Normal distribution question problem.

Question

The professor of a introductory calculus class has stated that, historically, the distribution of final exam grades in the course resemble a Normal distribution with a mean final exam mark of $μ=60\%$ and a standard deviation of $σ=10\%$.

Suppose this professor randomly picked $28$ final exams, observing the earned mark on each. What is the probability that $5$ of these have a final exam grade of less than $50\%$? Use four decimals in your answer.

I am not sure how to approach this.. I think I need to combine hypergeometric distribution with normal distribution but I have no clue how to do it.. Some help/tips would be greatly appreciated! :)

Answer 1

Let's call $X$ the r.v. associated with mark of one exam, with pdf $f_X(x) \sim \mathcal{N}(\mu = 0.6, \sigma^2 = 0.01)$. The probability that one exam has a lower mark than 0.5 is defined as:

$$ P(X<0.5) = \int_{-\infty}^{0.5}f_X(x)dx=F_X(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)=\Phi(-1)=0.1587$$

Where $F_X$ is the cumulative density function (cdf) of r.v. $X$, and $\Phi$ is the cdf of standard normal distribution (i.e. $\mathcal{N}(0,1)$). Values for $\Phi$ can be found in tables.

Now, lets define $M$ as the total number of picked exams, and $N$ as the number of "failed" (<0.5). Probability of having exactly $N$ exams with lower mark than 0.5 having picked $M$, is defined by a binomial, as follows:

$$P(N \vert M) = \binom{M}{N}P(X<0.5)^{N}(1-P(X<0.5))^{M-N}$$

Substituting numerical values leads to:

$$P(N=5 \vert M=28) = \binom{28}{5}0.1587^{5}0.8413^{23}=0.1859$$

Which is the result that you are looking for.

Remark: When computing $P(X<0.5)$, it has been assumed that marks below 0 are possible. By the way, same assumption has been done by your professor when saying that it is a normal distribution.

Answer 2

Each exam will have some chance $p$ of having score lower than $50\%$. You can find that $p$ by "inverting" the normal distribution - you either need to standardise your distribution and look for a table, or calculate this in Excel for example (NORM.INV).

Then, since each exam will have the same probability $p$ of having a mark lower than $50\%$ the total number of exams with marks lower than $50\%$ will be binomially distributed with parameters $p$ and $n=28$. Then it's simply the matter of calculating the probability of that number being equal to $5$.

Answer 3

What informations do we have?

They tell us they somehow got the true distribution of the random variable M:"Mark in final exam", and M ~ N(µ=0.6, σ^2=0.01).

Now they tell you that they are going to randomly sample 28 exams.

Then they ask, what is the probability that 5 of these exams have a grade of less than 50%?

And now what? We are totally lost! No, wait.

Can we think of the problem of figuring out the probability for 1 exam to have a grade of less than 50%? P(M<50%) = F(50%) = ...? M follows a Normal, and now we are stuck because we have to integrate this complicated Normal PDF from 0 to 50%, and that sound daunting. Oh wait again, I vaguely remember something about the Normal tables? The "Standard Normal" tables, yeah some tables that were made to ease in the computations. Ok then, let's standardize our variable: Z = (M-µ)/σ ~ N(0,1), so Z = (50%-60%)/10% = -10%/10% = -1, what is this number? It's called the z-score: it tells you how many standard deviations away from the true mean you are expecting your data to be. And since we are talking about standard deviations and not a specific standard deviation value, we can look up the standard normal tables.

What will we find in the tables? They tell you exactly what proportion of the population will have grades more extreme than 1 standard deviation to the left (because of the minus sign) of the true mean.

Ok I looked the tables for you, but you should try it by yourself. The table tells us that a proportion .15866 people will have a lower grade or equal to 50% if the true mean is 60% and the standard deviation is 10%. That is equivalent to say P(M<50%) = .15866

So we solved the problem for a single exam. But we were asked for 5 of them among 28. Remember that before we were told that 28 exams were chosen at random? Do you know what they mean by that? They mean that each exam before we look at the grade has exactly the same probability to have a grade smaller than 50%. Each exam is an independent Random Variable Mi, i=1,...28 that follows the Normal distribution above with the same parameters. We call that Independent and Identically Distributed random variables.

That is all fair, but does it help us in any way solve the problem of the 5 exams among 28? Could we maybe consider each of the 28 exams as an unfair coin toss, with probability of landing heads of p=.15866, and q=(1-p)=.84134? Let's try.

To do that we could use the Binomial distribution. What do you think? P(X=k) = nCkp^kq^(n-k), here : P(X=5) = 28C5*0.15866^5*.84134^23 = ?

It's not the hypergeometric solution that you suggested. Maybe everything I said is wrong. Sorry I have to go.