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anonymous
 3 years ago
I need help with problem 1 part (d)
anonymous
 3 years ago
I need help with problem 1 part (d)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Also, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 @Jemurray3

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2i cant make any sense of how they presented the information in the graph :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you attempt Q2 Part A? I think I got the other one.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2i recall Expected Value is something like np ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do i just add up em all up and divide by the number of games?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2it 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2thats my understanding of it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0for the first one, don't you add up the column?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2another 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but keep going with the other one... it's more confusing...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i am confused which one is it? 1d)? or 2?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 has it for 2 for sure

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Problem 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)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2i just noticed when I put my two thoughts together that they are really the same thought ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol how would i do the variance part of the question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am not sure your answer to 1b) is right though

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2find the difference that each row value is from the mean, square it, add up all the square and divide by ..... 500 i believe

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2ir is it divide by number of rows?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2standard 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the mean is the # of free throws made divided by 9 right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2hmm the mean is prolly a small number 105 * (11  mean)^2 is prolly more accurate to do

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2might be easier to do in excel fer sure :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you didnt answer my question lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the mean is the # of free throws made divided by 9 right? <<<<<

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the mean is the number of free throws per game for 500 games divided by 500

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2im still uneasy about the var and sd calculations for some reason.

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the mean is the expected value; yes. You expect the average results to happen in any given situation.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2if 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? \[Var=\frac{156(11\mu)^2+110(12\mu)^2+...+1(19\mu)^2}{500}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For Problem 3 A through C, are all 3 answers .10??

phi
 3 years ago
Best ResponseYou've already chosen the best response.0How 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

anonymous
 3 years ago
Best 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.

phi
 3 years ago
Best ResponseYou've already chosen the best response.0For 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.04 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?

phi
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0how did you get these numbers? those are the gpa for each box

phi
 3 years ago
Best ResponseYou've already chosen the best response.0First, you can figure out what the gpa would be for each little box?

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0Each box represents a possible grade assignment, and consequently a particular gpa

phi
 3 years ago
Best ResponseYou've already chosen the best response.0the expected value will be each gpa times probability of that gpa all added up

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so for the first one it would be: 4.0 * 0.10?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not sure if .10 is right though

phi
 3 years ago
Best ResponseYou've already chosen the best response.0I think you use the binomial distribution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0somewhat, 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.

phi
 3 years ago
Best ResponseYou've already chosen the best response.0notice 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)

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0p 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can you fill in the formula for part A so I can see it

phi
 3 years ago
Best ResponseYou've already chosen the best response.0wikipedia gives an example, scroll down...

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

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.0question 4 looks like you use the Poisson distribution

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok thanks for the help!!

phi
 3 years ago
Best ResponseYou've already chosen the best response.0For 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0o 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.

phi
 3 years ago
Best ResponseYou've already chosen the best response.0for 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
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