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"Given that boys' heights are distributed normally $\mathcal{N}(68$ inches, $4.5$ inches$)$ and girls are distributed $\mathcal{N}(62$ inches, $3.2$ inches$)$, what is the probability that a girl chosen at random is taller than a boy chosen at random?"

I was able to follow the problem in class, and apply the same method to solve another, but I don't really know what I'm doing. Can someone please explain why this works? I want to be able to apply these ideas to different scenarios as well, both questions we did in class were very similar.

First we found a new mean:
$U_{\text{Boys}-\text{Girls}} = U_{\text{Girls}}-U_{\text{Boys}} = 68-62 = 6$.

Then we obtained a new standard deviation from the variance of differences:
$\text{Var}_{\text{Boys}-\text{Girls}}=\text{Var}_{\text{Boys}}+\text{Var}_{\text{Girls}}=4.5^{2} + 3.2^{2} = 30.49$

So the new standard deviation is $30.49^{1/2} = 5.521$.

To find the probability of a girl being taller than a boy, we then found the area of the new normal distribution, $\mathcal{N}(6$ inches, $5.521$ inches$)$ with the upper bound of $0$.

$\text{normal cdf}(-\infty, 0, 6, 5.21) = 0.138$ or $13.8$% chance.

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  • $\begingroup$ "by using the difference of variances" -- look again. Looks like a sum to me $\endgroup$ – Glen_b Nov 19 '14 at 9:12
  • $\begingroup$ I was fairly sure the sum and difference of variances both equal the sum of their components. $\endgroup$ – ta3920 Nov 19 '14 at 9:31
  • $\begingroup$ You're confusing the phrase "difference of variances" with "variance of difference". The phrase "difference of variances" literally means to subtract one variance from another. The "variance of the difference" is indeed the sum of the variances. You should alter the wording to match your intended meaning. $\endgroup$ – Glen_b Nov 19 '14 at 9:42
  • $\begingroup$ I knew that trying to correct someone with 74k rep wasn't a good idea. Thanks for clearing it up, I appreciate it. $\endgroup$ – ta3920 Nov 19 '14 at 9:55
  • $\begingroup$ You did much better not to assume I was correct. I make more than my share of mistakes. $\endgroup$ – Glen_b Nov 19 '14 at 13:10
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Assume that $X$ is a random height of a boy and $Y$ is a random height of a girl and these variables are independent. Now, the probability you want to assess is $$ P(X < Y) = P(X - Y < 0). $$ Since $X$ and $Y$ are independent gaussian variables we use a theorem saying that a sum of independent gaussians is gaussian with a mean equal to sum of the means and the variance equal to sum of the variances. So $Z = X - Y$ has the distribution with parameters which you calculated.

So you found the parameters and, as long as you have the normalcdf function for any gaussian distribution, you are done. You may be also asked for solution when you are given just a standard gaussian table. Then you need to still simplify.

We are looking for $P(Z < 0)$. Let's say $Z \sim N(\mu,\sigma^2)$. Then $$ P(Z < 0) = P(Z - \mu < -\mu) = P\left(\frac{Z - \mu}{\sigma} < -\frac{\mu}{\sigma}\right). $$ Note that I transformed the inequality such that $$G = \frac{Z - \mu}{\sigma}$$ is standard gaussian. So we look for the probability $$ P\left(G < -\frac{\mu}{\sigma}\right), $$ where the left side is standard gaussian so we can find the result in a table.

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