly? Compute this probability two ways. First, use the rule for mutually exclusive events and the probabilities shown in the binomial probability distribution table. Then use the fact that P(r ≥ 1) = 1 − P(r = 0). Compare the two results. Should they be equal? Are they equal? If not, how do you account for the difference? (d) What is the probability that Richard will answer at least half the questions correctly?
A- .002 B- .410 C- .592 .590 They should be equal, but may differ slightly due to rounding error. D- .181
Sara is a 60-year-old Anglo female in reasonably good health. She wants to take out a $50,000 term (i.e., straight death benefit) life insurance policy until she is 65. The policy will expire on her 65th birthday. The probability of death in a given year is provided. x = age 60 61 62 63 64 P(death at this age) 0.00595 0.00914 0.00908 0.00951 0.01000 Sara is applying to Big Rock Insurance Company for her term insurance policy. (a) What is the probability that Sara will die in her 60th year? (b) Repeat part (a) for ages 61, 62, 63, and 64. What would be the total expected cost to Big Rock Insurance over the years 60 through 64? (c) If Big Rock Insurance wants to make a profit of $700 above the expected total cost paid out for Sara's death, how much should it charge for the policy? (d) If Big Rock Insurance Company charges $5000 for the policy, how much profit does the company expect to make?
A- .00595 $297.5 B- Age Expected Cost 61 $457 62 $454 63 $475.5 64 $500 $2184 C- $2884 D- $2816
The college student senate is sponsoring a spring break Caribbean cruise raffle. The proceeds are to be donated to the Samaritan Center for the Homeless. A local travel agency donated the cruise, valued at $2000. The students sold 2714 raffle tickets at $5 per ticket. (a) Kevin bought twenty-two tickets. What is the probability that Kevin will win the spring break cruise to the Caribbean? What is the probability that Kevin will not win the cruise? (b) Expected earnings can be found by multiplying the value of the cruise by the probability that Kevin will win. What are Kevin's expected earnings? Is this more or less than the amount Kevin paid for the twenty-two tickets?less How much did Kevin effectively contribute to the Samaritan Center for the Homeless?
A- .00811 .99189 B- $16.22 less $93.78
Jim is a 60-year-old Anglo male in reasonably good health. He wants to take out a $50,000 term (i.e., straight death benefit) life insurance policy until he is 65. The policy will expire on his 65th birthday. The probability of death in a given year is provided. x = age 60 61 62 63 64 P(death at this age) 0.01147 0.01435 0.01705 0.01972 0.02209 Jim is applying to Big Rock Insurance Company for his term insurance policy. (a) What is the probability that Jim will die in his 60th year? Using this probability and the $50,000 death benefit, what is the expected cost to Big Rock Insurance? (b) Repeat part (a) for years 61, 62, 63, and 64. What would be the total expected cost to Big Rock Insurance over the years 60 through 64? (c) If Big Rock Insurance wants to make a profit of $700 above the expected total cost paid out for Jim's death, how much should it charge for the policy? (d) If Big Rock Insurance Company charges $5000 for the policy, how much profit does the company expect to make?
A- .01147 $573.50 B- Year Expected Cost 61 $717.5 62 $852.5 63 $986 64 $1104.5 $4234 C- $4934 D- $766
Consider a binomial experiment with n = 6 trials where the probability of success on a single trial is p = 0.40 (a) Find P(r = 0). (b) Find P(r ≥ 1) by using the complement rule.
A- .047 B- .953
A tidal wave or tsunami is usually the result of an earthquake in the Pacific Rim, often 1000 or more miles from Hawaii. Tsunamis are rare but dangerous. Many tsunamis are small and do little damage. However, a tsunami nine meters or higher is very dangerous. Civil Defense authorities sound an alarm telling people near the beach to go to higher ground. About 30% of all recorded tsunamis have been nine meters or higher.† You are writing a report about six recent earthquakes in the Pacific Rim and you want to include a brief statistical profile of some possible events regarding tsunamis in Hawaii. Let r be the number of tsunamis nine meters or higher resulting from six randomly chosen earthquakes in the Pacific Rim. (a) What is the probability none of the tsunamis are nine meters or higher? (b) What is the probability at least one is nine meters or higher? (c) What is the expected number of tsunamis nine meters or higher? (d) What is the standard deviation of the r-probability distribution?
A- .118 B- .882 C- 1.8 D- 1.122
A fair quarter is flipped three times. For each of the following probabilities, use the formula for the binomial distribution and a calculator to compute the requested probability. (a) Find the probability of getting exactly three heads. (b) Find the probability of getting exactly two heads. (c) Find the probability of getting two or more heads. (d) Find the probability of getting exactly three tails.
A- .125 B- .375 C- .5 D- .125
A vending machine automatically pours soft drinks into cups. The amount of soft drink dispensed into a cup is normally distributed with a mean of 7.6 ounces and standard deviation of 0.4 ounce. Examine the figure below and answer the following questions. (a) Estimate the probability that the machine will overflow an 8-ounce cup (b) Estimate the probability that the machine will not overflow an 8-ounce cup (c) The machine has just been loaded with 875 cups. How many of these cups do you expect will overflow when served?
A- .16 B- .84 C- 140 cups
A tidal wave or tsunami is usually the result of an earthquake in the Pacific Rim, often 1000 or more miles from Hawaii. Tsunamis are rare but dangerous. Many tsunamis are small and do little damage. However, a tsunami nine meters or higher is very dangerous. Civil Defense authorities sound an alarm telling people near the beach to go to higher ground. About 30% of all recorded tsunamis have been nine meters or higher.† You are writing a report about five recent earthquakes in the Pacific Rim and you want to include a brief statistical profile of some possible events regarding tsunamis in Hawaii. Let r be the number of tsunamis nine meters or higher resulting from five randomly chosen earthquakes in the Pacific Rim. (a) What is the probability none of the tsunamis are nine meters or higher? (b) What is the probability at least one is nine meters or higher? (c) What is the expected number of tsunamis nine meters or higher? (d) What is the standard deviation of the r-probability distribution?
A- .168 B- .832 C- 1.5 D- 1.025
The one-time fling! Have you ever purchased an article of clothing (dress, sports jacket, etc.), worn the item once to a party, and then returned the purchase? This is called a one-time fling. About 15% of all adults deliberately do a one-time fling and feel no guilt about it! In a group of eight adult friends, what is the probability of the following? (a) no one has done a one-time fling. (b) at least one person has done a one-time fling. (c) no more than two people have done a one-time fling
A- .272 B- .728 C- .895
In Hawaii, January is a favorite month for surfing since 60% of the days have a surf of at least 6 feet.† You work day shifts in a Honolulu hospital emergency room. At the beginning of each month you select your days off, and you pick 7 days at random in January to go surfing. Let r be the number of days the surf is at least 6 feet. (a) What is the probability of getting 5 or more days when the surf is at least 6 feet? (b) What is the probability of getting fewer than 2 days when the surf is at least 6 feet? (c) What is the expected number of days when the surf will be at least 6 feet? (d) What is the standard deviation of the r-probability distribution? (e) Can you be fairly confident that the surf will be at least 6 feet high on one of your days off? Explain.
A- .420 B- .019 C- 4.2 D- 1.296 E- Yes , because the probability of at least 1 day with surf of at least 6 feet is .998 and the expected number of days when the surf will be at least 6 feet is greater than one.
Sociologists say that 95% of married women claim that their husband's mother is the biggest bone of contention in their marriages (sex and money are lower-rated areas of contention). Suppose that twelve married women are having coffee together one morning. Find the following probabilities. (a) All of them dislike their mother-in-law. (b) None of them dislike their mother-in-law. c) At least ten of them dislike their mother-in-law. (d) No more than nine of them dislike their mother-in-law.
A- .54 B- 0 C- .980 D- .020
A particular lake is known to be one of the best places to catch a certain type of fish. In this table, x = number of fish caught in a 6-hour period. The percentage data are the percentages of fishermen who caught x fish in a 6-hour period while fishing from shore. x 0 1 2 3 4 or more % 43% 36% 13% 7% 1% (a) Find the probability that a fisherman selected at random fishing from shore catches one or more fish in a 6-hour period. (b) Find the probability that a fisherman selected at random fishing from shore catches two or more fish in a 6-hour period. (c) Compute 𝜇, the expected value of the number of fish caught per fisherman in a 6-hour period (round 4 or more to 4).
A- .57 B- .21 C- 𝜇 = .87 fish
In Hawaii, January is a favorite month for surfing since 60% of the days have a surf of at least 6 feet.† You work day shifts in a Honolulu hospital emergency room. At the beginning of each month you select your days off, and you pick 5 days at random in January to go surfing. Let r be the number of days the surf is at least 6 feet. (a) What is the probability of getting 3 or more days when the surf is at least 6 feet? (b) What is the probability of getting fewer than 2 days when the surf is at least 6 feet? (c) What is the probability of getting fewer than 2 days when the surf is at least 6 feet? (d) What is the standard deviation of the r-probability distribution? (e) Can you be fairly confident that the surf will be at least 6 feet high on one of your days off? Explain.
A- .683 B- .087 C- 3 days D- 1.095 days E- Yes , because the probability of at least 1 day with surf of at least 6 feet is .99 and the expected number of days when the surf will be at least 6 feet is greater than one.
USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders. x 0 1 2 3 4 5 P(x) 0.223 0.378 0.203 0.169 0.026 0.001 (a) Find the probability that one or more of the five parolees will be repeat offenders How does this number relate to the probability that none of the parolees will be repeat offenders? (b) Find the probability that two or more of the five parolees will be repeat offenders. (c) Find the probability that four or more of the five parolees will be repeat offenders. (d) Compute 𝜇, the expected number of repeat offenders out of five. (e) Compute 𝜎, the standard deviation of the number of repeat offenders out of five.
A- .777 This is the complement of the probability of no repeat offenders. B- .399 C- .027 D- 𝜇 = 1.4 prisoners E- 1.09
USA Today reported that approximately 25% of all state prison inmates released on parole become repeat offenders while on parole. Suppose the parole board is examining five prisoners up for parole. Let x = number of prisoners out of five on parole who become repeat offenders. x 0 1 2 3 4 5 P(x) 0.220 0.363 0.222 0.165 0.029 0.001 (a) Find the probability that one or more of the five parolees will be repeat offenders. How does this number relate to the probability that none of the parolees will be repeat offenders? (b) Find the probability that two or more of the five parolees will be repeat offenders. (c) Find the probability that four or more of the five parolees will be repeat offenders. (d) Compute 𝜇, the expected number of repeat offenders out of five. (e) Compute 𝜎, the standard deviation of the number of repeat offenders out of five.
A- .78 This is the complement of the probability of no repeat offenders B- .417 C- .03 D- 𝜇 = 1.423 prisoners E- 1.1 prisoners
Let x = red blood cell (RBC) count in millions per cubic millimeter of whole blood. For healthy females, x has an approximately normal distribution with mean 𝜇 = 4.3 and standard deviation 𝜎 = 0.7. (a) Convert the x interval, 4.5 < x, to a z interval. (b) Convert the x interval, x < 4.2, to a z interval (c) Convert the x interval, 4.0 < x < 5.5, to a z interval. (d) Convert the z interval, z < −1.44, to an x interval. (e) Convert the z interval, 1.28 < z, to an x interval. (f) Convert the z interval, −2.25 < z < −1.00, to an x interval. (g) If a female had an RBC count of 5.9 or higher, would that be considered unusually high? Explain using z values.
A- 0.29 < z B- z < -0.14 C- -0.43 < z < 1.71 D- x < 3.3 E- 5.2 < x F- 2.7 < x < 3.6 G- No. A z score of 2.29 implies that this RBC is normal.
Do you tailgate the car in front of you? About 30% of all drivers will tailgate before passing, thinking they can make the car in front of them go faster. Suppose that you are driving a considerable distance on a two-lane highway and are passed by 7 vehicles. (a) Compute the expected number of vehicles out of 7 that will tailgate. (b) Compute the standard deviation of this distribution.
A- 2.1 vehicles B- 1.212 vehicles
Consider the probability distribution shown below. x 0 1 2 P(x) 0.25 0.70 0.05 Compute the expected value of the distribution. Compute the standard deviation of the distribution.
A- 4/5 B- .5099
Assuming that the heights of college women are normally distributed with mean 63 inches and standard deviation 2.1 inches, answer the following questions (a) What percentage of women are taller than 63 inches? (b) What percentage of women are shorter than 63 inches? (c) What percentage of women are between 60.9 inches and 65.1 inches? (d) What percentage of women are between 58.8 and 67.2 inches?
A- 50% B- 50% C- 68% D- 95%
(a) What percentage of the area under the normal curve lies to the left of 𝜇? (b) What percentage of the area under the normal curve lies between 𝜇 − 𝜎 and 𝜇 + 𝜎? (c) What percentage of the area under the normal curve lies between 𝜇 − 3𝜎 and 𝜇 + 3𝜎?
A- 50% B- 68% C- 99.7%
The incubation time for a breed of chicks is normally distributed with a mean of 21 days and standard deviation of approximately 2 days. Look at the figure below and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch in the following time periods? (a) in 17 to 25 days (b) in 19 to 23 days (c) in 21 days or fewer (d) in 15 to 27 days
A- 950 chicks B- 680 chicks C- 500 chicks D- 997 chicks
According to the college registrar's office, 35% of students enrolled in an introductory statistics class this semester are freshmen, 25% are sophomores, 5% are juniors, and 35% are seniors. You want to determine the probability that in a random sample of five students enrolled in introductory statistics this semester, exactly two are freshmen. (a) Describe a trial. Can we model a trial as having only two outcomes? If so, what is success? What is failure? What is the probability of success? (b) We are sampling without replacement. If only 30 students are enrolled in introductory statistics this semester, is it appropriate to model 5 trials as independent, with the same probability of success on each trial? Explain. (c) What other probability distribution would be more appropriate in this setting?
A- A trial consists of looking at the class status of a student enrolled in introductory statistics. Yes we can model this trial with "freshman" being a success and "any other class" as a failure. .35 B- No. These trials are not independent. C- hypergeometric
In an experiment, there are n independent trials. For each trial, there are three outcomes, A, B, and C. For each trial, the probability of outcome A is 0.60; the probability of outcome B is 0.30; and the probability of outcome C is 0.10. Suppose there are 10 trials. (a) Can we use the binomial experiment model to determine the probability of four outcomes of type A, five of type B, and one of type C? Explain. (b) Can we use the binomial experiment model to determine the probability of four outcomes of type A and six outcomes that are not of type A? Explain. (c) What is the probability of success on each trial?
A- No. A binomial probability model applies to only two outcomes per trial. B- Yes. Assign outcome A to "success" and outcomes B and C to "failure." C- .6
A certain newspaper reports that about 50% of all prison parolees become repeat offenders. Alice is a social worker whose job is to counsel people on parole. Let us say success means a person does not become a repeat offender. Alice has been given a group of four parolees. (a) Find the probability P(r) of r successes ranging from 0 to 4. (b) What is the expected number of parolees in Alice's group who will not be repeat offenders? (c) what is the standard deviation?
A- P(0)=.0623 P(1)=.25 P(2)=.375 P(3)=.25 P(4)=.0623 B- 2 parolees C- 1 parolee
(a) In general, when the probability of success p is close to 0.5, would you say that the graph is more symmetrical or more skewed? (b) In general, when the probability of success p is close to 0.5, would you say that the graph is more symmetrical or more skewed?
A- Symmetrical B- When p is close to 1 it will be skewed left. When p is close to 0 it will be skewed right.
Consider a binomial distribution with n = 5 trials. Either use SALT, or use the probabilities given in the Binomial Probability Distribution table in the Appendix to make histograms showing the probabilities of r = 0, 1, 2, 3, 4, and 5 successes for each of the following. Comment on the skewness of each distribution. (a) The probability of success is p = 0.50. (b) The probability of success is p = 0.25. (c) The probability of success is p = 0.75. (d) What is the relationship between the distributions shown in parts (b) and (c)? (e) If the probability of success is p = 0.73, do you expect the distribution to be skewed to the right or to the left? Why?
A- This distribution is symmetric B- This distribution is skewed to the right. C- This distribution is skewed to the left D- The distributions are mirror images of each other. E- Skewed to the left, since p> 0.50.
Consider a binomial distribution of 200 trials with expected value 80 and standard deviation of about 6.9. Use the criterion that it is unusual to have data values more than 2.5 standard deviations above the mean or 2.5 standard deviations below the mean to answer the following questions. (a) Would it be unusual to have more than 120 successes out of 200 trials? Explain. (b) Would it be unusual to have fewer than 40 successes out of 200 trials? Explain. (c) Would it be unusual to have from 70 to 90 successes out of 200 trials? Explain.
A- Yes. 120 is more than 2.5 standard deviations above the expected value B- Yes. 40 is more than 2.5 standard deviations below the expected value. C- No. 70 to 90 observations is within 2.5 standard deviations of the expected value.
What was the age distribution of nurses in Great Britain at the time of Florence Nightingale? Suppose we have the following information. Note: In 1851 there were 25,466 nurses in Great Britain. Age range (yr) 20-29 30-39 40-49 50-59 60-69 70-79 80+ Midpoint x 24.5 34.5 44.5 54.5 64.5 74.5 84.5 Percent of nurses 5.7% 9.1% 19.4% 29.2% 25.5% 9.2% 1.9% (a) Using the age midpoints x and the percent of nurses, do we have a valid probability distribution? Explain. (b) Find the probability that a British nurse selected at random in 1851 would be 60 years of age or older. (c) Compute the expected age 𝜇 of a British nurse contemporary to Florence Nightingale. (d) Compute the standard deviation 𝜎 for ages of nurses shown in the distribution.
A- Yes. The events are distinct and the probabilities sum to 1. B- .366 C- 53.99 year D- 13.65 year
What is the income distribution of super shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon. In the following table, income units are in thousands of dollars, and each interval goes up to but does not include the given high value. The midpoints are given to the nearest thousand dollars. Income range 5-15 15-25 25-35 35-45 45-55 55 or more Midpoint x 10 20 30 40 50 60 Percent of super shoppers 20% 15% 20% 16% 19% 10% (a) Using the income midpoints x and the percent of super shoppers, do we have a valid probability distribution? Explain. (b) Compute the expected income 𝜇 of a super shopper.
A- Yes. The events are distinct and the probabilities sum to 1. B- 𝜇 = 32.9 thousands of dollars
What is the age distribution of promotion-sensitive shoppers? A supermarket super shopper is defined as a shopper for whom at least 70% of the items purchased were on sale or purchased with a coupon. Age range, years 18-28 29-39 40-50 51-61 62 and over Midpoint x 23 34 45 56 67 Percent of super shoppers 10% 41% 25% 9% 15% For the 62-and-over group, use the midpoint 67 years. (a) Using the age midpoints x and the percentage of super shoppers, do we have a valid probability distribution? Explain. (b) Compute the expected age 𝜇 of a super shopper. (c) Compute the standard deviation 𝜎 for ages of super shoppers.
A- Yes. The events are distinct and the probabilities sum to 1. B- 𝜇 = 42.58 yr C- 13.25
Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (a) x 0 1 2 P(x) 0.20 0.61 0.19 (b) x 0 1 2 P(x) 0.20 0.61 0.29
A- Yes. The probabilities sum to 1. B- No. The probabilities do not sum to 1.
Consider each distribution. Determine if it is a valid probability distribution or not, and explain your answer. (a) x 0 1 2 P(x) 0.20 0.64 0.16 (b) x 0 1 2 P(x) 0.20 0.64 0.17
A- Yes. The probabilities sum to 1. B- No. The probabilities do not sum to 1.
Which of the following are continuous variables, and which are discrete? (a) speed of an airplane (b) age of a college professor chosen at random (c) number of books in the college bookstore (d) weight of a football player chosen at random (e) number of lightning strikes in Rocky Mountain National Park on a given day
A- continuous B- continuous C- discrete D- continuous E- discrete
Which of the following are continuous variables, and which are discrete? (a) number of traffic fatalities per year in the state of Florida (b) distance a golf ball travels after being hit with a driver (c) time required to drive from home to college on any given day (d) number of ships in Pearl Harbor on any given day (e) your weight before breakfast each morning
A- discrete B- continuous C- continuous D- discrete E- continuous
Consider a binomial distribution with 10 trials. Either use SALT, or look at a binomial probability distribution table showing binomial probabilities for various values of p, the probability of success on a single trial. (a) For what value of p is the distribution symmetric? (b) What is the expected value of this distribution? (c) Is the distribution centered over the value? (d) For small values of p, is the distribution skewed right or left? (e) For large values of p, is the distribution skewed right or left?
A- p=.50 B- 5 C- yes D- right E- left
Consider a binomial distribution with n = 10 trials and the probability of success on a single trial p = 0.85. (a) Is the distribution skewed left, skewed right, or symmetric? (b) Compute the expected number of successes in 10 trials. (c) Given the high probability of success p on a single trial, would you expect P(r ≤ 3) to be very high or very low? Explain. (d) Very low. The expected number of successes in 10 trials is more than 3, and p is so high that it would be unusual to have so few successes in 10 trials.
A- skewed left B- 8.5 C- Very low. The expected number of successes in 10 trials is more than 3, and p is so high that it would be unusual to have so few successes in 10 trials. D- Very high. The expected number of successes in 10 trials is more than 8, and p is so high that it would be common to have 8 or more successes in 10 trials.
(a) The video suggests several ways to keep non-cash transactions secure. Which of the following did they not recommend? (b) The video states that 78% of people carry less than $50 in cash, 40% carry less than $20 in cash, and 9% carry no cash. This suggests that if you randomly choose a person and ask him how much cash he is carrying, the probability that he is carrying $50 or more is which of the following?
A- using a PIN number with a debit card B- 0.22
Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean 𝜇 = 29.0 kilograms and standard deviation 𝜎 = 3.4 kilograms. Let x be the weight of a fawn in kilograms. Convert the following x intervals to z intervals. (a) x < 30 (b) 19 < x (c) 32 < x < 35 Convert the following z intervals to x intervals. (d) −2.17 < z (e) z < 1.28 (f) −1.99 < z < 1.44 (g) If a fawn weighs 14 kilograms, would you say it is an unusually small animal? Explain using z values and the figure above. (h) If a fawn is unusually large, would you say that the z value for the weight of the fawn will be close to 0, −2, or 3? Explain.
A- z < .29 B- -2.94 < z C- .88 < z < 1.76 D- 21.6 < x E- x < 33.4 F- 22.2 < x < 33.9 G- Yes. This weight is 4.41 standard deviations below the mean; 14 kg is an unusually low weight for a fawn. H- It would have a large positive z, such as 3
Tree-ring dates were used extensively in archaeological studies at Burnt Mesa Pueblo. At one site on the mesa, tree-ring dates (for many samples) gave a mean date of 𝜇1 = year 1294 with standard deviation 𝜎1 = 30 years. At a second, removed site, the tree-ring dates gave a mean of 𝜇2 = year 1149 with standard deviation 𝜎2 = 43 years. Assume that both sites had dates that were approximately normally distributed. In the first area, an object was found and dated as x1 = year 1175. In the second area, another object was found and dated as x2 = year 1200. (a) Convert both x1 and x2 to z values, and locate both of these values under the standard normal curve of the figure above. (b) Which of these two items is the more unusual as an archaeological find in its location?
A- z1 =-3.97 z2 =1.19 B- x1; the further a z value is from zero, the more unusual it is.
The quality-control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of six syringes taken from the batch. Suppose the batch contains 1% defective syringes. (a) Find 𝜇 (b) What is the expected number of defective syringes the inspector will find? (c) What is the probability that the batch will be accepted? (d) Find 𝜎
A- 𝜇 = .06 syringes B- .06 syringes C- .999 D- 𝜎 = .244 syringes
USA Today reported that about 20% of all people in the United States are illiterate. Suppose you take eight people at random off a city street. (a) Find the mean and standard deviation of this probability distribution. (b) Find the expected number of people in this sample who are illiterate.
A- 𝜇 = 1.6 people 𝜎 = 1.131 people B- 1.6 people
A company is in the business of finding addresses of long-lost friends. The company claims to have a 80% success rate. Suppose that you have the names of six friends for whom you have no addresses and decide to use the company to track them. (a) Find the mean and standard deviation of this probability distribution. (b) What is the expected number of friends for whom addresses will be found?
A- 𝜇 = 4.8 friends 𝜎 = .98 friends B- 4.8 friends
Suppose 25% of the area under the standard normal curve lies to the left of z. Is z positive or negative?
Negative
Suppose 90% of the area under the standard normal curve lies to the right of z. Is z positive or negative?
Negative
Consider two discrete probability distributions with the same sample space and the same expected value. Are the standard deviations of the two distributions necessarily equal? Explain.
No, the individual probabilities may differ in a way that produces the same expected value but a different standard deviation.
In a binomial experiment, is it possible for the probability of success to change from one trial to the next? Explain.
No. A binomial experiment requires that the probability of success be the same for each trial.
The college physical education department offered an advanced first aid course last summer. The scores on the comprehensive final exam were normally distributed, and the z-scores for some of the students are shown below. If the mean score was 𝜇 = 148 with standard deviation 𝜎 = 17, what was the final exam score for each student?
Robert 168.74 Juan 175.54 Haley 117.06 Joel 148 Jan 135.08 Linda 178.6
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = 0.31 and z = 1.84 is .3454
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = 1.32 and z = 2.15 is .0776
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = −1.46 and z = 1.93 is .9011
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = −2.29 and z = 1.33 is .8972
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = 0 and z = −1.97 is .4756
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area between z = 0 and z = 2.46 is .4931
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the left of z = −0.40 is .3446
Sketch the area under the standard normal curve over the indicated interval and find the specified area.
The area to the left of z = −1.32 is .0934
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the left of z = 0.54 is .7054
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the left of z = 0.78 is .7823
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the right of z = −1.14 is .8729
Sketch the area under the standard normal curve over the indicated interval and find the specified area.
The area to the right of z = 0 is .5 The area to the left of z = 0 is .5
Sketch the area under the standard normal curve over the indicated interval and find the specified area
The area to the right of z = 1.63 is .0516
What does the expected value of a binomial distribution with n trials tell you?
The average number of successes
Consider two binomial distributions, with n trials each. The first distribution has a higher probability of success on each trial than the second. How does the expected value of the first distribution compare to that of the second?
The expected value is higher for the first distribution
Find z such that 70% of the standard normal curve lies to the left of z.
z= .52
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 100; 𝜎 = 19 P(x≥90) =.7019
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 2.1; 𝜎 = 0.33 P(x≥2)=.6179
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 28; 𝜎 = 4.2 P(x≥30) =.3156
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 5.1; 𝜎 = 1.7 P(7≤x≤9) =.1204
Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.
𝜇 = 8; 𝜎 = 2 P(7 ≤ x ≤ 10) = 0.5328
Look at the normal curve below, and find 𝜇, 𝜇 + 𝜎, and 𝜎.
𝜇=27 𝜇 + 𝜎=29 𝜎=2
