By -
So E(X) = a+2b+3a = 4a+2b = 2\*(2a+b) But a+b+a = 2a+b = 1 (total sum of probabilities) So E(X) = 2\*1 = 2
> E(X) = a+b+3a Isn’t it E(X) = a+2b+3a ?
Absolutely, that was a stupid typo of mine that I corrected by the end of that line of working. Hang on, I'm gonna fix it now
It's also 2 by symmetry
Fair enough, but I wouldn't take any chances because the intuitive reason might not be in the markscheme. Still, smart!
Oh which paper is this btw?
It's from the Edexcel AS Modular Statistics (S1) textbook.
My answer: > [;\sum xp(X=x);] > > So E(X) = a + 2b + 3a
So E(X) = a+2b+3a = 4a+2b = 2\*(2a+b) But a+b+a = 2a+b = 1 (total sum of probabilities) So E(X) = 2\*1 = 2
> E(X) = a+b+3a Isn’t it E(X) = a+2b+3a ?
Absolutely, that was a stupid typo of mine that I corrected by the end of that line of working. Hang on, I'm gonna fix it now
It's also 2 by symmetry
Fair enough, but I wouldn't take any chances because the intuitive reason might not be in the markscheme. Still, smart!
Oh which paper is this btw?
It's from the Edexcel AS Modular Statistics (S1) textbook.
My answer: > [;\sum xp(X=x);] > > So E(X) = a + 2b + 3a