I need help with problem 1 part (d)

- anonymous

I need help with problem 1 part (d)

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- anonymous

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- anonymous

I'm thinking 2.5 ? ?

- anonymous

@AccessDenied @jim_thompson5910 @LoveYou*69 @Luis_Rivera @phi @robtobey

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- anonymous

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

- anonymous

@satellite73

- anonymous

@Hero @dumbcow

- anonymous

@amistre64 @Jemurray3

- anonymous

i think the answer for my first question that i asked above is 2.96...
any help with the second?

- amistre64

i cant make any sense of how they presented the information in the graph :/

- anonymous

can you attempt Q2 Part A? I think I got the other one.

- amistre64

i recall Expected Value is something like np ...

- anonymous

do i just add up em all up and divide by the number of games?

- amistre64

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

- amistre64

thats my understanding of it

- anonymous

for the first one, don't you add up the column?

- anonymous

oh just part d)

- amistre64

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

- anonymous

for the first one i did: (4.0*.41)+(3.0*.36)+(2.0*.12)+(0.0*.11)=2.96

- anonymous

but keep going with the other one... it's more confusing...

- amistre64

\[EV=avg.=\frac{105(11)+110(12)+167(13)+...+1(19)}{500}\]

- anonymous

sorry i am confused
which one is it? 1d)? or 2?

- anonymous

@amistre64 has it for 2 for sure

- anonymous

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)

- amistre64

i just noticed when I put my two thoughts together that they are really the same thought ;)

- anonymous

lol
how would i do the variance part of the question?

- anonymous

i am not sure your answer to 1b) is right though

- anonymous

1d lol

- amistre64

find the difference that each row value is from the mean, square it, add up all the square and divide by ..... 500 i believe

- amistre64

ir is it divide by number of rows?

- amistre64

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

- anonymous

the mean is the # of free throws made divided by 9 right?

- amistre64

hmm
the mean is prolly a small number
105 * (11 - mean)^2 is prolly more accurate to do

- amistre64

might be easier to do in excel fer sure :)

- anonymous

you didnt answer my question lol

- anonymous

the mean is the # of free throws made divided by 9 right? <<<<<

- amistre64

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- amistre64

the mean is the number of free throws per game for 500 games divided by 500

- amistre64

im still uneasy about the var and sd calculations for some reason.

- anonymous

wait isnt that what we did for the expected value?
105*11 + 110*12 + ...... / 500 ?? <

- amistre64

the mean is the expected value; yes. You expect the average results to happen in any given situation.

- amistre64

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?
\[Var=\frac{156(11-\mu)^2+110(12-\mu)^2+...+1(19-\mu)^2}{500}\]

- amistre64

thats better.

##### 1 Attachment

- anonymous

For Problem 3 A through C, are all 3 answers .10??

- phi

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

- anonymous

(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

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

- anonymous

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?

- phi

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

- phi

how did you get these numbers?
those are the gpa for each box

- anonymous

confused -_-

- phi

First, you can figure out what the gpa would be for each little box?

- anonymous

so you take each intersection, e.g A & B and take their GPAs and average them out? so, 4.0+3.0 / 2 ?

- phi

yes,

- phi

Each box represents a possible grade assignment, and consequently a particular gpa

- phi

the expected value will be each gpa times probability of that gpa
all added up

- anonymous

so for the first one it would be: 4.0 * 0.10?

- phi

yes

- anonymous

can you help with problem 3? i got .10 for A through C

- anonymous

not sure if .10 is right though

- phi

I think you use the binomial distribution

- phi

http://en.wikipedia.org/wiki/Binomial_distribution

- phi

does that sound familiar ?

- anonymous

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.

- phi

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)

- phi

for part A, I think you use
\[\Pr(k \text{ occurrences }) = \left(\begin{matrix}n \\ k\end{matrix}\right)p^k (1-p)^{n-k}\]

- phi

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

- phi

I think for part C you do it brute force for k= 1,3,5,7,9
part B will be 1 - part C

- anonymous

can you fill in the formula for part A so I can see it

- phi

wikipedia gives an example, scroll down...

- anonymous

wait i see it now

- anonymous

so it would be .05^0 * (1-.05) ^20-0
for 0
and then replace the zero with a 1 and then 2???

- phi

there is an n choose k out front or 20 choose 0 (which I think is 1)
but yes

- phi

question 4 looks like you use the Poisson distribution

- phi

But I have to go now.

- anonymous

ok thanks for the help!!

- phi

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).

- anonymous

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.

- anonymous

@phi

- phi

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

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