BANA Exam 2
General Hospital has noted that they admit an average of 8 patients per hour. Assuming a Poisson distribution, what is the probability that during the next two hours exactly 8 patients will be admitted?
0.012
General Hospital has noted that they admit an average of 8 patients per hour. Assuming a Poisson distribution, what is the probability that during the next hour less than 3 patients will be admitted?
0.014
The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. What is the probability that a randomly selected tire will have a life of of less than 30,000 miles?
0.023
The weight of football players for a team is normally distributed with a mean of 275 pounds and a standard deviation of 50 pounds. The probability of a player weighing more than 350 pounds is
0.067
An insurance company has determined that on average they receive nine claims per week at their Cincinnati office. Assume that the claims distribution can be described by a Poisson distribution. What is the probability that they will receive nine claims in a week?
0.132
The probability distribution for the number of goals the Lions soccer team makes per game is given below. Number of Goals Probability 0 0.05 1 0.15 2 0.35 3 0.30 4 0.15 What is the probability that in a given game the Lions will score less than 3 goals?
0.55
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample more than four are female?
0.594
The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. What proportion of the tires will have a life of 34,000 to 46,000 miles?
0.77
The number of customers that enter a store during one day is an example of
a discrete random variable
A normal distribution with a mean of 0 and a standard deviation of 1 is called
a standard normal distribution
A random variable that can assume only a finite number of values is referred to as a(n)
discrete random variable
A measure of the average value of a random variable is called a(n)
expected value
The probability distribution that can be described by just one parameter is the
exponential
To compute the probability that in a random sample of n elements, selected without replacement, we will obtain x successes, we would use the _____ probability distribution.
hypergeometric
The probability that a continuous random variable takes any specific value
is equal to zero
The function that defines the probability distribution of a continuous random variable is a
probability density function
The key difference between the binomial and hypergeometric distribution is that, with the hypergeometric distribution, the
probability of success changes from trial to trial.
Larger values of the standard deviation result in a normal curve that is
wider and flatter
The following table provides a probability distribution for the random variable y. y f(y) 2 0.30 4 0.10 7 0.40 8 0.20 (a) Compute E(y) E(y) = 5.4 (b) Compute Var(y) and σ. Var(y)=6.04 σ=2.46
y f(y) yf(y) 2 0.30 0.6 4 0.10 0.4 7 0.40 2.8 8 0.20 1.6 Total 1.0 5.4 E(y) = μ = 5.4 (b) y y − μ (y − μ)^2 f(y) (y − μ)^2f(y) 2 −3.4 11.56 0.30 3.468 4 −1.4 1.96 0.10 0.196 7 1.6 2.56 0.40 1.024 8 2.6 6.76 0.20 1.352 Total 6.04 Var(y) = 6.04 σ = radical 6.04 = 2.46
Males in the Netherlands are the tallest, on average, in the world with an average height of 183 centimeters (cm).† Assume that the height of men in the Netherlands is normally distributed with a mean of 183 cm and standard deviation of 10.5 cm. (a) What is the probability that a Dutch male is shorter than 176 cm? 0.2514 (b) What is the probability that a Dutch male is taller than 194 cm? 0.1469 (c) What is the probability that a Dutch male is between 175 and 191 cm? 0.5528 (d) Out of a random sample of 1,000 Dutch men, how many would we expect to be taller than 189 cm? 284
z= (x − μ)/σ= (176 − 183)/10.5 = −0.67 P(x ≤ 176) = P(z ≤ −0.67) = 0.2525 The probability that a Dutch male is shorter than 176 cm is 0.2525. (b) (194 − 183)/10.5 = 1.05 P(x ≥ 194) = 1 − P(x ≤ 194) = 1 − P(z ≤ 1.05) = 1 − 0.8526 = 0.1474 The probability that a Dutch male is taller than 194 cm is 0.1474. (c) For x = 191, (191 − 183)/10.5 = 0.76. For x = 175, (175 − 183)/10.5 = −0.76. P(175 ≤ x ≤ 191) = P(z ≤ 0.76) − P(z ≤ −0.76) = 0.7769 − 0.2231 = 0.5538 The probability that a Dutch male is between 175 cm and 191 cm is 0.5539. (d) (189 − 183)/10.5 = 0.57 P(x ≥ 189) = 1 − P(x ≤ 189) = 1 − P(z ≤ 0.57) = 1 − 0.7161 = 0.2839 So, we would expect about 0.2839(1,000) ≈ 284, Dutch men out of a random sample of 1,000, to be taller than 189 cm.
Suppose an electric-vehicle manufacturing company estimates that a driver who commutes 50 miles per day in a particular vehicle will require a nightly charge time of around 1 hour and 30 minutes (90 minutes) to recharge the vehicle's battery. Assume that the actual recharging time required is uniformly distributed between 70 and 110 minutes. (a) Give a mathematical expression for the probability density function of battery recharging time for this scenario. (b) What is the probability that the recharge time will be less than 101 minutes? (c) What is the probability that the recharge time required is at least 79 minutes? (Round your answer to four decimal places.) (d) What is the probability that the recharge time required is between 75 and 105 minutes?
(a) Length of Interval = 110 − 70 = 40 f(x) = 1/40 at a ≤ x ≤ b, and 0 elsewhere (b) P(x ≤ 101) = (101 − 70)(1/40)= 31/40= 0.775 (c) P(x ≥ 79) = (110 − 79)(1/40)=31/40=0.775 (d) P(75 ≤ x ≤ 105) = (105 − 75)(1/40)= 30/40= 0.75
Suppose that we have two events, A and B, with P(A) = 0.50, P(B) = 0.60, and P(A ∩ B) = 0.40. (a) Find P(A | B). (b) Find P(B | A). (c) Are A and B independent? Why or why not?
(a) P(A | B) = P(A ∩ B)/P(B)= 0.40/0.60 = 0.6667 (b) P(B | A) = P(A ∩ B)/P(A)= 0.40/0.50 = 0.80 (c) No, because P(A | B) ≠ P(A).
A recent survey examined the use of social media platforms. Suppose the survey found that there is a 0.64 probability that a randomly selected person will use Facebook and a 0.29 probability that a randomly selected person will use LinkedIn. In addition, suppose there is a 0.26 probability that a randomly selected person will use both Facebook and LinkedIn. (a) What is the probability that a randomly selected person will use Facebook or LinkedIn? (b) What is the probability that a randomly selected person will not use either social media platform?
(a) P(F ∪ L) = P(F) + P(L) − P(F ∩ L) = 0.64 + 0.29 − 0.26 = 0.67 (b) 1 − P(F ∪ L) = 1 − 0.67 = 0.33
A consulting firm submitted a bid for a large research project. The firm's management initially felt they had a 50-50 chance of getting the project. However, the agency to which the bid was submitted subsequently requested additional information on the bid. Past experience indicates that for 80% of the successful bids and 35% of the unsuccessful bids the agency requested additional information. (a) What is the prior probability of the bid being successful (that is, prior to the request for additional information)? (b) What is the conditional probability of a request for additional information given that the bid will ultimately be successful? (c) Compute the posterior probability that the bid will be successful given a request for additional information.
(a) P(S1) = 0.50 (b) P(B | S1) = 0.80 (c) P(S1 | B) = (0.50)(0.80)/ (0.50)(0.80) + (0.50)(0.35)= 0.4/0.575 = 0.70
Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. (a) The process standard deviation is 0.08, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.92 or greater than 10.08 ounces will be classified as defects. Calculate the probability of a defect. Calculate the expected number of defects for a 1,000-unit production run. (b) Through process design improvements, the process standard deviation can be reduced to 0.04. Assume the process control remains the same, with weights less than 9.92 or greater than 10.08 ounces being classified as defects. Calculate the probability of a defect. Calculate the expected number of defects for a 1,000-unit production run. (c) What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?
(a) P(defect) = 1 − P(9.92 ≤ x ≤ 10.08) = 1 − P(−1.00 ≤ z ≤ 1.00) = 1 − 0.6827 = 0.3173 expected number of defects = 1,000(0.3173) = 317 defects (b) P(defect) = 1 − P(9.92 ≤ x ≤ 10.08) = 1 − P(−2.00 ≤ z ≤ 2.00) = 1 − 0.9545 = 0.0455 expected number of defects = 1,000(0.0455) = 46 defects (c) Reducing the process standard deviation causes a substantial reduction in the number of defects.
The mean time between failures (MTBF) is a common metric used to measure the performance of manufacturing systems. MTBF is the elapsed time between failures of a system during normal operations. The failures could be caused by broken machines or computer errors, among other failures. Suppose that the MTBF for a new automated manufacturing system follows an exponential distribution with a mean of 12.4 hours. (a) What is the probability that the automated manufacturing system runs for more than 18 hours without a failure? 0.2342 (b) What is the probability that the automated manufacturing system runs for 9 or fewer hours before failure? 0.5161 (c) What is the probability that the automated manufacturing system runs for more than 6 hours but less than 11 hours before a failure? 0.2045
(a) P(x > 18) = 1 − P(x ≤ 18) = 1 − (1 − e^(−18/12.4)) = 1 − 0.7658 = 0.2342 (b) P(x ≤ 9) = 1 − e^(−9/12.4) = 0.5161 (c) P(x ≤ 6) = 1 − e^(−6/12.4) = 0.3836 P(x ≤ 11) = 1 − e^(−11/12.4) = 0.5882 The probability that the system runs between 6 and 11 hours before a failure is 0.5882 − 0.3836 = 0.2045.
Given that z is a standard normal random variable, compute the following probabilities. (a) P(z ≤ −2.0) 0.0228 (b) P(z ≥ −2) 0.9773 (c) P(z ≥ −1.6) 0.9452 (d) P(−2.6 ≤ z) 0.9953 (e) P(−3 < z ≤ 0) 0.4987
(a) P(z ≤ −2) = 0.0228 (b) P(z ≥ −2) = 1 − P(z < −2) = 1 − 0.0228 = 0.9772 (c) P(z ≥ −1.6) = 1 − P(z < −1.6) = 1 − 0.0548 = 0.9452 (d) P(−2.6 ≤ z) = 1 − P(z < −2.6) = 1 − 0.0047 = 0.9953 (e) P(−3 < z ≤ 0) = P(z ≤ 0) − P(z ≤ −3) = 0.5000 − 0.0013 = 0.4987
Given that z is a standard normal random variable, compute the following probabilities. (a) P(−1.98 ≤ z ≤ 0.45) 0.6498 (b) P(0.52 ≤ z ≤ 1.24) 0.1940 (c) P(−1.75 ≤ z ≤ −1.04) 0.1091
(a) P(−1.98 ≤ z ≤ 0.45) = P(z ≤ 0.45) − P(z < −1.98) = 0.6736 − 0.0239 = 0.6498 (b) P(0.52 ≤ z ≤ 1.24) = P(z ≤ 1.24) − P(z < 0.52) = 0.8925 − 0.6985 = 0.1940 (c) P(−1.75 ≤ z ≤ −1.04) = P(z ≤ −1.04) − P(z < −1.75) = 0.1492 − 0.0401 = 0.1091
Given that z is a standard normal random variable, find z for each situation. (a) The area to the left of z is 0.2119. -0.80 (b) The area between −z and z is 0.9232. 1.77 (c) The area between −z and z is 0.2128. 0.27 (d) The area to the left of z is 0.9948. 2.56 (e) The area to the right of z is 0.6179. -0.3
(a) The z value corresponding to a cumulative probability of 0.2119 is z = −0.80. (b) Compute 0.9232/2= 0.4616;z corresponds to a cumulative probability of 0.5000 + 0.4616 = 0.9616. So z = 1.77. (c) Compute 0.2128/2= 0.1064;z corresponds to a cumulative probability of 0.5000 + 0.1064 = 0.6064. So z = 0.27. (d) The z value corresponding to a cumulative probability of 0.9948 is z = 2.56. (e) The area to the left of z is 1 − 0.6179 = 0.3821. So z = −0.30.
Intensive care units (ICUs) generally treat the sickest patients in a hospital. ICUs are often the most expensive department in a hospital because of the specialized equipment and extensive training required to be an ICU doctor or nurse. Therefore, it is important to use ICUs as efficiently as possible in a hospital. Suppose that a large-scale study of elderly ICU patients shows that the average length of stay in the ICU is 3.8 days. Assume that this length of stay in the ICU has an exponential distribution. (a) What is the probability that the length of stay in the ICU is one day or less? (b) What is the probability that the length of stay in the ICU is between two and three days? (c) What is the probability that the length of stay in the ICU is more than five days?
(a) f(x) = (1/μ)e^(−x/μ) = (1/38)e^(−x/3.8) for x ≥ 0 P(x ≤ x0) = 1 − e(−x0/μ) = 1 − e^(−x0/3.8) P(x ≤ 1) = 1 − e^(−1/3.8) = 1 − 0.7686 = 0.2314 (b) P(x ≤ 3) = 1 − e^(−3/3.8) = 1 − 0.4541 = 0.5459 P(x ≤ 2) = 1 − e^(−2/3.8) = 1 − 0.5908 = 0.4092 P(2 ≤ x ≤ 3) = P(x ≤ 3) − P(x ≤ 2) = 0.5459 − 0.4092 = 0.1367 (c) P(x > 5) = 1 − P(x ≤ 5) = 1 − (1 − e^(−5/3.8)) = e^(−5/3.8) = 0.2683
Intensive care units (ICUs) generally treat the sickest patients in a hospital. ICUs are often the most expensive department in a hospital because of the specialized equipment and extensive training required to be an ICU doctor or nurse. Therefore, it is important to use ICUs as efficiently as possible in a hospital. Suppose that a large-scale study of elderly ICU patients shows that the average length of stay in the ICU is 3.3 days. Assume that this length of stay in the ICU has an exponential distribution. (a) What is the probability that the length of stay in the ICU is one day or less? 0.2614 (b) What is the probability that the length of stay in the ICU is between two and three days? 0.1426 (c) What is the probability that the length of stay in the ICU is more than five days? 0.2198
(a) f(x) = (1/μ)e^(−x/μ) = (1/3.3)e^(−x/3.3) for x ≥ 0 P(x ≤ x0) = 1 − e^(−x0/μ) = 1 − e^(−x0/3.3) P(x ≤ 1) = 1 − e^(−1/3.3) = 1 − 0.7386 = 0.2614 (b) P(x ≤ 3) = 1 − e^(−3/3.3) = 1 − 0.4029 = 0.5971 P(x ≤ 2) = 1 − e^(−2/3.3) = 1 − 0.5455 = 0.4545 P(2 ≤ x ≤ 3) = P(x ≤ 3) − P(x ≤ 2) = 0.5971 − 0.4545 = 0.1426 (c) P(x > 5) = 1 − P(x ≤ 5) = 1 − (1 − e^(−5/3.3)) = e(−5/3.3) = 0.2198
A random variable is normally distributed with a mean of μ = 50 and a standard deviation of σ = 10. (b) What is the probability the random variable will assume a value between 40 and 60? 0.683 (c) What is the probability the random variable will assume a value between 20 and 80? 0.997
(b) 0.683 since 40 and 60 are within plus or minus one standard deviation from the mean of 50. (Use the Standard Normal Distribution Table or the characteristic of the normal distribution that 68.3% of the values of a normal random variable are within plus or minus one standard deviation of its mean.) (c) 0.997 since 20 and 80 are within plus or minus three standard deviations from the mean of 50. (Use the Standard Normal Distribution Table or the characteristic of the normal distribution that 99.7% of the values of a normal random variable are within plus or minus three standard deviations of its mean.)
Most computer languages include a function that can be used to generate random numbers. In Excel, the RAND function can be used to generate random numbers between 0 and 1. If we let x denote a random number generated using RAND, then x is a continuous random variable with the following probability density function. f(x) = 1 for 0 ≤ x ≤ 1 0 elsewhere (b) What is the probability of generating a random number between 0.35 and 0.75? 0.4 (c) What is the probability of generating a random number with a value less than or equal to 0.20? 0.2 (d) What is the probability of generating a random number with a value greater than 0.70? 0.3
(b) P(0.35 < x < 0.75) = 1(0.40) = 0.40 (c) P(x ≤ 0.20) = 1(0.20) = 0.20 (d) P(x > 0.70) = 1(0.30) = 0.30
The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 10 seconds. (b) What is the probability that the arrival time between vehicles is 10 seconds or less? (c) What is the probability that the arrival time between vehicles is 6 seconds or less? (d) What is the probability of 34 or more seconds between vehicle arrivals?
(b) P(x ≤ 10) = 1 − e^−10/10 = 1 − 0.3679 = 0.6321 (c) P(x ≤ 6) = 1 − e^−6/10 = 1 − 0.5488 = 0.4512 (d) P(x ≥ 34) = 1 − P(x < 34) = 1 − (1 − e^−34/10) = 0.0334
The time between arrivals of vehicles at a particular intersection follows an exponential probability distribution with a mean of 12 seconds. (b) What is the probability that the arrival time between vehicles is 12 seconds or less? 0.6321 (c) What is the probability that the arrival time between vehicles is 6 seconds or less? 0.3935 (d) What is the probability of 30 or more seconds between vehicle arrivals? 0.0821
(b) P(x ≤ 12) = 1 − e^−12/12 = 1 − 0.3679 = 0.6321 (c) P(x ≤ 6) = 1 − e^−6/12 = 1 − 0.6065 = 0.3935 (d) P(x ≥ 30) = 1 − P(x < 30) = 1 − (1 − e^−30/12) = 0.0821
Delta Airlines quotes a flight time of 5 hours, 4 minutes for a particular flight. Suppose we believe that actual flight times are uniformly distributed between 5 hours and 5 hours, 16 minutes. (b) What is the probability that the flight will be no more than 4 minutes late? 0.5 (c) What is the probability that the flight will be more than 8 minutes late? 0.25 (d) What is the expected flight time, in minutes? 308
(b) P(x ≤ 308) = (1/16)(308 − 300) = 0.5 (c) P(x > 312) = (1/16)(316 − 312) = 0.25 (d) E(x) = (300 + 316)/2 = 308 min
(b) Compute the probability of one success, f(1). f(1) = 0.32 (c) Compute f(0). f(0) = 0.64 (d) Compute f(2). f(2) = 0.04 (e) Compute the probability of at least one success. 0.36 (f) Compute the expected value, variance, and standard deviation. expected value 0.4 variance 0.32 standard deviation 0.5657
(b) f(1) = 2 1 (0.2)^1(0.8)^1 = (2!)/(1!1!) (0.2)(0.8) = 0.32 (c) f(0) = 2 0 (0.2)^0(0.8)^2 = (2!)/(0!2!) (1)(0.64) = 0.64 (d) f(2) = 2 2 (0.2)^2(0.8)^0 = (2!)/(2!0!) (0.04)(1) = 0.04 (e) P(x ≥ 1) = f(1) + f(2) = 0.32 + 0.04 = 0.36 (f) E(x) = np = 2(0.2) = 0.4 Var(x) = np(1 − p) = 2(0.2)(0.8) = 0.32 σ = radical 0.32 = 0.5657
The random variable x is known to be uniformly distributed between 1.0 and 1.5. (b) Compute P(x = 1.25). (c) Compute P(1.0 ≤ x ≤ 1.25). (d) Compute P(1.3 < x < 1.5).
(b) 0, The probability of any single point is zero since the area under the curve above any single point is zero. (c) 0.5 2(0.25) = 0.50 (d) 0.4 2(0.20) = 0.40
The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product in less than 6 minutes is
0
A company that manufactures toothpaste is studying five different package designs. Assuming that one design is just as likely to be selected by a consumer as any other design, what selection probability would you assign to each of the package designs? We would assign a probability of _____ Design Number of Times Preferred 1 10 2 15 3 25 4 30 5 20 Do the data confirm the belief that one design is just as likely to be selected as another? Explain.
0.2 for all designs No, a probability of about 0.20 would be assigned using the relative frequency method if selection is equally likely.
Suppose you first randomly sample one card from a deck of 52. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that at least one ace will be in the three sampled cards? Four of the 52 cards in the deck are aces.
0.217
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample exactly four are female?
0.232
The time required to load a truck is exponentially distributed with a mean of 15 minutes. What is the probability that a truck will be loaded in 10 to 20 minutes?
0.25
The time required to load a truck is exponentially distributed with a mean of 15 minutes. What is the probability that a truck will be loaded in 20 minutes or more?
0.264
An experiment with three outcomes has been repeated 50 times, and it was learned that E1 occurred 15 times, E2 occurred 13 times, and E3 occurred 22 times. Assign probabilities to the outcomes. P(E1)= P(E2)= P(E3)= What method did you use?
0.3 0.26 0.44 Relative Frequency Method
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample at most four are female?
0.406
The time required to load a truck is exponentially distributed with a mean of 15 minutes. What is the probability that a truck will be loaded in 10 minutes or less?
0.487
The ages of students at a university are normally distributed with a mean of 21. What proportion of the student body is at least 21 years old?
0.5
The assembly time for a product is uniformly distributed between 6 to 10 minutes. The probability of assembling the product between 7 to 9 minutes is
0.5
The probability distribution for the daily sales at Michael's Co. is given below. (In 1000s) Probability 40 0.1 50 .04 60 .03 70 .23 The probability of having sales of at most $50,000 is
0.5
Forty percent of the students who enroll in a statistics course go to the statistics laboratory on a regular basis. Past data indicates that 65% of those students who use the lab on a regular basis make a grade of A in the course. On the other hand, only 10% of students who do not go to the lab on a regular basis make a grade of A. If a particular student made an A, determine the probability that she or he used the lab on a regular basis. Enter your answer rounded to four decimal places. Use a tree diagram to answer this question.
0.8125
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample at least four are female?
0.826
The life expectancy of computer terminals is normally distributed with a mean of 4 years and a standard deviation of 1 year. What is the probability that a terminal will last 5 years or less?
0.841
A class of 50 consists of 60% female students. A random sample of 8 students is selected. What is the probability that among the students in the sample at most four are male?
0.847
The weight of football players for a team is normally distributed with a mean of 275 pounds and a standard deviation of 50 pounds. The probability of a player weighing between 200 and 350 pounds is
0.866
The probability distribution for the number of goals the Lions soccer team makes per game is given below. Number of Goals Probability 0 0.05 1 0.15 2 0.35 3 0.3 4 0.15 What is the probability that in a given game the Lions will score at least 1 goal?
0.95
(a) What is the z-score for an upper-tail probability of 0.10? (b) What is the probability for the region −1.75 ≤ z ≤ 1.5?
1.282 0.893 (a) Enter the value of 0.10 into the box for p in the graphic. This will give you the z-score of −1.282 for a lower-tail probability of 10%. The z-score for the upper-tail probability is z = 1.282. Alternatively, you could enter the value of 1 − 0.1 = 0.9 into the box for p to find the positive z-score directly. (b) To find the probability for the region −1.75 ≤ z ≤ 1.5, first find the probability for z = −1.75 by entering this value into the box for z. You should get p = 0.04. Next, enter the value of z = 1.5 into the box for z. You will get p = 0.933. The probability for the total region is the difference: 0.933 − 0.04 = 0.893.
Do you think global warming will have an impact on you during your lifetime? A poll of 1,000 adults asked this question. Consider the responses by age groups shown below. Response Age 18-29 30+ Yes 140 287 No 137 426 Unsure 8 2 (a) What is the probability that a respondent 18-29 years of age thinks that global warming will not pose a serious threat during his/her lifetime? (Round your answer to four decimal places.) (b) What is the probability that a respondent 30+ years of age thinks that global warming will not pose a serious threat during his/her lifetime? (Round your answer to four decimal places.) (c) For a randomly selected respondent, what is the probability that a respondent answers yes? (d) Based on the survey results, does there appear to be a difference between ages 18-29 and 30+ regarding concern over global warming?
137/ (140 + 137 + 8) = 0.4807 (b) probability = 426/ (287 + 426 + 2) = 0.5958 (c) probability = (140 + 287)/ (285 + 715) = 0.427 (a) 0.4807 (b) 0.5958 (c) 0.427 (d) It appears that older respondents are less concerned about global warming being a threat in their lifetime than are younger respondents.
The weight of football players is normally distributed with a mean of 275 pounds and a standard deviation of 50 pounds. What is the minimum weight of the middle 95% of the players?
177
Random variable x has the probability function f(x) = X/6, for x = 1, 2 or 3 The expected value of x is
2.333
The probability distribution for the number of goals the Lions soccer team makes per game is given below. Number of Goals Probability 0 0.05 1 0.15 2 0.35 3 0.30 4 0.15 The expected number of goals per game is
2.35
How many ways can three items be selected from a group of six items?
20 6 3 = (6!)/ (3!3!)= (6 · 5 · 4 · 3 · 2 · 1)/ (3 · 2 · 1)(3 · 2 · 1) = 20
The student body of a large university consists of 60% female students. A random sample of 8 students is selected. What is the expected number of females selected?
4.8
An experiment has three steps with four outcomes possible for the first step, two outcomes possible for the second step, and five outcomes possible for the third step. How many experimental outcomes exist for the entire experiment?
40
"DRUGS R US" is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a mean of 6 ounces and a standard deviation of 0.3 ounces. Assume the contents of the bottles are normally distributed. Ninety-five percent of the bottles will contain at least how many ounces?
5.51
Simple random sampling uses a sample of size n from a population of size N to obtain data that can be used to make inferences about the characteristics of a population. Suppose that, from a population of 52 bank accounts, we want to take a random sample of four accounts in order to learn about the population. How many different random samples of four accounts are possible?
52 4= (52!)/ (4!48!)= (52 · 51 · 50 · 49)/ (4 · 3 · 2 · 1) = 270,725
In a binomial experiment consisting of five trials, the number of different values that x (the number of successes) can assume is 10 6 2 5
6
"DRUGS R US" is a large manufacturer of various kinds of liquid vitamins. The quality control department has noted that the bottles of vitamins marked 6 ounces vary in content with a mean of 6 ounces and a standard deviation of 0.3 ounces. Assume the contents of the bottles are normally distributed. Ninety-five percent of the bottles will contain less than how many ounces?
6.49
The assembly time for a product is uniformly distributed between 6 to 10 minutes. The expected assembly time (in minutes) is
8
A professor at a local community college noted that the grades of his students were normally distributed with a mean of 74 and a standard deviation of 10. The professor has informed us that 6.3 percent of his students received A's while only 2.5 percent of his students failed the course and received F's. What is the minimum score needed to earn an A?
89.3
A survey of magazine subscribers showed that 45.6% rented a car during the past 12 months for business reasons, 56% rented a car during the past 12 months for personal reasons, and 32% rented a car during the past 12 months for both business and personal reasons. (a) What is the probability that a subscriber rented a car during the past 12 months for business or personal reasons? (b) What is the probability that a subscriber did not rent a car during the past 12 months for either business or personal reasons?
Let M = rented a car for business reasons N = rented a car for personal reasons. (a) P(M ∪ N) = P(M) + P(N) − P(M ∩ N) = 0.456 + 0.56 − 0.32 = 0.696 (b) P(neither) = 1 − 0.696 = 0.304
Business Engineering Other Total Full Time 352 195 253 800 Part-Time 151 163 196 510 Totals 503 358 449 1,310 (b) Use the marginal probabilities of undergraduate major (business, engineering, or other) to comment on which undergraduate major produces the most potential MBA students. From the marginal probabilities, we can tell that business (c) If a student intends to attend classes full-time in pursuit of an MBA degree, what is the probability that the student was an undergraduate engineering major? 0.244 (d) If a student was an undergraduate business major, what is the probability that the student intends to attend classes full-time in pursuit of an MBA degree? 0.701 (e) Let A denote the event that the student intends to attend classes full-time in pursuit of an MBA degree, and let B denote the event that the student was an undergraduate business major. Are events A and B independent? Justify your answer. P(A)P(B) = 0.235 P(A ∩ B) = 0.269
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values. (a) Undergraduate Major Business Engineering Other Totals Full-Time 0.269 0.149 0.193 0.611 Part-Time 0.115 0.124 0.150 0.389 Totals 0.384 0.273 0.343 1.000 (b) P(business) = 0.384, P(engineering) = 0.273, P(other) = 0.343, business (c) P(engineering | full-time) = P(engineering ∩ full-time)/ P(full-time)= 0.149/0.611 = 0.244 (d) P(full-time | business) = P(full-time ∩ business)/ P(business)= 0.269/0.384 = 0.700 (e) For independence, we must have that P(A)P(B) = P(A ∩ B); from the joint probability table in part (a) of this problem, we have P(A) = 0.611 P(B) = 0.384. So P(A)P(B) = (0.611)(0.384) = 0.234, but P(A ∩ B) = 0.269. Because P(A)P(B) ≠ P(A ∩ B), the events are not independent.
How many permutations of three items can be selected from a group of six?
P3^6 = (6!)/ (6 − 3)! = (6)(5)(4) = 120
In the textile industry, a manufacturer is interested in the number of blemishes or flaws occurring in each 100 feet of material. The probability distribution that has the greatest chance of applying to this situation is the
Poisson distribution
Consider the experiment of selecting a playing card from a deck of 52 playing cards. Each card corresponds to a sample point with a 1/52 probability. (a) List the sample points in the event an ace is selected. (b) List the sample points in the event a club is selected. (c) List the sample points in the event a face card (jack, queen, or king) is selected. (d) Find the probabilities associated with each of the events in parts (a), (b), and (c). (Enter your probabilities as fractions.)
S = {ace of clubs, ace of diamonds, ace of hearts, ace of spades} S = {2 of clubs, 3 of clubs, ..., 10 of clubs, jack of clubs, queen of clubs, king of clubs, ace of clubs} S = {jack of clubs, jack of diamonds, jack of hearts, jack of spades, queen of clubs, queen of diamonds, queen of hearts, queen of spades, king of clubs, king of diamonds, king of hearts, king of spades} 4/52 0.25 12/52
An experiment consists of determining the speed of automobiles on a highway by the use of radar equipment. The random variable in this experiment is a
continuous random variable
The weight of an object is an example of
continuous random variable
A numerical description of the outcome of an experiment is called a
random variable
You may need to use the appropriate appendix table to answer this question. The average return for large-cap domestic stock funds over three years was 14.4%. Assume the three-year returns were normally distributed across funds with a standard deviation of 4.4%. (a) What is the probability an individual large-cap domestic stock fund had a three-year return of at least 19%? (b) What is the probability an individual large-cap domestic stock fund had a three-year return of 10% or less? (c) How big does the return have to be to put a domestic stock fund in the top 15% for the three-year period?
μ = 14.4 and σ = 4.4. (a) At x = 19, z = (19 − 14.4)/(4.4)= 1.05 P(z ≤ 1.05) = 0.8521 P(x ≥ 19) = 1 − 0.8521 = 0.1479 (b) At x = 10, z = (10 − 14.4)/(4.4)= −1.00. P(z ≤ −1.00) = 0.1587 So, P(x ≤ 10) = 0.1587. (c) A z-value of 1.04 cuts off an area of approximately 15% in the upper tail. x = 14.4 + 4.4(1.04) = 18.96 A return of 18.96% or higher will put a domestic stock fund in the top 15%.