In a market study for Target, a researcher found that 70% of customers are repeat customers. If 14 customers are selected at random, find the probability of getting,
(a) Exactly 9 of them are repeat customer
(b) At least 12 of them are repeat customers
(c) At most 12 of them are repeat customers.
Solution
Let $X$ denote the number of repeat customers.
70% of customers are repeat customers. So $p=0.70$
Given that $n = 14$, $p= 0.7$. $X\sim B(14, 0.7)$.
That is the random variable $X$ follows a Binomial distribution with $n = 14$ and $p= 0.7$.
The probability mass function of $X$ is
$$ \begin{aligned} P(X=x) &= \binom{14}{x} (0.7)^x (1-0.7)^{14-x}, \\ &\quad x=0,1,\cdots, 14 \end{aligned} $$
(a) The probability that exactly $9$ of them are repeat customers is
$$ \begin{aligned} P(X = 9) & =\binom{14}{9}(0.7)^{9}(1-0.7)^{14-9}\\\\ & = 0.1963 \end{aligned} $$
(b) The probability that at least $12$ customers are repeat customers is
$$ \begin{aligned} P(X\geq 12) & =\sum_{x=12}^{14} P(x)\\\\ & =\sum_{x=12}^{14}\binom{14}{x}(0.7)^x(1-0.7)^{14-x}\\\\ & = (0.11336)+(0.04069)+(0.00678) \\\\ & = 0.16084 \end{aligned} $$
(c) The probability that at most 12 of them are repeat customers is
$$ \begin{aligned} P(X\leq 12) & =1-\sum_{x=13}^{14} P(x)\\\\ & =1-\sum_{x=13}^{14}\binom{14}{x}(0.7)^x(1-0.7)^{14-x}\\\\ & = 1-(0.04069)+(0.00678) \\\\ & = 0.95252 \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
