Tugas Statistika 3 [PDF]

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STATISTIK 3 A. 5.39 In a study by Peter D. Hart Research Associates for the Nasdaq Stock Market, it was determined that 20% of all stock investors are retired people.In addition,40% of all U.S. adults invest in mutual funds. Suppose a random sample of 25 stock investors is taken. Stock investors p = 0.20 Sample n = 25 a. What is the probability that exactly seven are retired people? The question is the proability when stock investors X=7, P(x=7) Using the binomial probability distribusion table in the textbook, we can find when n = 25, P1= 0.20, X= 7, the probability is 0.111 or 11.1%. X



.1



.2



.3



.4



.5



0



.072



.004



.000



.000



.000



1



.199



.024



.001



.000



.000



2



.266



.071



.007



.000



.000



3



.226



.136



.024



.002



.000



4



.138



.187



.057



.007



.000



5



.065



.196



.103



.020



.002



6



.024



.163



.147



.044



.005



7



.007



.111



.171



.080



.014



8



.002



.062



.165



.120



.032



b. What is the probability that 10 or more are retired people? Add the probabilities from X=10 all the way to X =25.



X cannot be greater than 25, because you only have 25 people in your sample. So the equation is: P(x ≥ 10) = P(X=10) + P(X=11)+ ….+ P(x=25) Find the probabilities in the table: P(x ≥ 10) = 0.0012+0.004+0.001+0.000= 0.17 or 17% c. How many retired people would you expect to find in a random sample of 25 stock investors? If n = 25 and on average 20% are retired in the population, then n x p = 25 X 0.2 =5 Or you would expected to find 5 people out of your sample of 25 that are retired. d. Suppose a random sample of 20 U.S. adults is taken. What is the probability that exactly eight adults invested in mutual funds? P= 0.40 and sample n =20 Using the binomial probability distribution table in textbook, P(x=8) = 0.180 or 18%; e. Suppose a random sample of 20 U.S. adults is taken. What is the probability that fewer than six adults invested in mutual funds? P(X< 6) = P (x=0) + P(x= 1)+ …..+ P(x= 5) = 0.000+0.000+ 0.003+0.012 +0.035+ 0.075 =0.125 or 12,5%; Add to X=5 not X= 6 , because you want fewer than 6. f. Suppose a random sample of 20 U.S. adults is taken. What is the probability that none of the adults invested in mutual funds? P (x= 0) =0.000, there is some small probability, but its considered negligible g. Suppose a random sample of 20 U.S. adults is taken. What is the probability that 12 or more adults invested in mutual funds? P (x ≥ 12) = P (x= 12) + P (x= 11) + …..+ P (x= 25)



= 0.035+ 0.015+ 0.005 + 0.001 + 0.000 = 0.056 or 5,6%. h. For parts e–g, what exact number of adults would produce the highest probability? How does this compare to the expected number? In the table, the highest number of adults with the highest probability is 0.180. The highest number occurs when X = 8 The expected or mean number is n x p =20 x 0.4 = 8 out of 20 In this case the highest probability occurs at the expected or mean walue, but for more skewed distributions this might not always be the case. B. 5.42 Suppose that,for every lot of 100 computer chips a company produces, an average of 1.4 are defective. Another company buys many lots of these chips at a time, from which one lot is selected randomly and tested for defects. If the tested lot contains more than three defects, the buyer will reject all the lots sent in that batch.What is the probability that the buyer will accept the lots? Assume that the defects per lot are Poisson distributed. There are a large number of batches of computer chips and the probability of selecting a bad chip is 1.4/100, so it is p=0.014. we could consider this a binomial problem with p=0,014 and n=100, but problem working askk use to consider that selecting a bad computer chip is a rarea event and tells us to use the poisson approximation. ƛ = np =100 x 0.014=1.4 If there are more than 3 defects, the buyer will reject, so if ƛ =1.4 , what is the probability that the lot will be rejected? Using posission probabilities table in the textbook, we can find when ƛ =1.4, X



1.1



1.2



1.3



1.4



1.5



0



.3329



.3012



.2725



.2466



.2231



1



.3662



.3614



.3543



.3452



.3347



2



.2014



.2169



.2303



.2417



.2510



3



.0738



.0867



.0998



.1128



.1255



4



.0203



.0260



.0324



.0395



.0471



5



.0045



.0062



.0084



.0111



.0141



6



.0008



.0012



.0018



.0026



.0035



7



.0001



.0002



.0003



.0005



.0008



8



.0000



0000



.0001



.0001



.0001



9



.0000



0000



.0000



.0000



.0000



P(x>3) = 0.0395+ 0.0111 + 0.00026 + 0.0005 + 0.0000 =0.0538 or 5,8% So, There is a 5.38% chance that buyer will reject the lot and the probability that the buyer will accept the lots is: P(X ≤ 3) =1 -P(X > 3) =1-0.0538 = 0,9462 or 94.62% Best to do quality improvement and reduce the ƛ number of detect to be alot less than 1.4, or you will go out of business. C. 5.46 According to a recent survey, the probability that a passenger files a complaint with the Department of Transportation about a particular U.S.airline is .000014. Suppose 100,000 passengers who flew with this particular airline are randomly contacted. a. What is the probability that exactly five passengers filed complaints? b. What is the probability that none of the passengers filed complaints? c. What is the probability that more than six passengers filed complaints? The probability of passenger files a complaint is 0.000014, which is very rare, and the sample size is very large; n=100,000. So all the parts of the question fulfill the requirements for using the Poisson distribution. For the Poisson distribution, ƛ = np = 100000*0.000014 = 1.4



This is an easier problem because the lambda is in the table at the back of the book which says: ƛ = 1.4



X



P(X)



0



0.2466



1



0.3452



2



0.2417



3



0.1128



4



0.0395



5



0.0111



6



0.0026



7



0.0005



8



0.0001



9



0.0000



(Meaning less than 0.00005 - negligible a. What is the probability that exactly five passengers filed complaints? P(5) 0.0111 b. What is the probability that none of the passengers filed complaints? P(0) = 0,2466 c. What is the probability that more than six passengers filed complaints? The most simple way is to add P(7) + P(8) =0.0005+0.0001=0.0006 We could also use the formula to get more significant figures, but the above answer is sufficient for calss purposes and indicates that the answer is a very samll probability; D. 5.51 An office in Albuquerque has 24 workers including management. Eight of the workers commute to work from the west side of the Rio Grande River.Suppose six of the office workers are randomly selected.



N worker = 24 Number of worker from west side = 8 number of ways of selecting any 6 worker = C (24,6) = 134,596 a. What is the probability that all six workers commute from the west side of the Rio Grande? number of ways of selecting all 6 worker commute from west side = C(8 ,6) =28 required probability = 28/ 134,596 = 0.00021 b. What is the probability that none of the workers commute from the west side of the Rio Grande? number of ways of selecting all 0 worker commute from west side = C(8 ,0) = required probability = 1/ 134,596 = 0.0000075 c. Which probability from parts (a) and (b) was greatest? Why do you think this is? From probability parts a and b was great is probability a =0.00021 > probability b =0.0000075 d. What is the probability that half of the workers do not commute from the west side of the Rio Grande? number of ways of selecting all 12 worker do not commute from west side = C(16,12) =2730 required probability = 2730/ 134,596 = 0.02 E. 5.58 In one midwestern city, the government has 14 repossessed houses,which are evaluated to be worth about the same.Ten of the houses are on the north side of town and the rest are on the west side.A local contractor submitted a bid to purchase four of the houses. Which houses the contractor will get is subject to a random draw.



Total number of houses = 14 number of houses on north side = 10 number of houses on west side = 14 - 10 = 4 number of ways of selecting any 4 houses = C(14 ,4)=1001 a. What is the probability that all four houses selected for the contractor will be on the north side of town? number of ways of selecting all 4 houses on north side = C(10 ,4) =210 required probability = 210/ 1001 = 30 / 143 b. What is the probability that all four houses selected for the contractor will be on the west side of town? number of ways of selecting all 4 houses on west side = C(4 ,4) = 1 required probability = 1/ 1001 c. What is the probability that half of the houses selected for the contractor will be on the west side and half on the north side of town? number of ways of selecting half of the houses on the west side and half on the north side = C(10 ,2)( C(4 ,2)) =45 (6) = 270 required probability = 270/ 1001



TESTING YOUR UNDERSTANDING F. 6.38 The U.S. Bureau of Labor Statistics reports that of persons who usually work full-time, the average number of hours worked per week is 43.4.Assume that the number of hours worked per week for those who usually work full-time is normally distributed. Suppose 12% of these workers work more than 48 hours. Based on this percentage, what is the standard deviation of number of hours worked per week for these workers?



2



2



∑ X 1 − μ = √∑ 43.5 −12 % = 40.81 = 5.89 σ√ 6.93 √N



√ 48



G. 6.40 An entrepreneur opened a small hardware store in a strip mall. During the first few weeks, business was slow, with the store averaging only one customer every 20 minutes in the morning.Assume that the random arrival of customers is Poisson distributed. N= 20 minutes = 1/3 hour 1 minggu = 7 x 24 hour =168 a. What is the probability that at least one hour would elapse between customers? number of ways of selecting poison = C(3,1) = 3 required probability = 3/ 168= 0.018 b. What is the probability that 10 to 30 minutes would elapse between customers? number of ways of selecting poison = C(1/3, 1/6) + C (1/3, 1/2) =20+3 =23 required probability = 23/ 168= 0.137 c. What is the probability that less than five minutes would elapse between customers? number of ways of selecting poison = C(1/3, 1/12)= 660 required probability = 660/ 168= 3.928



H. 6.46 According toThe Wirthlin Report, 24% of all workers say that their job is very stressful.If 60 workers are randomly selected:



N = 60 P = 24% a. What is the probability that 17 or more say that their job is very stressful? Probability = 17/60 x24 %=0,068 b. What is the probability that more than 22 say that their job is very stressful? Probability = 22/60 x24 %=0.088 c. What is the probability that between 8 and 12 (inclusive) say that their job is very stressful? Probability between 8 and 12 = (8/60 x 24%) + (12/60 x 24%) /2 = (0.032+0.2)/2 =0.08 I.



6.50



The Conference Board published information on why companies expect to increase the number of part-time jobs and reduce full-time positions.Eighty-one percent of the companies said the reason was to get a flexible workforce. Suppose 200 companies that expect to increase the number of part-time jobs and reduce full-time positions are identified and contacted. What is the expected number of these companies that would agree that the reason is to get a flexible workforce? What is the probability that between 150 and 155 (not including the 150 or the 155) would give that reason? What is the probability that more than 158 would give that reason? What is the probability that fewer than 144 would give that reason? Given that p = 0.81 and n = 200 Since np = 200 x 0.81 = 162 and n (1 – p) = 200 x (1 – 0.81) = 38 both are greater than 5, we use normal approximation to binomial. Mean, µ = 200*0.81 = 162 Standard deviation, σ = √np (1 – p) = √200 x 0.81 x 0.19 = 5.548 The expected number of these companies that would agree that the reason is to get a flexible workforce is 162 Using continuity correction, P (150 < X < 155) = P (150.5 < X < 154.5) The respective Z-score with X = 150.5 is Z = (X - µ)/σ Z = (150.5 – 162)/5.548 Z = -2.07 The respective Z-score with X = 154.5 is Z = (X - µ)/σ Z = (154.5 – 162)/5.548 Z = -1.35 Using Z-tables, the probability is P [-2.07 < Z < -1.35] = 0.4808



J. 6.57 According to the Air Transport Association of America,the average operating cost ofan MD-80 jet airliner is $2,087 per hour.Suppose the operating costs of an MD-80 jet airliner are normally distributed with a standard deviation of $175 per hour. At what operating cost would only 20% of the operating costs be less? At what operating cost would 65% ofthe operating costs be more? What operating cost would be more than 85% of operating costs? ɥ = $2,087 per hour



dan σ =¿$175 per hour



 At what operating cost would only 20% of the operating costs be less?



At x amount, only 20% of the flights would operate cheaper than this, so solve for x by plugging given values to z formula: Note: The area between x and  must be 0.3 (or 30%), because half of the curve is 50% and x is 20%. x − μ x −2,087 Z= = = z for 0.30 area σ 175 So the z score above represents 0.30 in the Areas of the Standard Normal Distribution table. In the body of table, we find the closest value (percent) to 0.30 is 0.2995, So Z = 0.84. Z



0.00



0.01



0.02



0.03



0.04



0.0



0000



0040



0080



0120



0160



0.1



0398



0438



0478



0517



0557



0.2



0793



0832



0871



0910



0948



0.3



1179



1217



1255



1293



1331



0.4



1554



1591



1628



1664



1700



0.5



1915



1950



1985



2019



2054



0.6



2257



2291



2324



2357



0389



0.7



2580



2611



2642



2673



2704



0.8



2881



2910



2939



2967



2995



0.9



3159



3186



3212



3238



3264



Then plug Z = 0.84 into the formula to solve for X: Note: The area is on the left of the mean, so use negative (-) sign: x −2,087 , So 175 X= (-0.84) (175) + 2,087 = $1,90 So, only 20% of the planes cost less than $1940 to fly.  At what operating cost would 65% ofthe operating costs be more? Z = - 0.84 =



 At What operating cost would be more than 85% of operating costs? The area between X and is 65% - 50% =15%, so find Z associated with 15% or 0.15. Plug given into Z Formula:



Z=



x − μ x −2,087 = = z for 0.15 area σ 175



In the body of table, we find teh closest value to 0.15 is 0.1517. So Z= - 0.39 (negative since the boundary is left of the mean, Then plug Z = - 39 into formula to solve for X: Z = - 0.39 =



x −2,087 = $ 2018.75 175



So only 65% of pelanes would cost more than $ 2018.75 to fly.  What operating cost would be more than 85% of operating costs?



The area between X and μ is 50% -15% =35%, so now find Z associated with 35% or 0.35; plug given value into Z formula: Z=



x − μ x −2,087 = = z for 0.35 area σ 175



In the body of table, we can find the closest value (percent) to 0.35 is 0.3508. So Z =1.04, then plug Z = 1.04 into the formula to solve for x: Z = 1.04 =



x −2,087 = $ 2,269 175



So, 85% of the planes operate at a cost lower than this amount. Managerially, if you sell the average ticket for $227,00 and sell 100 tickets then you have covered your operating cost. Unless you intend to sell fewer tickets, it would be best to raise ticket prices.