Question

let x be the number of heads in two tosses of a biased coin with P(H)=.4. Work out the mean and variance of X?

Answer 1

It's a binomial distribution, right?

Answer 2

That would make the expected value \(np\) and the variance \(np(1-p)\)

Answer 3

In this case \(n=2\) and \(p=0.4\).

Do you want to do this manually?

Answer 4

yes, please

Answer 5

Okay, so there are only \(2^2=4\) outcomes.\[
HH, HT, TH, TT
\]

Answer 6

Remember that \(P(T)= 1-P(H) = 1-0.4=0.6\).
The respective probabilities are: \[
(0.4)(0.4), (0.4)(0.6), (0.6)(0.4),(0.6)(0.6)
\]

Answer 7

Do you understand these probabilities? Remember the flips are independent events so probability of getting heads twice is just \(\Pr(H)\times \Pr(H)\)

Answer 8

X=2 (prob=1/4)
x=1 (prob=1/2)
x=0 (prob=1/4)

Answer 9

oh. wait, i got how you got those probabilities. but how did you get 0.6?

Answer 10

\(1-0.4=0.6\)

Answer 11

got it! how do i solve it manually?

Answer 12

The respective value of \(x\) for these outcomes are \[
2, 1, 1, 0
\]

Answer 13

The expected value is:
The sum of:
The probability of each outcome times the value of that outcome

Answer 14

So \[
E[X] = 2(0.4)(0.4)+1(0.4)(0.6)+1(0.6)(0.4)+0(0.6)(0.6)
\]

Answer 15

Which simplifies to \(0.8\).
And checking it with the typical binomial distribution: \(n=2,p=0.4\): \[2\times 0.4=0.8\]

Answer 16

ok, that's what i got the first time! thank you!

Answer 17

For the variance, we have to use different values. We subtract the expected value from the normal value and square it.

Answer 18

so i would use the .16, .24, .24, .16 ?? multiplying it with 2, 1, 1, and 0?

Answer 19

I'm not sure where you are getting those numbers.

Answer 20

the probabilities of 0.4*0.4=.16???

Answer 21

The values \[
2,1,1,0
\]Need to be replaced with \[
(2-0.8)^2,(1-0.8)^2,(1-0.8)^2,(0-0.8)^2
\]

Answer 22

\((0.4)(0.4)=0.16\) But \((0.6)(0.6)=0.36\)

Answer 23

right!
(2*.16) + (1*.24) + (1*.24) + (0*.36)

Answer 24

Yes, that is for the expected value.

Answer 25

that would the sigma^2 answer correct?

Answer 26

how do i find the variance?

Answer 27

It's almost the same as expected value except for each value, you change it to \((x-\mu)^2\)

The values \[
2,1,1,0
\]Need to be replaced with \[
(2-0.8)^2,(1-0.8)^2,(1-0.8)^2,(0-0.8)^2
\]

Answer 28

why are you subtracting it by .8?

Answer 29

i can't get it to equal .48...

Answer 30

Because \(0.8\) is the expected value.

Answer 31

\[
(2−0.8)^2(0.4)(0.4)+(1−0.8)^2(0.4)(0.6)+(1−0.8)^2(0.6)(0.4)+(0−0.8)^2(0.6)(0.6)
\]

Answer 32

That comes out to \(0.48\) for me.