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

Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators