Expected value and variance of a single dice

[Mos5180d]

I’ve been asked to let the values of a roll on a single dice can take be a random variable X

State the function. Which I have as f(x) = 1/6 x + 1/6 x₂ + 1/6 x₃ + 1/6 x₄ + 1/6 x₅ + 1/6 x₆

Then calculate the expected value and variance of f(x)

As I understand expected value = summation of x * P(x)

but I have no numbers for x and so assumed it was 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 which doesn’t seem right as its basically the same as my formula. Or do I assign the values as 1,2,3...etc in which case my expected value is 3.5

as for the variance, would I use the formula for discrete random variables - sum of P_i * (x_i - u^2), where u = the mean. And if so is the mean 3.5?

 

[firemath]

By x2,x3, etc., do you mean x subscript 2, or 2x?

By the way, it is single die, not single dice.

 

[pka]

I’ve been asked to let the values of a roll on a single dice can take be a random variable X

State the function. Which I have as f(x) = 1/6 x + 1/6 x2 + 1/6 x3 + 1/6 x4 + 1/6 x5 + 1/6 x6

Then calculate the expected value and variance of f(x)

As I understand expected value = summation of x * P(x)

but I have no numbers for x and so assumed it was 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 which doesn’t seem right as its basically the same as my formula. Or do I assign the values as 1,2,3...etc in which case my expected value is 3.5

as for the variance, would I use the formula for discrete random variables - sum of Pi * (xi - u^2), where u = the mean. And if so is the mean 3.5?

\(\displaystyle Var(X) = E({X^2}) - {\mu ^2} = \frac{1}{6}\sum\limits_{k = 1}^6 {{k^2}} - {(3.5)^2} = \frac{{35}}{{12}}\)
\(\displaystyle SD(X)=\sqrt{Var}\)

 

[Mos5180d]

By x2,x3, etc., do you mean x subscript 2, or 2x?

By the way, it is single die, not single dice.

Yes sorry I mean subscript, I haven’t used the website much and I use my iPad so I’m not sure exactly how to use the correct notations. And thanks, I guess this is why I’m doing maths and not English haha

 

[firemath]

Try graphing it to visualize it.

 

[Steven G]

I’ve been asked to let the values of a roll on a single dice can take be a random variable X

State the function. Which I have as f(x) = 1/6 x + 1/6 x₂ + 1/6 x₃ + 1/6 x₄ + 1/6 x₅ + 1/6 x₆

Then calculate the expected value and variance of f(x)

As I understand expected value = summation of x * P(x)

but I have no numbers for x and so assumed it was 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 which doesn’t seem right as its basically the same as my formula. Or do I assign the values as 1,2,3...etc in which case my expected value is 3.5

as for the variance, would I use the formula for discrete random variables - sum of P_i * (x_i - u^2), where u = the mean. And if so is the mean 3.5?

I’ve been asked to let the values of a roll on a single dice can take be a random variable X

State the function. Which I have as f(x) = 1/6 x + 1/6 x₂ + 1/6 x₃ + 1/6 x₄ + 1/6 x₅ + 1/6 x₆

Then calculate the expected value and variance of f(x)

As I understand expected value = summation of x * P(x)

but I have no numbers for x and so assumed it was 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1 which doesn’t seem right as its basically the same as my formula. Or do I assign the values as 1,2,3...etc in which case my expected value is 3.5

as for the variance, would I use the formula for discrete random variables - sum of P_i * (x_i - u^2), where u = the mean. And if so is the mean 3.5?

You seem confused about what X and p(x) means. X is the set that has the possible outcomes on a roll of a die. So X = {1, 2, 3, 4,5, 6}. p(X=x) is the probability that the toss of a die results in x. So p(X=1) = p(X=2) = p(X=3) = p(X=4) = p(X=5) = p(X=6) = 1/6

I have no idea at all what you mean by State the function. Which I have as f(x) = 1/6 x + 1/6 x₂ + 1/6 x₃ + 1/6 x₄ + 1/6 x₅ + 1/6 x₆. What does for example f(4) = S = 1/6* 4 + 1/6 x₂ + 1/6 x₃ + 1/6 x₄ + 1/6 x₅ + 1/6 x₆ mean?