Question

A fair coin is tossed 3 times in a row. If X is the number of heads counted, what is the mean of the probability distribution of X?
1.0
1.5
2.0
2.5
3.0

Answer 1

@Michele_Laino

Answer 2

please wait a moment

Answer 3

please wait, your last equation is the mean value of x, so it can not be equal to 1

Answer 4

ok

Answer 5

I'm trying to solve your question, please wait...

Answer 6

Yea its all good

Answer 7

X can be 0, 1, 2, or 3

Answer 8

so we get the corresponding probabilities:
X=0--->p=0
X=1--->p=1/3
X=2--->p=2/3
X=3--->p=1

Answer 9

so it is 1.0?

Answer 10

why?

Answer 11

cause x=3 at p=1

Answer 12

we have to determine the mean value of X

Answer 13

so find the mean?

Answer 14

I get a 0.5 for the mean

Answer 15

yes! I got the same result

Answer 16

maybe the answer is 1

Answer 17

I think that we have to consider the binomial distribution.

Answer 18

now, if I toss my coin three times in a row, then we have N tests, and X is the number of success in our tests, furthermore the probability to get a tail is 1/2, and the probability to get a head is 1/2, so the probability to get X head is given by the binomial distribution, as follows:
\[\left( {\begin{array}{*{20}{c}}
3 \\
X
\end{array}} \right){\left( {\frac{1}{2}} \right)^3}\]

Answer 19

where:
\[\left( {\begin{array}{*{20}{c}}
3 \\
X
\end{array}} \right) = \frac{{3!}}{{X!\left( {3 - X} \right)!}}\]

Answer 20

I get (6 (3-X)!)/(X!)

Answer 21

no, please, the mean value is given by the subsequent formula:
\[\mu = \sqrt {Npq} = \sqrt {3 \times \frac{1}{2} \times \frac{1}{2}} = ...?\]

Answer 22

0.866

Answer 23

or sqrt(3)/2

Answer 24

Sorry I have made an error, the mean value of the binomial didtribution, is given by the subsequent formula:
\[\Large \mu = Np = 3 \times \frac{1}{2} = ...?\]

Answer 25

distribution*

Answer 26

1.5

Answer 27

that's right!

Answer 28

p is the probability to get a head, namely p=1/2, whereas q is the probability to get a tail, namely q=p=1/2

Answer 29

I also have more questions but I don't think you would want to help me again

Answer 30

thanks alot

Answer 31

I have to help another student, so you have to wait, please post your question

Answer 32

ok! I can help you now!

Answer 33

:)

Answer 34

What is the mean of this probability distribution?

Answer 35

P(X=xi) is for the decimals :0.2,0.3,0.2,0.1,0.1,0.1

Answer 36

1.9
2.9
3.1
3.2
3.4

are the answers

Answer 37

here we have to apply the classical definition of mean value, so we have to do this computation:
\[\large 1 \times 0.2 + 2 \times 0.3 + 3 \times 0.2 + 4 \times 0.1 + 5 \times 0.1 + 6 \times 0.1 = ...?\]

Answer 38

2.9

Answer 39

1*0.2+2*0.3+3*0.2+4*0.1+5*0.1+6*0.1=2.9

Answer 40

that's right!

Answer 41

cool

Answer 42

one last question

Answer 43

ok!

Answer 44

I think uniform means continuous

Answer 45

I think that your distribution is like this:

Answer 46

yea

Answer 47

namely:
p(x)=0, if x>b or x<a,
and
p(x)= b-a, if a<x<b

Answer 48

we can eliminate a

Answer 49

oops. I have made an error, here is the right graph: