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# Stats Help, Discrete Random Variable watch

1. n/a
2. Let X=price, so X=250, 400, 500 and 1000.
Each x-value has a probability from the distribution A~N(2, 9)
e.g. P(X=250) = P(A<= -2) ...which you can work out.

For E(X), using the formula.
3. (Original post by vc94)
Let X=price, so X=250, 400, 500 and 1000.
Each x-value has a probability from the distribution A~N(2, 9)
e.g. P(X=250) = P(A<= -2) ...which you can work out.

For E(X), using the formula.
I'm struggling to work this out? sorry to be a pain and thanks for the help so far

Also i'm a little confused by the A~N(2, 9) part?
4. A~N(2, 9) means that the A-content is normally distributed with mean 2 and variance 9 as the question says.

P(A<=-2) = P(Z<= (-2 - mean)/(std.Dev)) ...you know about standardising?
= P(Z<= -4/3) = P(Z<= -1.33) = 1 - P(Z<= 1.33) ...now use the tables.

Have you done much work in calculating Normal probabilities?

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Updated: March 29, 2011
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