Random Variable X Question (Statistics)

Aurelius032 said:

It’s known that a random variable X is distributed normally with E(X) = -1

and it’s also known that p(-4≤X≤-3)+p(1≤X≤2) = 0.3. Find p(-4≤X≤-3).

My work so far:

p(-4≤X≤-3) = p(1≤X≤2) , due to intervals (-4,-3) and (1,2) are symmetric...

Can anyone help guide me through this question? I' m having a tough time..

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Ishuda said:

You are almost there. As you point out p(-4\(\displaystyle \le\)x\(\displaystyle \le\)-3)=p(1\(\displaystyle \le\)x\(\displaystyle \le\)2) so
p(-4\(\displaystyle \le\)x\(\displaystyle \le\)-3) + p(1\(\displaystyle \le\)x\(\displaystyle \le\)2) = 2 p(-4\(\displaystyle \le\)x\(\displaystyle \le\)-3) = 0.3

So what is p(-4\(\displaystyle \le\)x\(\displaystyle \le\)-3)?

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Aurelius032 said:

Is it .3/2 = .15? Jeez..

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