Baggage Check
Thousands of travelers pass through Guadalajara airport each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs officials want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs officers claim that the probability that the light turns green on any press of the button is 0.70. You observe 100 passengers at the Customs “stop-light.” Only 65 get a green light. Does this give you reason to doubt the Customs officials?
a.) Use your calculator to simulate 50 groups of 100 passengers activating the Customs light. Use the command randBin(100, 0.7, 50)/100 store in L1.List 1 will contain 50 values of , the proportion of the 100 passengers who got a green light.
b.) Sort L 1 in descending order. In how many of the 50 simulations did you obtain a value of that is less than or equal to 0.65? Do you believe the Customs agents?
c.) Describe the shape, center, and spread of the sampling distribution of for samples of n = 100 passengers.
d.) Use the sampling distribution from part (c) to find the probability of getting a sample proportion of 0.65 or less if p=0.7 is actually true. How does this compare with the results of your simulation in part (b)?
e.) Repeat parts (c) and (d) for samples of size n = 1000 passengers. Report your findings as compared to n = 100.
Resources:
Beth Chance's Reese's Pieces applet and sampling pennies www.rossmanchance.com/applets/index.html
David Lane's Sampling Distribution
www.ruf.rice.edu/~lane/stat_sim/sampling_dist/
Central Limit Theorem
www.whfreeman.com/tps3e
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a. (in descending order)
ReplyDelete.79 .78 .77 .77 .76 .76 .75 .74 .74 .74 .73 .73 .73 .72 .72 .72 .71 .71 .71 .71 .71 .71 .7 .7 .7 .7 .7 .7 .7 .7 .69 .69 .69 .69 .68 .68 .68 .68 .68 .68 .67 .67 .66 .66 .65 .64 .63 .62 .61 .61
b. i do believe the customs agent. when i did the 1-Var Stats of L1, the x bar (average) came out to be .7014, which is basically a 70% chance.
c. if someone could help me remember how to graph this, please do. i keep having an error come up (ERR:STAT)
a. this is what i got beautiful :)
ReplyDelete.85 .79 .79 .79 .77 .77 .77 .76 .75 .75 .74 .74 .73 .73 .73 .72 .71 .71 .71 .71 .71 .7 .7 .7 .7 .7 .69 .69 .69 .69 .69 .69 .69 .68 .68 .68 .67 .67 .67 .67 .67 .67 .66 .66 .66 .65 .65 .64 .63 .62
b. um i got the average : .7058 which is remotely close to 70% dear:) and 5 stims were less then or equal to .65
c.yoo sandmann814 i gotcha back son! all you have to do is press graph i think :) and check if your stat plots are on
my graph looked symmetrical and my lovely center was about .7 and my spread was .85 to .62
ill do d and e later im too lazy!
c.) symmetrical/skewed slightly right
ReplyDeletespread is .61 to .815 (min =.79 max<.8157)
center is around .695 (min=.6871 max<.7128)
ok and
ReplyDeleted.) normalcdf(.65,.7,.7014,.0415) i got .3787
this means there is a 37.87% probability of getting a sample proportion of .65 or less
e.) the graph once again looks symmetrical/skewed slight right
ReplyDeletespread is .663 to .751
center is around .696
normalcdf(.65,.7,.69986,.0146) is .5035. there is a 50.35% probability of getting a sample proportion of .65 or less.
This comment has been removed by the author.
ReplyDeleteNo this doesnt give you any reason to doubt the officials because only a sample of 100 with one trial is being taken into consideration. If we had increased the sample size and had taken multiple trials of this same test, and then take the mean of all those trials, it will be more likely for the numbers to reach .7 for the green light turning on.
ReplyDeleteThis comment has been removed by the author.
ReplyDeletea.) .83,.8,.8,.8,.76,.76,.75,.75,.75,.74,.73,.73,.73,.73,.72,.72,.72,.72,.72,.72,.72,.71,.71,.71,.71,.71,.71,.7,.7,.7,.7,.69,.69,.69,.69,.69,.69, .69,.68,.68,.67,.66,.66,.66,.65,.65,.65,.64,.64,
ReplyDelete.58
b.) i recieved 6 values that are less than or equal to .65. the average of my list is .7082 which is very close to 70 % so i do believe the customs agents.
c.) my graph is slightly symmetrical possibly skewed right alittle :)
spread = .58 to .83
center = .687
d.) normalcdf(.65,.7,.7082,.04589) = .3267
there is a 32.67 % chance of getting less then .65 if the probability is .7 .
e.) spread = .669 to .731
center = .69773
skewed right
normalcdf(.65,.7,.69773,.01342) = .56697
56.697 % chance of getting less than or equal to .65 with the probability of .7.
a-b.)
ReplyDelete.82, .79, .78, .77, .77, .77, .76, .76, .75, .74, .74, .74, .74, .73, .72, .72, .72, .72, .72, .71, .71, .70, .70, .70, .70, .70, .70, .70, .70, .69, .69, .68, .68, .68, .68, .68, .67, .67, .67, .66, .65, .65, .65, .64, .64, .63, .62, .61, .61, .59
In the 50 simulations I obtained, there are 10 of them has the value of that is less than or equal to 0.65.
Yes, the custom's claimed percentage is posible because the average of the observed stimulation is .7004 (xbar=.7004) which matches the custom's average.
c.) still remembering how to graph.>.<
a. (descending order)
ReplyDelete.78 .78 .75 .75 .75 .75 .75 .74 .74 .74 .74 .73 .73 .73 .73 .72 .72 .72 .71 .71 .7 .7 .7 .7 .7 .7 .69 .69 .68 .68 .68 .67 .67 .67 .67 .66 .66 .66 .66 .66 .65 .65 .65 .65 .64 .64 .64 .63 .62 .62
b. 10 of my simulations were less than or equal to .65. The average of my simulation is .6952, which is almost 70% so yes I do believe the customs agents.
c. shape- symmetrical/ slightly skewed right
ReplyDeletespread- .62 - .80
center- .68
d.) normalcdf(.65,.7,.6952,.0421)= .4039
ReplyDeletethere is a 40.39% chance of getting .65 if the probability is less than .7
so what are we supposed to do after we answer? is that it?
ReplyDeleteno rickerd u gotta do more enj
ReplyDeletee.- my calculator keeps freezing and won't let me do n= 1000 !
ReplyDeleteok
ReplyDeleteC.
The graph is fairly symmetric, with
spread:X=.85-.59 and center:aprox. .70
D.
normalcdf(.65, .7, .7004,.049717)= .34143. This means that 34% probability of the sample are less than .65
E.
is taking forever..is it suppose to?
a) .81 .79 .78 .77 .76 .76 .76 .75 .75 .75 .75
ReplyDelete.74 .73 .73 .73 .72 .72 .72 .72 .71 .71 .71 .71
.70 .70 .70 .70 .69 .69 .68 .68 .68 .68 .68 .67
.67 .67 .66 .66 .66 .65 .64 .64 .62 .62 .62 .62
.61 .60
b) i got 9 values that are equal to or less than to .65. The average of my simulation is .6998 which is very close to .70 so i would believe the custome agent.
c)Shape-Mostly symetrical but slightly skewed right.
Spread-.60-.81
Center-.6885
d)normalcdf(.65,.7,.6998,.0499) = .3424 This means there is a 34.24% chance of getting less than .65 when the probability is .7.
e doesnt load???
ReplyDeleteE. ranBin(1000,.7,500)/1000 is this right??
ReplyDeletepart e took a while to do it all out, it's supposed to
ReplyDeleted. you gots to do the normcdf(.62,.65,.71) which equals .011934 , which means the prob that you will get a sample proportion less then or equal to .65
ReplyDeletee. the calcualtor wont load!!!
ReplyDelete56.57.6.61.63.64.64.65.66.67.67.67.67.68.68.68.68.69.69.69.69.69.69.7.7.7.7.7.71.71.71.71.72.72.73.74.74.74.74.74.74.75.75.75.76.76.76.77.77.78.81
ReplyDeleteaverage around .7 so i believe the guy
center is .7 spread is .56 to .81
8 below .65 and skewed right
.0497 for D
i can't do E
ReplyDeletei did randombin(1000, .7, 50)
first of all i don't know if that is right. second my calculator says domain error
E.
ReplyDeletenormalcdf(.65, .7, .7006, .01457)= .48331
48% chance of where samples coulbe be less than or equal to .65 in 1000 passengers with probability of .70
Graph is symmetric as well, with spread: .65-.74 and center in about .699