Exam 2

For this question, use a Binomial Probability Distribution. Assume all conditions are met. According to data from the U.S. Social Survey, the probability that an adult was never in a museum is 18%. You may safely assume that one randomly selected adult having never been in a museum does not change the chances of another adult having never been in a museum. (a) If 12 adults are randomly surveyed, what is the probability that 4 of them have never been in a museum? Type your answer as a proportion (not a percentage) rounded to the third decimal place. (b) If 12 adults are randomly surveyed, what is the probability that 3 or fewer of them have never been in a museum? Type your answer as a proportion (not a percentage) rounded to the third decimal place. (c) What is the expected value (mean) of the number of adults who have never been in a museum out of the 12 adults surveyed? Type your answer as a value rounded to the first decimal place.

(a) To compute an "exact" probability for a binomial distribution, use binompdf with trials: 12, p: 0.18, x value: 4, which gives a probability of 0.106 (rounded to the third decimal place). (b) To compute a probability for "3 or fewer," use binomcdf, which computes the probability of "x or fewer." The inputs for the binomcdf command are trials: 12, p: 0.18, x value: 3, which gives a probability of 0.845 (rounded to the third decimal place). (c) The formula for the expected value (mean) of a binomial distribution is given on the formula sheet = np = 12(0.18) = 2.16 (which rounds to 2.2 to the first decimal place).

The data in the table are based on the results of a survey comparing the commute time of adults to their score on a​ well-being test. Commute Time (in minutes) 7 16 24 37 51 70 97 Well-Being Score 69.5 68.2 67.5 67.1 66.5 65.2 63.8 (a) Find the linear correlation coefficient for the two variables, Commute Time (x) and Well-Being Score (y): (round the answer to the third decimal place) (b) Find the correct critical value to determine if there is a significant linear relationship between the two variables: (round the answer to the third decimal place) (c) Since |r| is (type "less" or "greater") than CV, there (type "is" or "is not") a significant linear relationship between Commute Time (x) and Well-Being Score (y). (d) Using Commute Time (x) and Well-Being Score (y), find the equation of the least-squares regression: y^y^ = x + . (round each answer to the third decimal place) (e) If a randomly selected commuter has a Commute Time of 15 minutes, the least-squares regression equation predicts that the Well-Being Score will be: (round the answer to the third decimal place)

(a) To compute the correlation coefficient, use Stat>Edit to enter the Commute Time values in L1 and the Well-Being Scores in L2. Then calculate Stat>Calc>LinReg(ax+b) with L1 as the XList, L2 as the YList, and leaving FreqList blank. The output will give you the correlation coefficient r = -0.986. (b) The critical value comes from the table on the Formula Sheet in the "Linear Correlation and Regression" section. Since there are n=7 pairs of data values, CV=0.754. (c) There is a significant linear relationship between X and Y if the correlation coefficient is stronger (closer to +1 or -1) than the critical value. That's why we look at the ABSOLUTE VALUE of r, not just the r value itself. Since |r| = +0.986 is greater than 0.754, the critical value, there is a significant linear relationship. -2 if work shown and less, is not matches (just misunderstood negative) (d) The LinReg(ax+b) output also gives you the values for slope (a = -0.058) and intercept (b = 69.346). (e) To predict, plug the given value for X (Commute Time = 15) into the regression equation: y = -0.058(15) + 69.346 = 68.476.

Determine if the random variable from the experiment follows a Binomial Distribution. Three cards are selected from a standard​ 52-card deck without replacement. The number of clubs selected is recorded. 1. There there are two mutually exclusive outcomes (success/failure). [ Select ] ["FALSE", "TRUE"] 2. Since a sample size of 3 is more than 5% of the cards in the deck, the trials are independent. [ Select ] ["FALSE", "TRUE"] 3. There is a fixed number of trials. [ Select ] ["TRUE", "FALSE"] 4. The probability of success is the same for each trial of the experiment. The random variable follows a Binomial Distribution. [ Select ] ["TRUE", "FALSE"]

1. TRUE--Since we are just checking whether a card is a club, there are two mutually exclusive outcomes (club/not a club). 2. TRUE--If n ≤ (LESS than or equal to) 0.05N, we can assume the trials are independent. Since 3 is more than 5% of 52, the trials are NOT independent. 3. TRUE--Since n=3 cards are selected, the number of trials is fixed. 4. TRUE--Since the cards are being selected WITHOUT replacement, the probability of drawing a club will change each time you select a card, so the probability will not be the same from card to card. Since the 2nd and 4th requirements are not met, this is NOT binomial.

The Environmental Protection Agency must inspect 11 factories for complaints of water pollution. How many different itineraries could an inspector use to visit 5 of these factories this week to investigate?

55,440-- An itinerary is a schedule in a particular order. Since order matters, use a permutation.

An insurance company earns $530 for each individual who purchases its life insurance policy. If the individual dies, the insurance company must pay out $250,000. The probability that an individual will survive is 0.99862. Fill in the probability distribution with the correct values. events are Individual survives or dies x=Value of the company $530 or ________ P(x)=_________ or 0.00138 What is the expected value of the policy to the company? [ Select ] ["$185", "0.99862", "0.000138", "$250000", "-$249470", "$530", "$9280.68"] What is the standard deviation of the value of the policy to the company?

In the x column, you put the values to the company. If an individual survives, the company gets $530 and doesn't have to pay anything, so the value is $530. If an individual dies, the company gets $530, but has to pay $250000, so the value = $530 - $250000 = -$249470. In the P(x) column, you put the probabilities associated with each event. The probability that an individual survives is given as 0.99862. The probability than an individual dies is the same as P(NOT survives) = 1 - 0.99862 = 0.00138. To compute the expected value, enter the x values in L1 and the P(x) values in L2. Then compute Stat>Calc>1-Var Stats(List:L1, FreqList:L2). This will give you the expected value (mean="x-bar") of $185 and standard deviation (Sx) of $9280.68.

High Uinta Bicycles makes three mountain bike models that each come in four colors. The following table shows the production volumes for last week: Blue Frame 22 12 4 Green Frame 28 18 10 Black Frame 16 22 8 Red Frame 32 14 6 Duchesne 22 28 16 32 Summit 12 18 22 14 Kamas 4 10 8 6 Find the marginal frequency for Summit bicycles: Find the marginal frequency for bikes with green frames: What proportion of bikes with green frames are the Summit style? Type your answer as a proportion (not a percent) rounded to the third decimal place.

Marginal frequencies are row or column totals (whole numbers) for the given category, not proportions (decimal values between 0 and 1). For Summit bicycles, add up all the values in the "Summit" row = 12+18+22+14 = 66. For bikes with green frames, add up all the values in the "Green frame" column = 28+18+10 = 56. The key word "of" tells you which category to condition on. To find the proportion OF BIKES WITH GREEN FRAMES that are the Summit style, look in just the "Green frame" column. In that column, 18 are the Summit style and a total of 56 in the column, so compute 18/56 = 0.321.

Because colas tend to replace healthier beverages and colas contain caffeine and phosphoric​ acid, researchers wanted to know whether cola consumption is associated with lower bone mineral density in women. Assume that the least-squares regression equation for bone mineral density (yy), based on the average number of colas per week (xx), is y^=−0.025x+0.900y^=−0.025x+0.900. What is the appropriate interpretation of the slope of the least-squares regression line? [ Select ] ["For each additional cola per week, we would expect the average bone density to decrease by 0.025.", "If a woman drank no colas, on average she would expect a bone density of about 0.9.", "For each additional cola per week, the average bone density will increase about 0.9.", "Because the value of the slope is negative, we expect a weak linear relationship."] What is the appropriate interpretation of the yy-intercept of the least-squares regression line? [ Select ] ["Because the value of the intercept is small, we will expect a weak correlation as well.", "For each additional cola per week, we would expect the average bone density to decrease by 0.025.", "For women who do not drink cola, the average bone density is about 0.9.", "If a woman drank no colas, on average she would expect a bone density of about -0.025."]

Slope is interpreted as the average, or expected change in Y, when X increases by one unit. The slope is the coefficient of the x term in the equation, so it's -0.025 here, which means Y=bone mineral density is expected to decrease (since it's negative) by 0.25 whenever X=the number of colas per week increases by one. The y-intercept gives the average, or expected, value for Y when X=0. For this regression equation, the y-intercept is 0.900, which is the average value for Y=bone mineral density when X=number of colas per week is zero.

You have a bag filled with 6 yellow tulip bulbs, 4 pink tulip bulbs, and 5 red tulip bulbs. i. What is the probability of selecting a yellow or red bulb? In other words, find P(YelloworRed)P(YelloworRed)? ii. What is the probability of selecting two bulbs; first you select a yellow bulb, plant it, then select a pink bulb and then plant it? In other words, find P(YellowandPink)P(YellowandPink)? iii. Find P(PinkgivenRed)P(PinkgivenRed). In other words, assume that a red bulb has already been planted. Out of the bulbs remaining, what is the probability that you will select a pink bulb? [ Select ]

i. For "or" probabilities," use the Addition Rule. Since "yellow" and "red" are disjoint and there are 15 total bulbs, we can just add P(Yellow) + P(Red) = 6/15 + 5/15 = 0.73. ii. For "and" probabilities, use the Multiplication Rule. Since you are planting the first bulb, this is selecting WITHOUT replacement, which means the event of selecting a yellow and then a pink bulb are dependent. We must compute P(yellow)*P(pink|yellow), where P(pink|yellow) is the probability of selecting one of the 4 pink bulbs from the remaining 14 bulbs after a yellow is already planted. The answer is (6/15)(4/14) = 0.11. iii. If a red bulb has already been planted, we know that there are only 14 remaining bulbs, but all 4 pink bulbs are still available. The probability is 4/14 = 0.29.

The General Social Survey asked, "Are you more likely to be satisfied with your automobile purchase when it is new or used?" The results are given in the table below. Not too satisfied 11 25 36 Pretty satisfied 72 58 130 Extremely satisfied 120 80 200 TOTAL 203 163 366 New cars 11 72 120 203 Used cars 25 58 80 163 TOTAL 36 130 200 366 i. What is the probability that an individual answered "pretty satisfied," GIVEN that he/she purchased a new car? [ Select ] ["0.71", "0.20", "0.55", "0.91", "0.35"] ii. What is the probability that an individual answered "pretty satisfied" OR purchased a new car? [ Select ] ["0.35", "0.20", "0.91", "0.71", "0.55"] iii. What is the probability than an individual answered "extremely satisfied" AND purchased a new car? [ Select ] ["1.10", "0.60", "0.77", "0.33", "0.59"]

i. The keyword "given" tells you to compute a conditional probability, where you are restricted to only the given category. In the "New cars" column, there are 72 out of 203 individuals who were pretty satisfied, so the probability is 72/203 = 0.35. ii. The key word "or" tells you to use the General Addition Rule, which is on the Formula Sheet: P(pretty satisfied or new car) = P(pretty satisfied) + P(new car) - P(pretty satisfied AND new car) = 130/366 + 203/366 - 72/366 = 0.71. iii. Since we have all the data, we know there are 120 individuals who answered "extremely satisfied" and purchased a new car, so the probability is 120/366 = 0.33.