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mariomintchev
I need help with problem 1 part (d)
I'm thinking 2.5 ? ?
@AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey
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
@amistre64 @Jemurray3
i think the answer for my first question that i asked above is 2.96... any help with the second?
i cant make any sense of how they presented the information in the graph :/
can you attempt Q2 Part A? I think I got the other one.
i recall Expected Value is something like np ...
do i just add up em all up and divide by the number of games?
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
thats my understanding of it
for the first one, don't you add up the column?
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
for the first one i did: (4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96
but keep going with the other one... it's more confusing...
\[EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}\]
sorry i am confused which one is it? 1d)? or 2?
@amistre64 has it for 2 for sure
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)
i just noticed when I put my two thoughts together that they are really the same thought ;)
lol how would i do the variance part of the question?
i am not sure your answer to 1b) is right though
find the difference that each row value is from the mean, square it, add up all the square and divide by ..... 500 i believe
ir is it divide by number of rows?
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
the mean is the # of free throws made divided by 9 right?
hmm the mean is prolly a small number 105 * (11 - mean)^2 is prolly more accurate to do
might be easier to do in excel fer sure :)
you didnt answer my question lol
the mean is the # of free throws made divided by 9 right? <<<<<
the mean is the number of free throws per game for 500 games divided by 500
im still uneasy about the var and sd calculations for some reason.
wait isnt that what we did for the expected value? 105*11 + 110*12 + ...... / 500 ?? <<That's the mean and expected value?
the mean is the expected value; yes. You expect the average results to happen in any given situation.
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 (11-mean)^2 (11-mean)^2 (11-mean)^2 ... 156 times (12-mean)^2 (12-mean)^2 (12-mean)^2 ... 110 times and for 500 games. adding up all those squared parts and dividing by 500 gives us variance right? \[Var=\frac{156(11-\mu)^2+110(12-\mu)^2+...+1(19-\mu)^2}{500}\]
For Problem 3 A through C, are all 3 answers .10??
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
(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.
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
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?
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
how did you get these numbers? those are the gpa for each box
First, you can figure out what the gpa would be for each little box?
so you take each intersection, e.g A & B and take their GPAs and average them out? so, 4.0+3.0 / 2 ?
Each box represents a possible grade assignment, and consequently a particular gpa
the expected value will be each gpa times probability of that gpa all added up
so for the first one it would be: 4.0 * 0.10?
can you help with problem 3? i got .10 for A through C
not sure if .10 is right though
I think you use the binomial distribution
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.
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)
for part A, I think you use \[\Pr(k \text{ occurrences }) = \left(\begin{matrix}n \\ k\end{matrix}\right)p^k (1-p)^{n-k}\]
p is the probability of an A 0.05 1-p= 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
I think for part C you do it brute force for k= 1,3,5,7,9 part B will be 1 - part C
can you fill in the formula for part A so I can see it
wikipedia gives an example, scroll down...
wait i see it now
so it would be .05^0 * (1-.05) ^20-0 for 0 and then replace the zero with a 1 and then 2???
there is an n choose k out front or 20 choose 0 (which I think is 1) but yes
question 4 looks like you use the Poisson distribution
ok thanks for the help!!
For problem 3 I hope you caught my mistake. p (chance of an A) is 20% so p= 20/100= 0.2 and (1-p) is 0.8 (somehow I got 0.05 which is wrong).
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.
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