Question

I need help with problem 1 part (d)

Answer 1

I'm thinking 2.5 ? ?

Answer 2

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

Answer 3

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

Answer 4

@satellite73

Answer 5

@Hero @dumbcow

Answer 6

@amistre64 @Jemurray3

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

i recall Expected Value is something like np ...

Answer 11

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

Answer 12

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

Answer 13

thats my understanding of it

Answer 14

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

Answer 15

oh just part d)

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

@amistre64 has it for 2 for sure

Answer 22

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)

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

1d lol

Answer 27

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

Answer 28

ir is it divide by number of rows?

Answer 29

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

Answer 30

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

Answer 31

hmm

the mean is prolly a small number

105 * (11 - mean)^2 is prolly more accurate to do

Answer 32

might be easier to do in excel fer sure :)

Answer 33

you didnt answer my question lol

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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?

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

Answer 40

thats better.

Answer 41

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

Answer 42

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

Answer 43

(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.

Answer 44

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

Answer 45

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?

Answer 46

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

Answer 47

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

Answer 48

confused -_-

Answer 49

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

Answer 50

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

Answer 51

yes,

Answer 52

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

Answer 53

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

Answer 54

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

Answer 55

yes

Answer 56

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

Answer 57

not sure if .10 is right though

Answer 58

I think you use the binomial distribution

Answer 59

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

Answer 60

does that sound familiar ?

Answer 61

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.

Answer 62

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)

Answer 63

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

Answer 64

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

Answer 65

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

Answer 66

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

Answer 67

wikipedia gives an example, scroll down...

Answer 68

wait i see it now

Answer 69

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

Answer 70

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

but yes

Answer 71

question 4 looks like you use the Poisson distribution

Answer 72

But I have to go now.

Answer 73

ok thanks for the help!!

Answer 74

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).

Answer 75

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.

Answer 76

@phi

Answer 77

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