Statistics

http://mathsathawthorn.pbworks.com/f/S2.pdf
Q3b?

What is the probability that 1 egg weighs less than 49 grams? Call this probability $p$. Then the probability that exactly three eggs weight less than 49 grams is $\mathbb{P}(X = 3)$ where $X \sim \text{B}(6, p)$.

Can i show what ive done first?

Why didn't you put that in the OP?

So ive done P(0>3Y-X)
3Y-X -N(54-50, 1.2^2 +2^2)
E(3Y-X)=(4,5.44)
Is that right so far?

Ah, my bad - I was answering (a) (ii) instead.

What you've done is fine - yes. Although I have no clue why you're finding E(3Y-X) in the last line.


So you have (I'll call it W to make it more obvious) $3Y - X = W \sim N(4, 1.2^2 + 2^2)$ and you want $\mathbb{P}(3Y-X < 0) = \mathbb{P}(W < 0)$.

Well, that's a simple normal distribution problem now. Can you see what to do from here?

The thing is in the mark scheme it says E(3Y-X)= (4,16.96) Where do they get 16.96 from? And do you use z=x-mean/ (standard deviation)

If it says that, then it's wrong. $E(3Y-X) = 3E(Y) - E(X) = 4$

.

So $W \sim N(4, 19.69)$. Then yeah, use z = x - mean / standard dev.

Okay thanks