According to national data, about 10% of American college students earn a graduate degree. Using this estimate, what is the probability that exactly 25 undergraduates in a random sample of 200 students will earn a college degree?

Solution

Let $X$ be the number of American college students earn a graduate degree and $p$ be the probability that American college students earn a graduate degree.

Given that $n =200$ and $p=0.1$. Thus $X\sim B(200, 0.1)$.

Since $n$ is large and $p$ is neither too small nor too large, the distribution of X is approximately Normal with mean = $\mu= n*p = 200 \times 0.1 = 20$ and standard deviation $\sigma= \sqrt{n*p*(1-p)}$.

The standard deviation is

$$ \begin{aligned} \sigma&= \sqrt{n*p*(1-p)}\\ &= \sqrt{200 \times 0.1 \times (1- 0.1)}\\ &=4.24 \end{aligned} $$

The probability that exactly 25 undergraduates in a random sample of 200 students will earn a college degree is

$$ \begin{aligned} P(X= 25) & = P(X< 25.5)-P(X< 24.5)\\ &\quad \quad (\text{ using continuity correction})\\ &= P\big(\frac{X-\mu}{\sigma}< \frac{25.5-20}{4.24}\big)-P\big(\frac{X-\mu}{\sigma}< \frac{24.5-20}{4.24}\big)\\ &= P(Z< 1.3)-P(Z< 1.06)\\ & = 0.047\\ \end{aligned} $$

normal area
normal area

Further Reading