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 Subject: Statistics homework Category: Reference, Education and News > Homework Help Asked by: as3411-ga List Price: \$2.50 Posted: 02 Nov 2002 15:15 PST Expires: 02 Dec 2002 15:15 PST Question ID: 96808
 ```I need help with some statistics homework...about 25-33 problems...I'll pay \$2.50 a question for this help....please!!``` Clarification of Question by as3411-ga on 02 Nov 2002 18:12 PST ```I need help with these problems... 1. The student body of a large univeristy consists of 60% female students. A random sample of 8 is selected. What is the probability that among students in the sample exactly two are female? 2. In a binomial experiment the probability of succes is 0.06. What is the probability of 2 successes in seven trials? 3. The random variable x is the # of occurences of an event over an interval of 10 minutes. It can be assumed that the probability of an occurence is the same in any two time periods of an = length. It is known that the mean # of occurences in 10 minutes is 5.3. The probaility that there are less than 3 occurences is? 4. The probaility that Pete will catch fish on a particular day when he goes fishing is 0.8. Pete is going fishing 3 days next week. The probability that Pete will catch fish on exactly 1 day is ? 5. Only 0.02% of credit card holders of a company report the loss or theft of thei credit cards each month. The company has 15,000 credit cards in the city of Memphis. What is the probability that during the next month in the city of Memphis a. No one reports the loss of theft of their credit cards? b. every credit card is lost or stolen? c.6 people report the loss or theft of their cards? d. at lesat 9 people report the loss of theft of their cards? e. Determine the expected # of reported lost or stolen cc's f. Determine the standard deviation for the # of reported lost or stolen cc's. \$2.50 for each question \$5.00 for the last problem \$15 total.```
 Subject: Re: Statistics homework Answered By: bobby_d-ga on 02 Nov 2002 21:46 PST Rated:
 ```Hi, and thanks for the questions... 1_) (Assuming the amount of students at the university is large enough that when less than ten students are removed, the probability remains the same. IE - we are assuming this is an independent event...) For each person in the sample, there is a 60/100 (or 3 in 5) chance that this person is female. Conversely, there is a 40/100 (or 2 in 5) chance that this person is male (assuming everyone is either male or female). The combination for you require would be 2 females and 6 males. or 3/5 * 3/5 * 2/5 * 2/5 * 2/5 * 2/5 * 2/5 * 2/5 = 0.00147456 probability (= 0.147456 % chance that this will occur) 2_) (http://www.uwm.edu/~stockbri/08bin.pdf - the page that shows what a binomial experiment is...) This question can be interpreted in two different ways - AT LEAST two successful trials, or exactly two successful trials. I have answered both ways. AT LEAST TWO SUCCESSES IN SEVEN TRIALS... Probability of At least two successes = P(2 successes in 7) The probabality of less than two success = P(less than 2 successes) P(2 successes in 7) = 1 - P(less than 2 successes) P(less than 2 successes) = P(0 successes) + (P 1 success) (Are you following?????) P(0 successes) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94 = 0.64847759419264 P(1 success) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.06 = 0.04139218686336 P(1 succces) + P(0 successes) = P(less than 2 successes) = 0.689869781056 Therefore P(AT LEAST two successes) = 1 - 0.689869781056 = 0.310130218944 = 31.0130218944 % chance EXACTLY 2 SUCCESSES... P(2 successes) = 0.94 * 0.94 * 0.94 * 0.94 * 0.94 * 0.06 * 0.06 = 0.00264205448064 = 0.264205448064 2 down... 3_) OK, this one is a bit strange, and involves the even stranger constant "e" - refer to (It's a cached page by google, so it also looks a bit strange...): http://216.239.33.100/search?q=cache:GHlKuJsQoJ0C:www.mis.coventry.ac.uk/jtm/slides/19/sld19p7.pdf+maths+probability+interval&hl=en&ie=UTF-8 This page seems to make it a bit clearer: Poisson Distribution http://www.utdallas.edu/~ammann/cs3341/node29.html By looking at the formula derived at the bottom of the page, we can sub our values... n = 0 and 1 and 2, as we want "not more than 2 cars every 10 minutes". u = 5.3, the average amount of events in the timeframe. so, P(N < 3) = P(n = 0) + P(n = 1) + P(n = 2) = 5.3^0 5.3^1 5.3^2 _____ -5.3 _____ -5.3 _____ -5.3 e e e 0! 1! 2! = (1 + 5.3 + 14.045) -5.3 ( ) * e = 0.101553978 = 10.1553978 % chance phew... (that was the best I could do for fractions and powers!) 4_) P(exactly 1 day out of three) = P(catch a fish) * P(won't catch a fish) * P(won't catch a fish) = 0.8 * 0.2 * 0.2 = 0.032 = 3.2 % chance 5_) P(a person claims theft/loss on card) = 0.0002 (converting from percentage). QUESTION A... P(a person doesn't claim theft/loss) = 0.9998 Therefore, P(15000 people don't claim theft/loss) = 0.9998^15000 = 0.049772132 = 4.9772132 % chance QUESTION B... P(everyone claims theft/loss) = 0.0002^15000 (Here, my calculator rounded off to 0, so the number is very small) = 1 _________ 5000^15000 QUESTION C... P(Six people claim loss/theft) = 0.0002^14994 * 0.9998^6 (Again, my calculator wouldn't have it!) = 4999^6 ________ 5000^15000 QUESTION D... P(AT LEAST 9 people claim) = 1 - ( < 9 people claim) P( < 9 claim theft/loss) = P(0 claims) + P(1 claim) + .... + P(8 claims) From looking at the result in question C, the 4999^6 can have its power changed for the amount of people claiming... Therefore, 4999^0 + 4999^1 + 4999^2... + 4999^8 all divided by 5000^15000 should give the answer we seek. Doing this, we get: ( 3.900784686 * 10^29 ) 1 - ( ___________________ ) P(AT LEAST 9 CLAIMS) = ( ) ( 5000^15000 ) QUESTION E... This one is easier. 0.02% of 15000 is the average, and hence how much we would expect... N * P = 0.0002 * 15000 = 3 people per month QUESTION F... Refer to http://engineering.uow.edu.au/Courses/Stats/File38b.html for formula for standard deviation formula. ****(Many, many thanks secre901 for helping find this formula!)**** 2 SD = N * P * Q Where N is the number of trials (15000) P is the probability of success (0.0002) Q is the probability of failure (0.9998) 2 Therefore, SD = 15000 * 0.0002 * 0.9998 = 2.9994 SD = + or - 1.731877594 There they all are! If you need any clarification on working, or are unhappy with any work that I have done, please feel free to ask for clarification! Also, you have already paid \$2.50 for posting the question. Could you please leave a tip of \$12.50 as payment for the remainder of the questions as stated in your clarification? Thanks again! bobby_d search strategy maths probability maths probability interval maths probability time``` Clarification of Answer by bobby_d-ga on 04 Nov 2002 00:00 PST ```I'm VERY sorry that these are incorrect, and thanks rbnn for pointing this out. It seems my maths is a little rusty... But I think I have ironed most of them out - again, please accept my apologies! 1_) I only took one possible combination, that is P(FFMMMMMM). I excluded P(FMFMMMMM), P(FMMFMMMM), P(FMMMFMMM), etc. There are, in total, 28 of these possibilities, and, fortunately, each has equal probability (as I said in my original answer) of 0.00147456 So, times this by 28, and the answer is 0.04128768. I believe this is your answer B 2_) Same problem again... EXACTLY 2 SUCCESSES... In total there are 6 + 5 + 4 + 3 + 2 + 1 possible combinations, which totals 21. Times 21 by my result: 0.05548314409344 I believe this is your answer B. I'm assuming the question does mean EXACTLY, as it fits in with the answers you provided...) For interest sake, I've corrected my answer to AT LEAST TWO as well: P(At least 2) = 1 - ( P(0) + P(1) ) I was correct with P(0) = 0.64847759419264 P(1), there are 7 possible combinations, therefore times my answer by 7: = 0.04139218686336 * 7 = 0.28974530804352 Therefore, P(At least 2) = 1 - (0.64847759419264 + 0.28974530804352) = 0.06177709776384 NB: I don't believe that AT LEAST 2 SUCCESSES is the answer they are looking for. 4_) The same error once more: Consider the following possible combinations for 1 success: SFF, FSF, FFS Therefore, times by answer by 3, and you get: P(exactly one success) = 0.096 I believe this equates to your answer C. 5_) Yes, rbnn is quite right in his comments below - this same error has affected the answers to question 5... Question A + Question B are still correct QUESTION C... We can use the COMBINATIONS function to determine how many different combinations of 6 reports out of 15000 card-holders there are, then ties this by my result. nCr gives the number of ways of choosing r items from n items. So here, 15000C6 = 1.580449816 * 10^22 Times with my answer... 1.580449816 * 10^22 * 4999^6 = ______________________________ 5000^15000 2.4665 * 10^44 = ________________ 5000^15000 (My calculator still won't so this number for me) QUESTION D... And, finally, the same mistake again... Let's use Poisson's discoveries to make this a little bit easier: http://www.utdallas.edu/~ammann/cs3341/node29.html n = 15000 P = 0.0002 np = 3 P (N equal to or more than 9) = 1 - P(N = 0) - P(N = 1) - .... - P(N = 8) -3 = 1 - e (1 + 3 + 4.5 + 4.5 + 27/8 + 81/40 + 81/80 + 729/680 + 729/4480) = 1 - 0.977558235 = 0.022441764 Question e seems fine. Question f should not be expressed as a + or - according to rbnn, but remember that it refers to both directions (towards negative and positive) from the mean. I hope this all works out now, and I'm very sorry for those mistakes - I was assuming permutation (ordered group) instead of a combination (unordered group). Thanks again, and good luck! bobby_d (rbnn - have a left any other errors?!?!?)```
 as3411-ga rated this answer: and gave an additional tip of: \$13.00 ```Excellent help...but we had a choice of 4 answers for question #1...a.0.2936 b.0.0413 c.0.0896 and d. 0.0007 For question #2 the possible answers were a.0.06 b. 0.0554 c. 0.28 d. 0.0036 for #4 the answers were a. .104 b..008 c. .096 and d. .8 Please advise.```

 Subject: Re: Statistics homework From: tutuzdad-ga on 02 Nov 2002 15:34 PST
 ```I recommend that you break the questions into groups, price them accordingly and post them as small groups of multiple questions. You may want to have smaller groups for questions requiring greater detail or more extensive research.```
 Subject: Re: Statistics homework From: as3411-ga on 02 Nov 2002 16:47 PST
 ```For those that know their stats...this is easy stuff. If you can help me with this stuff then you either know the material or you don't...but thank you in advance for the person who takes this job!```
 Subject: Re: Statistics homework From: haversian-ga on 02 Nov 2002 16:59 PST
 ```What tutuzdad is trying to say is that you don't really have a question here. Rather than post a "question" saying "I need someone to help" you should post a question saying "Here are my problems, please someone help" and if you've got 10 problems, that would make it a \$25 question. You don't need to post a separate question asking if anyone is willing to help you, just write out the problems and if someone is willing, they'll answer the question for you. There have been many stats questions asked here, so I'm sure someone will answer them for you (could even be me!). If you'd like to save yourself \$0.50 (the posting fee for a new question), post the 10 problems as a clarification to this question and raise the value.```
 Subject: Re: Statistics homework From: answerguru-ga on 02 Nov 2002 17:00 PST
 ```Hi as3411-ga, Just to let you know, this is not a place where you "hire" someone beforehand. Post your question(s) and the price you want to pay and whoever is knowledgeable about the topic of your question will answer for you. The suggestion that you combine groups of questions is a good one...it would allow someone with understanding of a specific topic to answer all such questions. For example, you could post a \$10 question with 5 similar stats problems (if you feel \$2.50/problem is fair). Obviously if you underprice you will have a hard time finding someone to answer so that's worth keeping in mind. answerguru-ga```
 Subject: Re: Statistics homework From: rbnn-ga on 03 Nov 2002 02:00 PST
 ```Although much of the analysis in the answer is correct, my analysis differs from that of the answerer in some other parts. Specifically, according to my own analysis the answer is incorrect for at least the following questions: Incorrect for question 1. Incorrect for question 2 under both interpretations. Incorrect for question 4. Incorrect for question 5 (c) Incorrect for question 5 (d) You can use the "request clarification" button to ask the answerer for the correct anser to these questions.```
 Subject: Re: Statistics homework From: rbnn-ga on 03 Nov 2002 02:02 PST
 `For "anser" in my last comment, please read "answer".`
 Subject: Re: Statistics homework From: rbnn-ga on 03 Nov 2002 02:57 PST
 `Also, in 5(f), a standard deviation cannot be negative.`
 Subject: Re: Statistics homework From: rbnn-ga on 04 Nov 2002 07:58 PST
 ```Well, I think the standard deviation is always taken to be nonnegative, but I cannot prove this.```