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mariomintchevBest ResponseYou've already chosen the best response.0
I'm thinking 2.5 ? ?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
Also, for problem 2 part (a)... what formulas would i use to calculate the problems? E(x+y)=E(x)+E(y) ??? For the expected value and V(x+y)=V(x)+V(y)+2COV (x,y) ??? For the variance
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
@amistre64 @Jemurray3
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
i think the answer for my first question that i asked above is 2.96... any help with the second?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
i cant make any sense of how they presented the information in the graph :/
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
can you attempt Q2 Part A? I think I got the other one.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
i recall Expected Value is something like np ...
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
do i just add up em all up and divide by the number of games?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
it looks to me like the number of games that the number of FreeThrows were made in. so in 105 games, each game had 11 FreeThrows made. giving us 105 * 11 FTs for the first row but what is the expected number of Free Throws with the given data? 105/500 percent of the time they made 11 shots per game. 110/500 percent of the time they made 12 shots per game etc ... then you add up all the EV values that each row represents
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
thats my understanding of it
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
for the first one, don't you add up the column?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
another thought I have is to just multiply across the rows, that will give you the total number of FTs for those games, then add up all the FTs that were made and divide it by 500 for an average
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
for the first one i did: (4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
but keep going with the other one... it's more confusing...
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
\[EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
sorry i am confused which one is it? 1d)? or 2?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
@amistre64 has it for 2 for sure
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
Problem 2 Part A (is what Amistre is helping me with right now) the other one was Problem 1 Part D (which I got 2.96 for)
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
i just noticed when I put my two thoughts together that they are really the same thought ;)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
lol how would i do the variance part of the question?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
i am not sure your answer to 1b) is right though
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
find the difference that each row value is from the mean, square it, add up all the square and divide by ..... 500 i believe
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
ir is it divide by number of rows?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
standard deviation is the average deviation that we expect to get huddled about the mean. (105(11)  mean)^2 +(110(12) mean)^2 + ......................... summed value of squares divide that by 500 to get the variance
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
the mean is the # of free throws made divided by 9 right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
hmm the mean is prolly a small number 105 * (11  mean)^2 is prolly more accurate to do
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
might be easier to do in excel fer sure :)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
you didnt answer my question lol
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
the mean is the # of free throws made divided by 9 right? <<<<<
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
the mean is the number of free throws per game for 500 games divided by 500
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
im still uneasy about the var and sd calculations for some reason.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
wait isnt that what we did for the expected value? 105*11 + 110*12 + ...... / 500 ?? <<That's the mean and expected value?
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
the mean is the expected value; yes. You expect the average results to happen in any given situation.
 one year ago

amistre64Best ResponseYou've already chosen the best response.2
if we had a list of ALL the games in a one to one list: it would look like: 1, 11 2, 11 3, 11 4, 11 ... 156, 11 157, 12 158, 12 .... etc for all 500 games played. if we take this list to find the variance, we have to subtract the mean from each data point, and square the reults (11mean)^2 (11mean)^2 (11mean)^2 ... 156 times (12mean)^2 (12mean)^2 (12mean)^2 ... 110 times and for 500 games. adding up all those squared parts and dividing by 500 gives us variance right? \[Var=\frac{156(11\mu)^2+110(12\mu)^2+...+1(19\mu)^2}{500}\]
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
For Problem 3 A through C, are all 3 answers .10??
 one year ago

phiBest ResponseYou've already chosen the best response.0
How did ou get 2.96 for question 1 d ? I would multiply the probabilities by these numbers and add them up 4 3.5 3 2 3.5 3 2.5 1.5 3 2.5 2 1 2 1.5 1 0
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
(4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96 is what i did. i added all the columns up and then multiplied by the different GPAS.
 one year ago

phiBest ResponseYou've already chosen the best response.0
For Q 2 I would get the total number of games and then create a weight # of games/total and use that to weight the number of free throws e.g. for the first row, 105/total * 11 do that for each row, and add up the results
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
4 3.5 3 2 3.5 3 2.5 1.5 3 2.5 2 1 2 1.5 1 0 how did you get these numbers?
 one year ago

phiBest ResponseYou've already chosen the best response.0
i added all the columns up and then multiplied by the different GPAS. that does not sound right. The expected value is the sum of the outcome * probability of the outcome. For example, a 4.0 (2 A's) has prob 0.1 multiply to get 0.4 do that for each box, and add up the numbers
 one year ago

phiBest ResponseYou've already chosen the best response.0
how did you get these numbers? those are the gpa for each box
 one year ago

phiBest ResponseYou've already chosen the best response.0
First, you can figure out what the gpa would be for each little box?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so you take each intersection, e.g A & B and take their GPAs and average them out? so, 4.0+3.0 / 2 ?
 one year ago

phiBest ResponseYou've already chosen the best response.0
Each box represents a possible grade assignment, and consequently a particular gpa
 one year ago

phiBest ResponseYou've already chosen the best response.0
the expected value will be each gpa times probability of that gpa all added up
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so for the first one it would be: 4.0 * 0.10?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
can you help with problem 3? i got .10 for A through C
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
not sure if .10 is right though
 one year ago

phiBest ResponseYou've already chosen the best response.0
I think you use the binomial distribution
 one year ago

phiBest ResponseYou've already chosen the best response.0
http://en.wikipedia.org/wiki/Binomial_distribution
 one year ago

phiBest ResponseYou've already chosen the best response.0
does that sound familiar ?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
somewhat, yes. i just started the semester recently so i'm not too keen on all of this. i haven't taken a lot of stats courses.
 one year ago

phiBest ResponseYou've already chosen the best response.0
notice part B and part C have to add to 100%. either an odd number or an even number got A's, and the two conditions have to add to probability 1: It is 100% certain either an odd number or an even # got an A (counting zero as even)
 one year ago

phiBest ResponseYou've already chosen the best response.0
for part A, I think you use \[\Pr(k \text{ occurrences }) = \left(\begin{matrix}n \\ k\end{matrix}\right)p^k (1p)^{nk}\]
 one year ago

phiBest ResponseYou've already chosen the best response.0
p is the probability of an A 0.05 1p= 0.95 n is 20 we have to do this computation 3 times, for k=0,1, 2 ) and add up the results k is the # of A's in the group of 20
 one year ago

phiBest ResponseYou've already chosen the best response.0
I think for part C you do it brute force for k= 1,3,5,7,9 part B will be 1  part C
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
can you fill in the formula for part A so I can see it
 one year ago

phiBest ResponseYou've already chosen the best response.0
wikipedia gives an example, scroll down...
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
wait i see it now
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
so it would be .05^0 * (1.05) ^200 for 0 and then replace the zero with a 1 and then 2???
 one year ago

phiBest ResponseYou've already chosen the best response.0
there is an n choose k out front or 20 choose 0 (which I think is 1) but yes
 one year ago

phiBest ResponseYou've already chosen the best response.0
question 4 looks like you use the Poisson distribution
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
ok thanks for the help!!
 one year ago

phiBest ResponseYou've already chosen the best response.0
For problem 3 I hope you caught my mistake. p (chance of an A) is 20% so p= 20/100= 0.2 and (1p) is 0.8 (somehow I got 0.05 which is wrong).
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.0
o my god. now i gotta go back and re do everything. ahhhh. haha can you help with problem 4 while re calculate everything from prob 3. i told you what i was having issues with in the message i sent you yesterday after you left.
 one year ago

phiBest ResponseYou've already chosen the best response.0
for problem 4, use \[ Pr(k)= f(k,\lambda) =\frac{λ^k e^{λ}}{k!} \] where λ is the expected number of events in an interval and k is the number of events that occur for part (a) λ is 2.5 for part (b) λ is 2.5*8= 20
 one year ago
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