I need help with some statistics homework...about 25-33
problems...I'll pay $2.50 a question for this help....please!!

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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.

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

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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?!?!?)

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.

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.

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!

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.

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

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.

For "anser" in my last comment, please read "answer".

Also, in 5(f), a standard deviation cannot be negative.

Well, I think the standard deviation is always taken to be
nonnegative, but I cannot prove this.