Class Test Scores

From the information above, if we let $G$ represent girls and $B$ represent boys, we have:

- $\mu_{G} = 86, \sigma_{G}=4$

- $\mu_{B} = 81, \sigma_{B} = 3$

The question is asking us for $P(G>B)$ , or the probability of a girl test score being greater than the boy test score. This is equivalent to finding $P(G-B>0)$ , or the probability of a girl test score MINUS the boy test score being greater than 0. And, since the test scores are Normally distributed, we can thus standardize this distribution.

$z = \dfrac{x-\mu}{\sigma} \implies z_{G-B} = \dfrac{x-\mu_{G-B}}{\sigma_{G-B}}$

We can find $\mu_{G-B}$ and $\sigma_{G-B}$ using rules for means and variances.

- $\mu_{G-B} = \mu_{G} - \mu_{B} = 86 - 81 = 5$

- $\sigma_{G-B}^2 = \sigma_{G}^2 + \sigma_{B}^2 = (4)^2 + (3)^2 = 25 \implies \sigma_{G-B} = \sqrt{25} = 5$

Using the above z-score formula, we can standardize this distribution with $x=0$ :

$z_{G-B} = \dfrac{0-5}{5} = -1$ .

Therefore, we have $P(G-B>0)=P(z>-1)$ . Using a z-score table, we have that the area to the left of a z-score of $-1$ is approximately $0.1587$ . However, we want the area GREATER than the z-score of $-1$ , so it would be to the right. Therefore,

$P(G-B>0)=P(z>-1)=1-(0.1587)=0.8413$ .

Rounded to the nearest percent, the probability that a randomly selected girl in your class has a higher test score than a randomly selected boy in your class is $\boxed{84\%}$ .