# Thread: so lost ;(

1. ## so lost ;(

let Y be a random vairable giving the no. of heads minus the no. of tails in 2 tosses of a coin. assuming that the coin is biased sot aht the head is twice as likely to occur as the tail,
calculate the mean and standard deviation.
then calculate the cumulative distribution function.

i just learnt this chapter and am completely lost. can someone help me?

these are my working:

P(-1) = 6/27 where there are 2 tails and 1 head
P (1) = 18/27 where there are 2 heads and one tail
P(-3) = 1/27 where there are 3 tails
P( 3) 8/27 wherer there are 3 heads
thus the mean = 1.2222 and the standard deviation = 1.547

how do i go about doing the last part? i know that to find the cumulative distribution function, i have to find the area under p(y) but i do not have an equation to integrate.

did i do something wrong or is there another method to solve this?

thank you!

2. Originally Posted by alexandrabel90
let Y be a random vairable giving the no. of heads minus the no. of tails in 2 tosses of a coin. assuming that the coin is biased sot aht the head is twice as likely to occur as the tail,
calculate the mean and standard deviation.
then calculate the cumulative distribution function.
I think that you have not understood this problem.
The coin is tossed only twice.
Thus your random variable has only three values: -2,two tails; 0, one of each; 2, two heads.
The probabilities of these are: $P(\{T,T\}=\left(\frac{1}{3}\right)^2,~ P(\{H,H\}=\left(\frac{2}{3}\right)^2,~\&~ P(\{H,T\}=2\left(\frac{2}{3}\right) \left(\frac{1}{3}\right)$ .

3. SORRY! that was a typo. it should be taht the coin was tossed 3 times instead.

apart form that typo mistake, how do i go about solving for the cumulative distribution function?

4. Would the cdf be equivalent to finding the probability as in part b? Thanks

5. Originally Posted by alexandrabel90
how do i go about solving for the cumulative distribution function?
Let $X$ be number of heads. Then $X=0,1,2,3$.
$P(X=k)=\binom{3}{k}\left(\frac{2}{3}\right)^k\left (\frac{1}{3}\right)^{3-k}$.

Now that is the Pdf, you use that to write the Cdf.

6. am i right to say taht that is the probability distribution function and the CDF is the summation of that from X = 0 to X= 3 since this is a discrete random variable?

7. wont the summation of it be equals to 1? sorry im not really sure with how to make use of cdf and pdf except that cdf is the summation of pdf

8. Originally Posted by alexandrabel90
am i right to say taht that is the probability distribution function and the CDF is the summation of that from X = 0 to X= 3 since this is a discrete random variable?
If P(x) is the PDF and C(x) is the CDF, then C(0)= P(0), C(1)= P(0)+ P(1), C(2)= P(0)+ P(1)+ P(2), and C(3)= P(0)+ P(1)+ P(2)+ P(3)= 1.