Question

someone be kind enough to help? :)

Answer 1

heres the question :)

Answer 2

anyyyyone?

Answer 3

@amistre64

Answer 4

i'd help but i suck at math D:

Answer 5

:/ thanks anyway :)

Answer 6

@thomaster

Answer 7

@modphysnoob

Answer 8

@ash2326

Answer 9

number of bases, divided by number of times

Answer 10

each?

Answer 11

the number of bases is the sum total of bases:

0(450)+1(190)+2(32)+3(10)+4(10)

all out of 692 attempts

Answer 12

sooo i dont divide?

Answer 13

of course you do; this is a weighted average

Answer 14

instead of counting out: 490 zeros, 190 "1s", 32 "2s" etc ... can multiply top to bottom to determine the total number of basses

Answer 15

since it tells you the total number of attempts in the problem is 692 .... thats your divider

Answer 16

E[X] = x1p1 +x2p2+x3p3...+xnpn

Answer 17

which simplifies to that setup as well

Answer 18

can you find p(x)?

Answer 19

im soo confused :(

Answer 20

\[\frac{0(450)+1(190)+2(32)+3(10)+4(10)}{692}\]
\[\frac{0(450)}{692}+\frac{1(190)}{692}+\frac{2(32)}{692}+\frac{3(10)}{692}+\frac{4(10)}{692}\]
\[0\frac{450}{692}+1\frac{190}{692}+2\frac{32}{692}+3\frac{10}{692}+4\frac{10}{692}\]

Answer 21

can you tell what is probability of hitting 0?

Answer 22

since the weighted average is simpler to compute; id just go with it

Answer 23

so i add 450 + 190 + 32 + 10 + 10 ?

Answer 24

not according to the format i presented ....

Answer 25

but then thats just adds to 692

Answer 26

thats why i asked, what am i supposed to do first then?

Answer 27

no need to, they already give you that information in the problem setup

Answer 28

@dianadelucio you know what Expectation is?

Answer 29

@ankit042 sort f yea.

Answer 30

of

Answer 31

lol, thought maybe that was a swear word :)

Answer 32

haha nooo, why would i swear when you guys are trying to help me lol @amistre64

Answer 33

haha I hope it was not the other f :P
anyways .
E[x] = x1p1 +x2p2+...+xnpn
Now from the table you can easily find probability of different events

Answer 34

he gets 324 bases out of 692 hits

Answer 35

p(0) = 450/692

Answer 36

what does that meann? E[x] = x1p1 +x2p2+...+xnpn

Answer 37

i broke down ankits method from the weighted average im using .... i beleive i did it correctly that is

Answer 38

sooo.. what am i supposed to do with the equation?

Answer 39

let me quote from Wikipedia
"Expected value refers to value of a random variable one would "expect" to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained."

Answer 40

the expected value formula is a "simplification" of the weighted average

since itll take extra steps to determine the probabilities of all the hits, i suggest just taking the weighted average approach is all

Answer 41

and how do i find that?

Answer 42

\[\frac{\#bases}{\#at-bats}\]
\[\frac{0(450)+1(190)+2(32)+3(10)+4(10)}{692}\]
\[\frac{324}{692}\]

Answer 43

its the same method of been doing since the start :)

Answer 44

reduce the top and bottom as needed

Answer 45

So in our sample space events(x) are 0,1,2,3,4
now we know the likelihood of an event(probability) from the table

Answer 46

81/173 ?

Answer 47

thats what i get, yes. it just seems weird in context to me ....

Answer 48

on average, he gets 81/173 bases per hit

Answer 49

yes, correct

Answer 50

the probability of him getting to a base is about 46%

Answer 51

thank you ! @amistre64 @ankit042 @ash2326 :)

Answer 52

youre welcome, and again ... good luck :)