[Math] Normal distribution problem with 2 normal distributions

normal distributionstatistics

I have this problem:

A professor of statistics noticed that the marks in his course are normally distributed. He also noticed that his morning classes average 70% with a standard deviation of 13% on their final exams. His afternoon classes average 79% with a standard deviation of 9%.

What is the probability that a randomly selected student in the morning class has a higher final exam mark than a randomly selected student from an afternoon class?

I've been trying to solve this problem myself for quite awhile and not really sure what is wrong with my approach.

I let the morning class = X1 and the afternoon class = X2. So E(X1) = 70, σ(X1) = 13, E(X2) = 79, and σ(X2) = 9. Then I let the variable Y = X1-X2, and since X1 and X2 are both normally distributed, then Y will be as well. So I have the following for Y:

$$E(Y) = E(X_1)-E(X_2) = 70-79 = -9$$
$$σ(Y) = \sqrt{σ(X_1)^2 + σ(X_2)^2} = \sqrt{13^2+9^2} = \sqrt{250}$$

In oder to find the probability that a randmly selected student in the morning class has a higher final exam mark than one from the afternoon class, we want P(X1 > X2) = P(X1 – X2 > 0) = P(Y > 0). So I converted this to z-score…

$$Z = \frac{X-E(Y)}σ = \frac{0-(-9)}{\sqrt{250}} = \frac 9{\sqrt{250}} \approx 0.57$$

So that we are looking for P(Z > 0.57) or 1-P(Z$\leq$0.57).

I looked at my z-score table and found my answer to be 0.2843, which is marked as wrong. I've tried doing different things for rounding etc., and still can't seem to find the right answer. Any insight is greatly appreciated – maybe I made some dumb error that I can't seem to see. Thank you! 🙂

Best Answer

Your approach is perfectly fine, but you've rounded too much precision off your $z$-score, which is why your answer is off. The $z$-score is $0.56921$, thus the probability is $0.284607$.

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Using the table of the standard normal, we find that the probability that the age is $\frac{1.3}{3.7}\approx 0.351$ standard units above the mean or less is approximately $0.637$.

The probability that we are below the mean by $\frac{0.2}{3.7}$ or more standard deviation units is about $0.48$. So our required probability is about $0.637-0.48$, some distance from your number.

For the next problem, if we take $15$ ages and average them, the resulting random variable has normal distribution with mean $36.2$ and standard deviation $\frac{3.7}{\sqrt{15}}\approx 0.955$. Now one needs to essentially repeat the first calculation, but with $0.955$ replacing $3.7$. The probabilities will change dramatically.

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For reasonable positive values of $z$, $\Pr(Z\le z)$ can be looked up approximately in a table of the standard normal.

For negative $z$, we have to work a bit. If $z$ is negative, then $\Pr(Z\le z)=1-\Pr(Z\le |z|)$.

Tables of the standard normal are available on the web. Typically they are a single page, you can download one and print it. Such a table is still often found at the back of introductory statistics books.

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[Math] Normal Distribution, finding the minimum mark

You have percentage scores $X \sim \mathsf{Norm}(\mu = 65,\, \sigma=6)$ and you seek $c$ such that $P(X \le c) = .35.$ This can be solved by standardizing and using printed tables of the standard normal distribution:

$$.35 = P(X < 35) = p\left(\frac{X - \mu}{\sigma} < \frac{c - 65}{6}\right) = P\left(Z < \frac{c - 65}{6}\right),$$

where $Z$ is standard nomal. From tables you should be able to find that $(c-65)/6 \approx -0.38,$ and then it is a simple matter to solve for $c,$ and round up to the nearest integer (percentage) and thus get the lowest passing score.

If you have access to statistical software or a statistical calculator, you may be able to solve the question directly, without standardizing. In R statistical software qnorm is the quantile function (inverse CDF) of a normal distribution. Thus the lowest passing score is 63.

qnorm(.36, 65, 6)
## 62.84925

I find that drawing a sketch is insurance against getting tangled up reading printed normal tables and finding an "answer" that is obviously wrong. By hand, you can't make a sketch as accurate as the plot below, but you can make a sketch that is better than nothing.

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