X is normally distributed if y=3+4X and if E(X)=6 and Var(X)=4 what is E(Y) and the Var(Y) what is the probability of Y<10
\[E(a+bX)=a+bE(X)\] \[Var(a+bX)=b^2Var(X)\]
okay I got that
If X is normal then Y is also normal
and you apperently already have the mean and variance
yes of x so would that be the same as y?
no
both normal, but they have different means and variances
okay so there both normal, but how do I figure out the expected value of y and the variance only given the values of x?
I gave you the formulas above
y=3+4(6)=42
3+4*6=3+24=27
E[Y]=3+4*E[X] = 3 + 4*6 = 27 V[Y]=4^2V[X] = 16*4 = 64
P(Y<10) = P(3+4X<10) = P(4X<7) = P(X<7/4)
... which is P(X<1.75) = Phi((1.75-6)/2) = Phi(-2.125). This can be found in a table of values for Phi.
I think the whole point of the exercise was to use the information we obtained about Y so we wouldn't need to convert back to X.
It's about 0.0168.
okay the probability formula to me I am only given a t-table and z-table can I figure it out on that table
The z-table is the one you want. You should get 0.0168 at -2.125.
okay yes I get that thank you very much!
in the probability model where did the 7 come from?
You mean where I wrote "P(3+4X<10) = P(4X<7)"? Or somewhere else?
no thats it
Oh. If you take 3+4X < 10 and subtract 3 from both sides, you get 4X < 7.
okay got it and then in the end you divide 7/4 because the var is 4 or because its y=3+4x sorry I i just want to make sure I get it right the numbers are confusing
No problem. It's like this: We want P(3+4X<10). Looking inside the parentheses, we have: 3 + 4X < 10 Subtract 3 from both sides to get: 4X < 7 Divide both sides by 4 to get: X < 7/4
okay awesome can you break down the variance formula for me?
Var Y = Var (4X+3) The +3 part has no effect. The mult by 4 causes the variance to mult by 4 squared. So Var Y = 4^2 Var X. (like b=4 in formula near top of this thread)
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