Question

does anyone know how to calculate expected value? i have some problems that i can't solve.

Answer 1

go for it

Answer 2

1000 chances are sold at $5 a piece for a raffle. Grand prize of $650, two prizes of $275, and 5 prizes of $100. first, calculate the expected value. if it is not a fair game, find the fair price for the game.

Answer 3

E[ X ] = sum of probabilities * values

E[ X ] = (1/1000)*650 + (2/1000)*275 + (5/1000)*100 + (992/1000)*(-5)

E[ X ] = ???

Answer 4

-3.26?

Answer 5

you got it, so does that mean we have a fair game?

Answer 6

no because fair game would be zero. so how exactly do I figure out the fair price?

Answer 7

let p = price

you want E[ X ] = 0 and you want to find the value of p that makes it zero

Answer 8

E[ X ] = (1/1000)*650 + (2/1000)*275 + (5/1000)*100 + (992/1000)*p

0 = (1/1000)*650 + (2/1000)*275 + (5/1000)*100 + (992/1000)*p

solve for p

Answer 9

oh sry, it should be

0 = (1/1000)*650 + (2/1000)*275 + (5/1000)*100 + (992/1000)*(-p)

solve for p to get your fair price

Answer 10

-1.71?

Answer 11

sorry i mean $1.71? per ticket

Answer 12

good, I'm getting the same pretty much

Answer 13

so if the price is $5 a ticket, then you expect to lose $3.26 on average

if the price is around $1.71 a ticket, then you won't lose or gain any money. So it's a fair game for both parties.

Answer 14

thanks! i get it now. i have two other questions, can you help me with those?

Answer 15

a student is taking a standardized test consisting of multiple choice questions. one point is awarded for each correct answer. questions left blank do not get or lose points. if there are 8 options for each question and the student is penalized 1/2 point for each wrong answer, how many options must the student be able to rule out before the expected value of guessing is zero?

Answer 16

is that question similar to the other?

Answer 17

Does it say how many questions total there are?

Answer 18

no

Answer 19

ok let me think for a sec

Answer 20

I think it's

E[ X ] = (1/8)*1 + (7/8)*(-1/2)

but I'm still figuring out how to factor in the ones leave blank...hmm

Answer 21

*you leave blank*

Answer 22

i got -.3125 for that so far

Answer 23

maybe it's a case where we don't leave any blank

Answer 24

let's assume that

Answer 25

so if

n = number of options you rule out

then

8 - n is the number of options left over

8 - n - 1 = 7 - n is the number of wrong options

so this means

E[ X ] = (1/(8-n))*1 + ((7-n)/(8-n))*(-1/2)

0 = (1/(8-n))*1 + ((7-n)/(8-n))*(-1/2)

solve that for n to get your answer

Answer 26

tell me what you get

Answer 27

im having trouble calculating it but it could be -5??

Answer 28

it's 5 actually, so you lost a sign somewhere

Answer 29

so if you rule out 5 options, then randomly guessing on every question you attempt will net you 0 points on average

Answer 30

ok got it

Answer 31

i have one more

Answer 32

this means that you will have 8-5 = 3 options to choose from and the expected value is

E[ X ] = (1/3)*1 + (2/3)*(-1/2)

E[ X ] = 0

which confirms the answer

Answer 33

assume you have a laptop worth $2,500 and there is a 4% chance that it will be lost or stolen during the next year. what would be a fair premium for the insurance? (assuming the insurance provider does not make a profit)

Answer 34

so the number of options the student must be able to rule out before the expected value of guessing becomes zero is 5?

Answer 35

yes

Answer 36

ok, thank you

Answer 37

Let p = premium amount

E[ X ] = (0.04)*(2500) + (0.96)*(-p)

0 = (0.04)*(2500) + (0.96)*(-p) ... Fair premium is when E[X] = 0

solve for p and round to the nearest cent

Answer 38

so would i divide 100 by -0.96 to get p?

Answer 39

correct, you would get

0 = (0.04)*(2500) + (0.96)*(-p)

0 = 100 + (0.96)*(-p)

-100 = (0.96)*(-p)

0.96p = 100

p = 100/0.96

p = 104.166666666667

p = 104.17

Answer 40

so the fair premium amount (for both parties) is $104.17

Answer 41

okay, thanks so much for the help!

Answer 42

np