4.2 Binomial Distribution
The following activities and questions relate to material covered in Chapter 4.2 Binomial Distribution in Introductory Business Statistics (OpenStax).
Data sets for the following questions are available in Excel: 4.2 Data Sets [XLSX].
Questions
- When a patient calls the medical advice line saying that they have the flu, the probability of him or her testing positive for the flu is 4%. Of the next 25 patients who call in claiming to have the flu, we are curious to see how many test positive for the flu. Find the probability that at exactly four of the 25 patients test positive flu.
- People visiting a northern outdoor gear rental shop during the summer often rent camping and backpacking gear. The probability distribution for gear rentals per customer is given in the table below. There is a five-gear rental limit per customer at this shop.
X P(x) 0 0.05 1 0.40 2 0.25 3 4 0.10 5 0.07 - Describe the random variable X in words.
- Find the probability that a customer rents three gear items.
- Find the probability that a customer rents at least four gear items.
- Find the probability that a customer rents at most 2 gear items.
- In a northern Alberta community, 46% of the children are involved in an organized sport. Suppose you randomly pick 20 children.
- Find the probability that at least 12 are enrolled in an organized sport.
- Is it more likely that 11 or that 12 will be enrolled in an organized sport?
- Suppose about 70% of youth (age 12 – 19 years) living in a northern Canadian town attend local music festivals. A group of 15 youths is randomly chosen.
- In words, define the random variable X.
- List the values that X may take on.
- Give the distribution of X. X ~ ( , )
- Find the probability that 7 or 8 will attend.
Solutions
- This is binominal distribution as there are two outcomes; test positive for the flu (success), test negative for the flu (failure).
[latex]\begin{array}{ll}p(x=4, n=25, p=0.04)&=\dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}\\&=\dfrac{25!}{4!25-4!}0.04^4{(1-0.04)}^{25-4}\\&=0.0137\\&=1.37\%\end{array}[/latex]
There is a 1.37% probability that exactly 4 out of 25 patients who call in claiming the flu, test positive for the flu. -
- The random variable X is an example of discrete random variables. X has a countable number of possible values. In the above problem X represents the number of items a customer will rent.
X = the number of northern outdoor gear items a customer will rent. - 1− 0.05 + 0.4 + 0.25 + 0.10+ 0.07 = 0.13
The probability that a customer rents three gear items is 13%. - P (X ≥ 4) = P (0.10) + P (0.07) = 0.17
The probability that a customer rents at least four gear items is 17%. - P (X ≤ 2) = P (0.05) + P (0.40) + P (0.25) = 0.70
The probability that a customer rents at least 2 gear items is 70%.
- The random variable X is an example of discrete random variables. X has a countable number of possible values. In the above problem X represents the number of items a customer will rent.
-
- P(x > = 12) = 0.1511 = 15.11%
- P(x = 11) = 0.1280, P(x = 12) = 0.0817. It is more likely that 11, rather than 12, children will be enrolled in an organized sport.
-
- Let X = the number of youths who would attend a local music festival
- X = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
- X ~ B(15, 0.70)
- P(X = 7 or X = 8) = 0.0348 + 0.0811 = 0.1159 = 11.59%