In a survey, 58% of American adults said they had never heard of the internet. If 10 Americans adults are selected at random, find the probability
(a) that exactly 3 will say they never heard of the internet.
(b) Find the mean of the distribution.
(c) Find the variance of the distribution.
Solution
Let $X$ denote the number of American adults who never heard of the internet.
58% of adult American said that they had never heard of the internet. So $p=0.58$
Given that $n = 10$, $p= 0.58$. $X\sim B(10, 0.58)$.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{10}{x} (0.58)^x (1-0.58)^{10-x},\\ &\quad x=0,1,\cdots, 10. \end{aligned} $$
(a) The probability that exactly $3$ will say they never heard of the internet is
$$ \begin{aligned} P(X = 3) & =\binom{10}{3}(0.58)^{3}(1-0.58)^{10-3}\\\\ & = 0.054 \end{aligned} $$
(b) The mean of $X$ is $E(X)=np=10\times 0.58 =5.8$.
(c) The variance of $X$ is $V(X) = np(1-p) = 10\times 0.58 \times 0.42 =2.436$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
