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S1 Question

Q) The random variable X has a binomial distribution with parameters n=25, p=0.8. The random variable Y is defined by Y=aX+b, where a, b > 0.

The mean of Y= 65
SD of Y=6

My question is how do you know which value to put in for x in the variance formula for x=Sum of xi^2pi-E(x)^2

What should I put in for xi^2
(i is subscript)
Original post by Laughh3
Q) The random variable X has a binomial distribution with parameters n=25, p=0.8. The random variable Y is defined by Y=aX+b, where a, b > 0.

The mean of Y= 65
SD of Y=6

My question is how do you know which value to put in for x in the variance formula for x=Sum of xi^2pi-E(x)^2

What should I put in for xi^2
(i is subscript)


We have that E(Y)=a E(X)+b \mathbb{E}(Y) = a \, \mathbb{E}(X) + b and Var(Y)=a2Var(Y) \text{Var}(Y) = a^2 \text{Var}(Y) . From these we can form two set of equations to solve for a,ba,b . Note that since X∼B(25,0.8) X \sim B(25,0.8) we have that E(X)=(25)(0.8) \mathbb{E}(X) = (25)(0.8) and Var(X)=(25)(0.8)(0.2) \text{Var}(X) = (25)(0.8)(0.2). Hope this helps!
(edited 6 years ago)
Reply 2
Avatar for Laughh3
Laughh3
OP
11
Original post by ccharlie97
We have that E(Y)=a E(X) \mathbb{E}(Y) = a \, \mathbb{E}(X) and Var(Y)=a2Var(Y) \text{Var}(Y) = a^2 \text{Var}(Y) . From these we can form two set of equations to solve for a,ba,b . Note that since X∼B(25,0.8) X \sim B(25,0.8) we have that E(X)=(25)(0.8) \mathbb{E}(X) = (25)(0.8) and Var(X)=(25)(0.8)(0.2) \text{Var}(X) = (25)(0.8)(0.2). Hope this helps!


Thanks :biggrin: So, we take the value of n for x?
(edited 6 years ago)
Original post by Laughh3
Thanks :biggrin: So, we take the value of n for x?


You don't need to use E(X−EX)2 \mathbb{E}(X - \mathbb{E}X)^2 or any other formula for expectation/variance to do this question. The values are given, you just need to find the two equations! Note that I made an error before, it should have said E(Y)=a E(X)+b \mathbb{E}(Y) = a \, \mathbb{E}(X) + b (I edited this in my post).
(edited 6 years ago)
Reply 4
Avatar for Laughh3
Laughh3
OP
11
How did you get Var(x)=(25)(0.8)(0.2) from the formula var(x)=Sum of xi^2pi-E(x)^2
Thanks
Reply 5
Avatar for Laughh3
Laughh3
OP
11
Original post by ccharlie97
You don't need to use
Unparseable latex formula:

\mathbb{E}(X - \bb{E}X)^2

or any other formula for expectation/variance to do this question. The values are given, you just need to find the two equations! Note that I made an error before, it should have said E(Y)=a E(X)+b \mathbb{E}(Y) = a \, \mathbb{E}(X) + b (I edited this in my post).


Thanks, that's really helpful!
Original post by Laughh3
How did you get Var(x)=(25)(0.8)(0.2) from the formula var(x)=Sum of xi^2pi-E(x)^2
Thanks


It's just the formula for the Binomial distribution, for X∼B(n,p) X \sim B(n,p) we have that E(X)=np \mathbb{E}(X) = np and Var(X)=np(1−p) \text{Var}(X) = np(1-p) . This can be proved, it's not hard, but for A-Level I think you just have to know it! Have you not seen it before? :smile:
Reply 7
Avatar for Laughh3
Laughh3
OP
11
Original post by ccharlie97
It's just the formula for the Binomial distribution, for X∼B(n,p) X \sim B(n,p) we have that E(X)=np \mathbb{E}(X) = np and Var(X)=np(1−p) \text{Var}(X) = np(1-p) . This can be proved, it's not hard, but for A-Level I think you just have to know it! Have you not seen it before? :smile:


Yeah, I get it now :smile:
Thanks
Original post by Laughh3
Yeah, I get it now :smile:
Thanks


No worries :smile:

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