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I'm thinking 2.5 ? ?

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?

thats my understanding of it

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

oh just part d)

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

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

1d lol

ir is it divide by number of rows?

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 ?? <

thats better.

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

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?

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

confused -_-

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 ?

yes,

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?

yes

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

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

does that sound familiar ?

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

But I have to go now.

ok thanks for the help!!