# Normal Distribution and Conditional Probability

Suppose that exam scores were distributed normally. Let the mean be 80 and standard deviation be 8. If it is known that a student's score is greater than 75, what is the probability that his score is greater than 90?

I am a bit confused by the wording of the question. Should this be considered a case of conditional probability? Is my approach correct?

This is know I thought about it:

$$P(Y > 90 | Y >75) = \frac{P(Y > 90 \cap Y > 75)}{P(Y > 75)} = \frac{P(Y > 90)}{P(Y > 75)}.$$

• $P(Y>90)\cap P(Y>75)$ is wrong. It should say $P(Y>90 \cap Y>75)$. The "$\cap$" belongs between two events, not between two numbers. Mar 30, 2012 at 4:26
• @MichaelHardy: Once I fix that, would my method work? Mar 30, 2012 at 4:27
• Yes, it is correct. Mar 30, 2012 at 4:35

Because the problem now reduces to $$\frac{P(Y\gt 90)}{P(Y\gt 75)}$$ I'll continue from thereon.
To compute $P(Y\gt 90)$ you first compute $P(Y\lt 90)$. To do his you first figure out how many standard deviations away from the mean is $90$. This is very often called the z-score, which you are aware of. So, $90$ is $10$ points away from your mean ($80$), this implies that 90 is $\frac{10}{8}=0.8 \sigma$s away from your mean. If you look up the probability of this particular z- score on a table, you'll see that $P(Y\lt90)=\Phi(0.8)=0.788.$ So
$P(Y\gt90)=1-\Phi(0.8)=0.212.$
Similarly, $P(Y\lt75)=\Phi(-0.625)=.266$. So $P(Y\gt75)=1-\Phi(-0.625)=.266=0.734$
So $$\frac{P(Y\gt 90)}{P(Y\gt 75)}=\frac{0.212}{0.734}=0.289$$