Thread: 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.

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



Can you take it from here?

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

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).