# Thread: Phi of normal distribution

1. ## Phi of normal distribution

Hi everyone!

So, my problem is:

Let X be birth weight of a randomly chosen child in Norway, given in gram, g. Assume that
X is normally distributed, where E(X) = 3315 and Var(X) = 5752
.
a) Calculate the following probabilities,

1) P(X > 3000) 2) P(3000 < X < 3500) 3) P(X > 3500 | X > 3000)

What I have done to solve it:

So, I managed to get the right answer on 1, and I thought I would get the right answer on number 2.

I thought: P(3000 < X < 3500) = P(X > 3000) ∩ P(X < 3500). In 1) I calculated that P(X > 3000) is 0,709, this was correct. I calculated that P(X < 3500) = 0,6235. So 0,709 x 0,6235 = 0,4421 , but the right answer is: 0,335. What am I doing wrong?

Thanks for taking a look!

2. ## Re: Phi of normal distribution

hello,

P(a < X < b), equations like this generally written as
P(X < b) - P(X < a).

hope this helps

3. ## Re: Phi of normal distribution

Let me see, I tried taking P(x < b) - P(x < a), which in this case would be 0,6235 - 0, 709, but then the answer is negative and way off. Not sure if this is what you meant I should do

4. ## Re: Phi of normal distribution

Ok, $\displaystyle P(x<b) = P(x<a \mbox{ AND } x<b) + P(x \ge a \mbox{ AND } x < b)$. This is the sum principle. If $\displaystyle a<b$ and $\displaystyle x<a$, then it is guaranteed that $\displaystyle x<b$. So, $\displaystyle P(x<a \mbox{ AND } x<b) = P(x<a)$. Now, $\displaystyle P(x<b) = P(x<a) + P(x \ge a \mbox{ AND }x < b) \ge P(x<a) + 0 = P(x<a)$. If $\displaystyle P(x<b) \ge P(x<a)$, then your data is wrong. If $\displaystyle P(x<b) = 0.6235$, then $\displaystyle P(x<a) \le 0.6235$.

Oh, I see. For your answer in part 1, you have $\displaystyle P(x>3000) = 0.709$. What is $\displaystyle P(x<3000)$? It is the compliment of $\displaystyle P(x>3000)$. So, $\displaystyle P(x<3000) = 1 - 0.709 = 0.291$. Now, $\displaystyle 0.6235 - 0.291 = 0.3325$ as you wanted.