Need help with Statistics. Dollar for each correct question.

Logey.

Logey.

Active member
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
 
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
 
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
 
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
 
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 :-)
 
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.
 
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
 
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