# Normal distribution question

• Nov 9th 2009, 06:49 AM
vexiked
Normal distribution question
Scores on an examination are assumed to be normally distributed with a mean 78 and variance 36. If it is known that a students score exceeds 72, what is the probability that her score exceeds 84?

Let Y denote the score, then P(Y>84 | Y>72) = P(z > .16 | z > -.16)
I am stuck here and not sure what to do.
• Nov 9th 2009, 07:20 AM
Chris L T521
Quote:

Originally Posted by vexiked
Scores on an examination are assumed to be normally distributed with a mean 78 and variance 36. If it is known that a students score exceeds 72, what is the probability that her score exceeds 84?

Let Y denote the score, then P(Y>84 | Y>72) = P(z > .16 | z > -.16)
I am stuck here and not sure what to do.

This is conditional probability, so we see that

$\mathbb{P}\!\left(z>.16|z>-.16\right)=\frac{\mathbb{P}\!\left((z>-.16)\cap (z>.16)\right)}{\mathbb{P}\!\left(z>-.16\right)}=\frac{\mathbb{P}\!\left(z>.16\right)}{ \mathbb{P}\!\left(z>-.16\right)}$

Can you take it from here?
• Nov 9th 2009, 07:38 AM
vexiked
So looking up .16 in the table gives .4364/.4364 giving us a value of 1. This is where I am confused.
• Nov 9th 2009, 07:13 PM
mr fantastic
Quote:

Originally Posted by vexiked
So looking up .16 in the table gives .4364/.4364 giving us a value of 1. This is where I am confused.

The denominator is NOT Pr(Z > 0.16), it's Pr(Z > -0.16).