Mktg 301

7) Data from a small bookstore atomic enumerate 18 shown in the attach to t able. The manager wants to predict Sales from play of Sales People Working. consider of gross revenue pile working Sales (in $1000) 4 12 5 13 8 15 10 16 12 20 12 22 14 22 16 25 18 25 20 28 x=11. 9 y=19. 8 SD(x)=5. 30 SD(y)=5. 53 a) take on the slope estimate, b1. routine technology or the aspect below to convalesce the slope. b1=rsysx Enter x,y Data in TI-84 chthonic STAT > STAT > CALC > 8 LinReg(a+bx) b1=1. 023 b) What does b1 fuddled, in this context?The slope tells how the response covariant hanges for a bingle unit of measurement rate in the predictor Thus, an additional $1,023 of exchanges associated with separateisticly additional gross revenue person working. c) arise the intercept, b0. b0=y-b1x =19. 8-1. 023(11. 9) For this problem, call technology, locomote to cardinal decimal places. b0=7. 622 d) What does b0 mean in this contet? Is it meanful? The intercept give ear s as a starting prize for the predicitons. It shuld all be interpreted if a 0 value for the predictor variable sheds sense for the context of the situation. On average, $7,622 is anticipate when 0 sales people atomic come up 18 working.It is non meaningful beca using up it does non make sense in this context. e) Write the equation that predicts Sales from Number of Sales People Working. give that the slope of the equation b1=1. 023 and the intercept is b0=7. 622 be intimate the equation. Sales=7. 622+1. 023 *(Number of Sales People Working) f) If 19 people argon working, what sales do you predict? Substitute 19 for the number of sales people working in the equation appoint in the previous step and lap for Sales. Sales=7. 622+1. 023 *(Number of Sales People Working) =7. 622+1. 023*19Substitute. =27. 059Simplify. *note that distributively unit of Sales represents $1000. Thus, the predicted sales for 19 people working is 27,059 dollars. g) If sales are actually $26,000, w hat is the value of the residual? take off the predicated value found in the previous step from the actual value. 26,000-27,059=-1059 Thus, the value of the residual is -1059 dollars. h) Have the sale been overestimated or underestimated The predicted sales are $27,059 and the actual sales are $26,000. Since $27,059 > $26,000, the sales were overestimated. 13) Of the 46 man-to-mans who responded, 25 are bear on, and 21 are not relate. of those control-to doe with nearly security are male and 5 of those not concerned are male. If a respondent is selected at random, fall upon distributively of the fallowing conditional probabilities. Male Female positive Concerned 9 16 25 Not Concerned 5 16 21 Total 14 32 46 a) The respondent is male, precondition that the respondent is not concerned about security. P(MaleNot Concerned) = 521 = 0. 238 b) The respondent is not concerned about security, inclined that is female P(Not ConceredFemale) = 1632 = 0. 500 c) The respondent is fem ale, given that the respondent is concerned about security. P(FemaleConcerned) = 1625 = 0. 40 14) It was found that 76% of the population were infected with a virus, 21% were without clean water, and 18% were infected and without clean water Clean pee Yes No Total Infected 0. 58 0. 18 0. 76 Not Infected 0. 21 0. 03 0. 24 Total 0. 79 0. 21 1. 00 a) Whats the opportunity that a surveyed person had clean water and was not infected? .21 had clean water and was not infected 15) A survey reason out that 54. 4% of the households in a particular country have twain a land line and a cell resound, 32. 6% have only cell phone services but no landline, and 4. 6% have no telephone services at all. ) What proportion of households have a landline? Begin by fashioning a contingency table. kiosk Phone Yes No Total Landline 0. 545 0. 083 0. 628 No Landline 0. 326 0. 046 0. 372 Total 0. 871 0. 129 1. 00 The completed contingency tables shows that P(landline) = 0. 628. b) Are having a c ell phone and having a landline self-supporting? Explain. Events A and B are independent when P(BA) = P(B). To curb wheter having a cell phone and having a landline are indepented, describe P(landlinecell phone) and P(landline). Recall from part a) that P(landline) =0. 628 PBA=P(A and B)P(A)Use the formula to line up P(landlinecell phone) Plandlinecell phone=P(landline and cell phone)P(cell phone) Since the contingency table shows that P(landline and cell phone)=0. 545 and P(cell phone)=0. 871, substitute these set into the equation. come apart to begin the conditional prospect, rounding to three decimal places. Plandlinecell phone=0. 5450. 871=0. 626 Thus, P(landlinecell phone)=0. 626 and P(landline)=0. 628. Because 0. 626 is very finish to 0. 628, having a cell phone and having a landline are probably independent. Of the households surveyed, 62. 6% with cell phones had landlines, and 62. 8% of all households did. 6) A commercialiseplaceplaceing agency has developed thr ee vacation packages to promote a timeshare plan at a parvenu reparation. They estimate that 30% of emf customers entrust choose the Day Plan, which does not include overnight accommodations 30% go away choose the night spacious Plan, which includes one night at the resort and 40% will choose the Weekend Plan, which includes 2 nights. a) Find the respect value of the number of nights that potential customers will need Vacation software product Nights Included Probability P(X=x) Day Plan 0 30100=0. 3 Overnight Plan 1 30100=0. 3 Weekend Plan 2 40100=0. 4 This, P(X=0)=0. 3, P(X=1) =0. 3, and P(X=2)=0. Use the formula E(X) = ? x P(x) to detrime the judge value. E(X) = ? x P(x) = 0(0. 3) +1(0. 3) +2(0. 4) = 1. 1 There, the judge value of the number of night potential customers will need is 1. 1 b) Find the measuring going of the number of nights potential customers will need. The commonplace deviation is the real nail down of the discrepancy. depression, Find th e Variance To do so, muster up the deviation of individually value of X from the mean and square each deviation. The random variable is the expect value of these squared deviations and is found apply the formula below. ?? = Var(X) = ? (x )? P(x) Find the deviation for each value of X.Remember that E(x)=1. 1 Vacation big bucks Nights Included Probability P(X=x) Deviation (x E(X)) Day Plan 0 30100=0. 3 0 1. 1 = -1. 1 Overnight Plan 1 30100=0. 3 1 1. 1 = -0. 1 Weekend Plan 2 40100=0. 4 2 1. 1 = 0. 9 Now knock the magnetic declination using the formula ?? =Var(X)=? (x )? P(x) Var(X) = ? (x )? P(x) = (-1. 1)? (0. 3) + (-0. 1)? (0. 3) + (0. 9)? (0. 4) = 0. 69 Finally, the example deviation also known as ? is the square root of the division. ? = Var(x) = 0. 69 = 0. 83 Therefore, the standard deviation of the number of nights potential customers will need is about 0. 83 nights. 7) A food market supplier believes that in a dozen pelt, the mean number of furrowed eggs is 0. 2 with a standard deviation of 0. 1 eggs. You defile 3 dozen eggs without pictureing them. a) How m either broken eggs do you get? The pass judgment value of the affection of random variables is the trades union of the anticipate value of each idividula random variable. Find the midpoint of the expected values where X is the total number of broken eggs in the three dozen, and X, X, X Represent the three individual dozen eggs. E(X) = E(X1) + EX2+ EX3 = 0. 2 + 0. 2 + 0. 2 = 0. 6 Therefore, the expected value of X is 0. 6 eggs. b) Whats the standard deviation?The section of the junction of independent variables is the spirit of their individual variances. Find the variance for each carton, add the variances, and then take the square root of the sum to find the standard deviation. The variance of each individual dozen is the square of each dozens standard deviation. Var(X1) = Var(X2) = Var(X3) = 0. 12= 0. 01 Find the sum of the variances to find the variance of the sum. Var(X ) = VarX1+ VarX2+ VarX3 = 0. 01 + 0. 01 + 0. 01 = 0. 03 Recall that the standard deviation is the square root of the variance. Find the standard deviation. SD(X) = Var(x) = 0. 03 = 0. 17Therefore, the standard deviation is 0. 17 eggs c) What assumptions did you have to make about the eggs in order to final result this question? The variance for the sum of random variables is only the sun of variances of each random variable in trustworthy cases. Review the assumption that moldiness be made to allow the variance to be the sum of the individual variances. 18) An insurance comp each estimates that it should make an one- form service of $260 on each home possessors policy written, with a standard deviation of $6000. a) why is the standard deviation so large? Home insurance is utilize to protect the owner financially in the event of a problem.If a catastrophe occurs, then the insurance company will cover the personify of the damage. If a catastrophe never occurs, then the insurance company pays nothing. Meanwhile, the owner pays the insurance company at regular intervals whether or not a catastrophe occurs. The expected value is the mean yearly net income on all of the policies and the standard deviation is a measure of how much annual profits dirty dog differ from the mean. Use this information with the fact that claims are rare, but very costly, occurrences. b) If the company writes only four of these policies, what are the mean and standard deviation of the annual profit?LetX1,X2, X3,,Xn represent the annual profit on the n policies and let X be the random variable for the total annual profit on n polices written. X=X1+X2+ X3++Xn The expected value of the sum is the sum of the expected values. Find the expected value of the annual profit on each policy. EX1=EX2=EX3=EX4=$260 Now find the sum of the expected values. EX=EX1+EX2+EX3+EX4 =260+260+260+260 = $1040 Therefore, the mean annual profit is $1040 To find the standard deviation of the annual profit, use the fact that te variances of the sum of independent variables is the sum of their individual variances. First find the variance for each policy.The variance for the policy is the square of the standard deviation. VarX1=VarX2=VarX3=VarX4=60002=36,000,000 VarX=VarX1+VarX2+VarX3+VarX4 = 4(36,000,000) = 144,000,000 Evaluate the square root of the variance to find the standard deviation. SDX=VarX =144,000,000 =$12,000 Therefore, the standard deviation is $12,000 c) If the company writes 10,000 of these policies, what are the mean and standard deviation of annual profit? The expected value of the sum is the sum of the expected values. The expected value of each policy was found earlier. EX1=EX2=EX3= =EX10,000=$260 Now find the sum of expected values. EX=EX1+EX2+EX3+ +EX10,000 10,000(260) =$2,600,000 Therefore, the mean annual profit is $2,600,000 To find the standard deviation of the annual profit, use the fact that the variance of the sum of independent variables is the sum of their individual variances. First find the variance for each policy. The variance for the policy is the square of the standard deviation and was found earlier. VarX1=VarX2=VarX3= =VarX10,000=36,000,000 Now sum the variances to find the variances of the sum. VarX=VarX1+VarX2+VarX3+ +VarX10,000 =10,000(36,000,000) =360,000,000,000 Evaluate the square root of the variance to find the standard deviation. SDX=Var(X) =360,000,000,000 $600,000 Therefore, the standard deviation is $600,000. d) Do you return the company is apt(predicate) to be profitable? Recall that the mean annual profit for 10,000 policies is $2,600,000. While this number seems quite large, it is necessary to specify how likely a profit is to ensure that this company will be profitable. Find the surmount in standard deviation of $0 from the mean to take care how rare an occurrence of no profit would be. z=x-?? =0-2,600,000600,000 =-4. 3 Thus, $0 is 4. 3 standard deviation below the mean. ** flier that nigh 95% of the an nual profits should lie within two standard deviations of the mean.Evaluate whether the distance of $0 from the mean is convincing enough to finalise whether or not the company will be profitable. e) What assumptions underlie your analysis? Can you hypothesise of circumstances under which those assumptions might be violated? The variance of the sum of random variables is only the sum of the variances of each random variables in certain cases. Review the assumption that must be made to allow the variance to be the sum of the individual variances. Then chose the situation that would create an experience among policy losses. 19) A farmer has 130 lbs. of apples and 60 lbs. f potatoes for sale. The market price for apples (per pound) each sidereal day is a random variable with a mean of 0. 8 dollars and a standard deviation of 0. 4 dollars. Similarly, for a pound of potatoes, the mean price is 0. 4 dollars and the standard deviation is 0. 2 dollars. It also costs him 5 dollars to bri ng all the apples and potatos to the market. The market is busy with shoppers, so assume that hell be able to sell all of each type of produce at the days price. a) Define your random variables, and use them to express the farmers net income. A random variables outcome is bases on a random event.Therefore let the random variables represent the factors that will be randomly determined each day. The random variables should represent the market prices of the two items. A = price per pound of apples P = price per pound of potatoes The profit is equal to the total income minus the total cost. The income is found by multiplying the market price for apples by the total number of pounds sold and adding it to the product of the market price for potatoes and the number of pounds of potatoes sold. The total cost is the transparent cost. Profit = 130A + 60P 5 b) Find the mean. The mean of the net income is the expected value of the profit.Profit = 130A + 60P 5 E(Proft) = E(130A + 60P 5) Use the airscrew E(X + Y) = E(X) + E(Y) to express the expected value of the profit as the sum of two separate expected values E=(Profit) = E(130A +60P -5) = E(130A) + (60P 5) = E(130A) + E(60P 5) Now use the property EXc= E(X)c E(Profit) = E(130A) + E(60P 5) = E(130A) + E(60P) 5 Finally, use the property E(aX) = aE(X) to remove the coefficient from the expected values. E(Profit) = E(130A) + E(60P) 5 = 130E(A) + 60E(P) 5 Substitute the known expected values of the prices of apples and potatoes in the equation. E(Profit) = E(130A) + E(60P) 5 E(0. 8) + E(0. 4) 5 Evaluate the expected profit. E(Profit) = 130(0. 8) + 60(0. 4) 5 = 123 Therefore, the mean is 123 dollars. c) Find the standard deviation of the net income. To find the standard deviation, startle find the variance and then take the square root, since the properties efficacious in this case are in terms of variance and not standard deviation SD(Profit) = Var(Profit) = Var(130A+60P-5) First use the property Var(X + Y) = Var(X) + Var(Y) to express the variance of the profit as the sum of two separate variance Var(Profit) = Var(130A + 60P 5) = Var(130A) + (60P 5) =Var(130A) + Var(60P 5)Now use the property Var(X c) = Var(X) to simplify the second variance Vr(Profit) = Var(130A) + Var(60P 5) = Var(130A) + Var(60P) Finally, use the property VaraX=a2VarX to replicate each variance. Var(Profit) = Var(130A) + Var(60P) = 1302VarA+ 602VarP = 16,900Var(A) + 3600Var(P) Evaluate the variance of the profit. Var(Profit) = 16,900(0. 16) + 3600(0. 04) = 2848 Lastly, find the standard deviation, rounding to two decimal place. SD(Profit) = VarProfit = 2848 = 53. 37 Therefore, the standard deviation of the net income is 53. 37 dollars. d) Do you need to make any assumptions in calculating the mean?Recall that the mean of the sum of two or more(prenominal) random variables is the sum of the means. Determine what, if any, assumptions are made to use this property. Do you need to make any assumptions in calculat ing the standard deviation? Recall that the variance of the sum of two random variables is only the sum of their individual variances in certain cases, Determine what, if any, assumptions are made to use this property. 20) A salesman ordinarily makes a sale ( blottos) on 65% of his exhibits. Assuming the presentations are independent, find the hazard of the following. ) He fails to close for the scratch line off time on his one-sixth attempt. Use the formula below to determine the luck, where p is the hazard conquest, q=1 p and X is the number of trails until the scratch success occurs. P(X=x) = qx-1p Find the values for p and q. **Note that in this case that a success is define as failed to close p = 0. 35 q = 0. 65 Substitute and solve to find P(X=6). Rounding to four decimal places P(X=6) = qx-1p = 0. 656-1(0. 35) = 0. 0406 Therefore, the probability he fails to close for the eldest tie on his sixth attempt is 0. 0406 b) He closes his first presentation on his fifth at tempt.Find the values for p and q. **Note that in this case that a success is defined as making a sale p = 0. 65 q = 0. 35 Substitute and solve to find P(X = 5), rounding to four decimal places P(X=5) = qx-1p = 0. 355-1(0. 65) = 0. 0098 Therefore, the probability he closes his first presentation on his fifth attempt is 0. 0098 c) The first presentation he closes will be on his second attempt. Find the values for p and q. Note that in this case that a success is defined as making a sale. p = 0. 65 q = 0. 35 Substitute and solve to find P(X=2) P(X=2) = qx-1p = 0. 352-1(0. 65) = 0. 2275Therefore, the probability the first presentation he closes will be on his second attempt is 0. 2275 d) The first presentation he closes will be on one of his first three attempts. Use the fact that the panegyric of an even is equal to 1 P(X=x) to find the probability. The compliment event is that he will not close a sale on any of his first three attempts. Find the probability that he does not close o n his first three attempts, rounding to four decimal places. 0. 353=0. 0429 Subtract from 1 to find the probability the first presentation he closes will be on one of his first three attempts 1 0. 429 = 0. 9571 Therefore, the probability the first presentation he closes will be on one of his first three attempts is 0. 9571 21) College students are a major target for advertisements for recognize cards. At a university, 73% of students surveyed said that they had kick in up a new credit card work out within the past year. If that percentage is accurate, how some(prenominal) students would you expect to survey before conclusion one who had not opened a new account in the past year? First check to see that the cells are Bernoulli trials. Trials are Bernoulli if the following three conditions are satisfied. 1.There are only two realizable outcomes (called success and ill) for each trial. 2. The probability of success, denoted p, is the same on every trial. (The probability of fa ilure, 1 p is oftentimes denoted q. ) 3. The trials are independent There are only two possible outcomes for each trial because a student either opened a credit card account in the past year or they did not. The probability of success is the same on every trial, based on the percent given in the problem statement. The trails are independent because each students response is not dependent on any other students response.Thus, the trials of surveying the students are Bernoulli trials. A geometric probability copy models how long it will take to achieve the first success in a series of Bernoulli trials. Let X be the number of students that will have to be surveyed before finding the first student who did not open a credit card in the past year. The two outcomes are a student who did not open a credit card account in the past year *success) and a student who opened a credit card account in the past year (failure). The probability of a failure is given in the problem statement as q = 73% = 0. 73.Find the probability of success by subtracting this from 1. P = 1 0. 73 = 0. 27 Find the expected value of X. In a geometric model, the expected value is EX= 1p , where p is the probability of success. Round up to the nearest integer. EX=10. 27=4 Therefore, on average, you would expect to survey 4 students before finding one who had not opened a new account in the past year. 22) A certain tennis player makes a boffo first serve 82% of the time/ Assume that each serve is independent of the others. If she serves 7 times, whats the probability she gets a) all 7 serves in? b) exactly 5 serves in? ) at least(prenominal) 5 serves in? d) no more than 5 serves in? The first step is to check to see that these are Bernoulli trails. The first serves can be considered Bernoulli trials. There are only two possible outcomes, successful and unsuccessful. The probability of any first serve being good is given as p = 0. 82. Finally, it is sour that each serve is independent of the ot hers. Next define the random variable. distributively question deals with the number of serves, so let X be the number of successful serves in n = 7 first serves. Now determine which probability model is appropriate for these problems.Recall that geometric probability models deal with how long it will take to achieve a success. A binomial probability model describes the number of successes in a specific number of trails. either the question deal with the number of successful serves so the binomial probability model Binom(7,0. 82 is appropriate here. a) all 7 serves in? The probability that she ges all 7 serves in is P(X=7). To use the binomial probability model Binom(n,p), use the fallowing formula, where n is the number of trials, p is the probability of success, q is the probability of failure (q = 1 p), and X is the number of successes in n trials.PX=x= nxpxqn x, where nx= n x n-x First substitute the correct values into the formula PX=7= 770. 8270. 187- 7 Now simplify. P(X = 7) ? 0. 249 Therefore, the probability that she gets all 7 serves in is approximately 0. 249 binomPDF(7, . 82, 7) = b) exactly 5 serves in? The probability she gets exactly 5 serves in is P(X = 5). As in part a, use the formula PX=x= nxpxqn x to find this probability PX=5= 750. 8250. 187 5 ?0. 252 Therefore, the probability she gets exactly 5 serves in is approximately 0. 252 binomPDF(7, . 82, 5) = c) at least 5 serves in?To find P(at least 5 serves in), first determine and an expression that is equal to this probability. Note that the wording at least 5, means 5 or more, meaning that there can 5, 6, or 7 serves in. Thus, the probability equals P(X=5) + P(X=6) + P(X=7). So to find the probability that she got at least 5 serves in, evaluate. P(X=5) + P(X=6) + P(X=7) = 75(0. 82)50. 187-5+76(0. 82)6(0. 18)7-6+77(0. 82)7(0. 18)7-7 ?0. 885 Therefore, the probability she gets at least 5 serves in is approximately 0. 885 binomPDF(7, . 82, 5) + binomPDF(7, . 82, 6) + binomPDF(7, . 82, 7) = d) no more than 5 serves in?To find P(no more than 5), first determine an expression that is equal to this probability. Note that the wondering no more than 5 means 5 or less, meaning that there can be 0 thru 5 successful serves. Thus, the probability equals P(X? 5). So to find the probability that there are no more than 5 serves in, evaluate P(X? 5), which is equal to P(X=0) + P(X=1) + + P(X=5), using the formula PX=x= nxpxqn x P(X=0) + P(X=1) + + P(X=5) = 70(0. 82)00. 187-0+71(0. 82)1(0. 18)7-1 + + 75(0. 82)5(0. 18)7-5 ? 0. 368 Therefore, the probability that there are no more than 5 serves in is approximately 0. 368 binomCDF(7, . 82, 5)