This is tough to type with an I(dot) pad, but the prob of abs y greater than 1.7 is twice that y is greater than 1.7
I've found this forum to be a lot of help. The answers provided to my last question were helpful and I'm hoping I can get similar help for this question.
Y has the standard normal distribution.
1. Determine the distribution function of |Y|, the absolute value of Y.
2. Determine the probability density function of |Y|
3. Find P(|Y|>1.70).
For part 1, I did this:
P(|Y|) ≤ x)= P(−x ≤ Y≤ x) = ϕ(x) − ϕ(-x)
= ϕ(x) -(1 − ϕ(x) ) = 2ϕ(x) − 1 .
Does that look correct?
For 2, this is my work,
f(x)= d P(|Y|) ≤ x)= 2ϕ'(x)
Which is equal to 2*(1/√2π)*e^(-x^2/2)
Is my work right for this?
I'm not sure about how I would go about doing 3, so all tips are appreciated. Thanks.