Suppose that 20% of the college seniors support an increase in federal funding for care of the elderly. If 20 college seniors are randomly selected, what is the probability that at most 3 of them support the increased funding?
(a) The expected number of college seniors, in a random sample of 20, supporting the increased funding.
(b) The probability that the number of sampled college seniors supporting the increased funding equals the expected number.
Solution
Here $X$ denote the number of college seniors who suport an increase in federal fund for care of the elderly out of 20 college seniors.
Probability that college seniors support an increase in federal funding for care of the elderly is $p=0.20$.
Given that $X\sim B(20, 0.2)$.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{20}{x} (0.2)^x (1-0.2)^{20-x},\\ &\quad x=0,1,\cdots, 20. \end{aligned} $$
The probability that at most 3 of the college seniors support the increased funding is
$$ \begin{aligned} P(X\leq 3) &= \sum_{x=0}^{3} P(x)\\ & = P(X=0) + P(X=1) + P(X=2) + P(X=3)\\ &= 0.0115+0.0576+0.1369+0.2054\\ &= 0.4114 \end{aligned} $$
(a) The expected number of college seniors, in a random sample of 20, supporting the increased funding = $E(X)= n*p = 20 * 0.2=4$.
(b) The probability that the number of sampled college seniors supporting the increased funding equals the expected number is
$$ \begin{aligned} P(X = 4) &= \binom{20}{4} (0.2)^{4} (1-0.2)^{20-4}\\ &= 0.2182 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
