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! 🙂
Best Answer
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.