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Need help with Statistics. Dollar for each correct question.

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Logey.
Posts: 2996 - Karma: 457
Just started this stupid ass hybrid class and the teacher thought it would be funny not to teach us anything that is on the quiz. To anyone that can come up with the right answer and explain how they got it will receive 1 dollar per question via paypal.

Question 1
A fisherman has a .01 probability of catching a fish on each cast he makes. What is the number the fisherman should expect to catch in the next 8 casts? (round to 2 decimals)

Question 2
Each person walking into a Cell Phone store has a probability of .2 of signing up for a new line of service. What is the probability that the first customer to get a new line of service will be in the first 3 customers to come to the store? (round to 3 decimal places)

Question 3
At a busy hotel, an average of 1.3 customers arrive each minute in the hour between 5PM and 6 PM on Monday nights. What is the standard deviation of the number of customers arriving in a minute? (round to 2 decimal places)

Question 4
Each person walking into a Cell Phone store has a probability of .2 of signing up for a new line of service. What is the probability that the first customer to sign up for a new line of service at a particular location will be the 5th one to come in the store that day? (Round to 2 decimal places)

Question 5
A student receives a box of chocolates from a friend. Reading the back, it seems that 4 of the 15 have coconut in them. The problem is that they all look identical. What is the expected number that should contain coconut from the 6 sampled? (round to 1 decimal place)

FUCK THIS SHIT
Mar 30 2015 9:04PM
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J_MAC
Posts: 572 - Karma: 127
question 5 = 1.6
Mar 31 2015 1:43AM
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J_MAC
Posts: 572 - Karma: 127
question 1 = 0.08?
Mar 31 2015 1:45AM
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J_MAC
Posts: 572 - Karma: 127
gime 2$ and youl get the rest mawwfuckaaa
Mar 31 2015 1:46AM
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ScubaSteveFather
Posts: 555 - Karma: 574
ur wrong on number one its actually .077250

Well at least thats the probability he will catch one fish within those tries.

You have to do .01+.01 - .01*.01 for the first 2 catches
then same thing but use ur result from first one so
.0199+.0199-.0199*.0199
same thing again
.039404+.039404-.039404*.039404
this gives a prob of .077250 of catching one fish in eight tries

Ill post answers with explanation to the rest in a bit
Mar 31 2015 2:42AM
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ScubaSteveFather
Posts: 555 - Karma: 574
Question 2

So for first 2 customers it would be .2+.2-.2*.2 = 0.36

Then to do first 3 customers we do
.2+.36-.2*.36=0.488

The main concept here is using the union of events to calculate the probability
P(C) = 0.2 (Probability of customer getting free line)
P(C U C) = P(C) + P(C) - P(C ∩ C)
For 2 independent events P(C ∩ C) = P(C)*P(C) (intersection of events)

Thus
P(C U C U C)= P(C U C) U P(C) Thats basically what I used

A more direct approach is to do all 3 at once but it gets a bit confusing. I'll show that here

P(C U C U C) = P(C) + P(C) +P(C) - P(C ∩ C) - P(C ∩ C) - P(C ∩ C) +(C ∩ C ∩ C)

or P(C) + P(C) +P(C) - P(C*P(C) - P(C*P(C) - P(C*P(C) +P(C)*P(C)*P(C)

= 0.2+0.2+0.2-0.2*0.2-0.2*0.2-0.2*0.2 + P(C)*P(C)*P(C)

=0.488

If you need clarification on that rule here is a good link

http://w3.uwyo.edu/~dlegg/union2.html
Mar 31 2015 2:57AM
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ScubaSteveFather
Posts: 555 - Karma: 574
Question 3 is impossible to answer

Right here you have mean (1.3) and sample size (one hour so 60 minutes)

It is not possible to calculate standard deviation (also known as variance) with only those 2 values. You also need the number for each data point in this case each minute you need to know how many people come in.

This is reference that shows that well
http://math.stackexchange.com/questions/248826/is-it-possible-to-calculate-the-standard-deviation-with-a-given-mean-and-sample
Mar 31 2015 3:03AM
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ScubaSteveFather
Posts: 555 - Karma: 574
Question 4

THis is a bit trickier but also uses the concept developed in 1 and 2

P(C) = 0.2 (probability customer will get a new line)
1-P(C) = P(NC) = 0.8 (probability he wont)

Thus assuming each event is independent we get P(NC U NC U NC U NC U C)
4 customers without getting one need to come in and one with needs to come in.

Here is solution of that union in two steps but easiest way to work it out is by doing P(NC U NC) and then putting in the others one by one

since P(NC U NC) = 0.8 +0.8 -0.8*0.8 = 0.96

Say P(LC) = 0.96

then P(NC U NC U NC U NC U C) = P(LC U LC U C) (which is much easier to solve)

so

P(LC U LC U C) = P(LC)+P(LC)+P(C)-P(LC)*P(LC)-P(LC)*P(C)-P(LC)*P(C)+P(LC)*P(LC)*P(C)

= 0.96+0.96+0.2-0.96*0.96-0.96*0.2-0.96*0.2+0.96*0.96*0.2

=0.63008

So in all probability that the fifth customer gets a new line and not the other 4 before him is 0.63008
Mar 31 2015 3:13AM
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ScubaSteveFather
Posts: 555 - Karma: 574
Question 5 is real easy just a proportion problem

Sooo

4/15 = X/6

solve for x

4/15*6=x

x= 1.6


Should expect 1.6 of them to be coconut (I'd normally round up because above .5 and cant really have .6 of one be coconut but it depends on ur teacher)

GL HF in prob stats :-)
Mar 31 2015 3:16AM
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theBearJew
Posts: 10951 - Karma: 15,633
Question 1. 7

Question 2. 7

Question 3. 7

Question 4. 7

Question 5. 7


The probability that any of these answers are correct, but I like my odds.
Mar 31 2015 8:01AM
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ScubaSteveFather
Posts: 555 - Karma: 574
I fucked up on question 4

after re reading it and checking my work it should be

for this its actually just
P(NC)*P(NC)*P(NC)*P(NC)*P(C)

where P(NC) = 0.8 and P(C) = 0.2

thus the answer is 0.08192 which actually makes more sense since it should be under 20% chance because the 4 before need to not be the ones to get a new line and we r looking at only the fifth event not all 5.

The previous answer i posted for question 4 actually was for 4 of them not getting a new line and one of them getting a new line (not in any order) and just the chance of that happening.

Answer for #4 is thus 0.08192
Mar 31 2015 10:19AM
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