Respuesta :
Answer:
a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.
b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.
c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.
Step-by-step explanation:
There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.
Binomial probability
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And [tex]\pi[/tex] is the probability of X happening.
In this problem, we have that:
10 student are sampled, so [tex]n = 10[/tex]
34% of college students say they use credit cards because of the rewards program, so [tex]\pi = 0.34[/tex]
(a) exactly two
This is P(X = 2).
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873[/tex]
There is a 18.73% probability that exactly two students use credit cards because of the rewards program.
(b) more than two
This is [tex]P(X > 2)[/tex].
Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:
[tex]P(X \leq 2) + P(X > 2) = 1[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2)[/tex]
In which
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
So
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157[/tex]
[tex]P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808[/tex]
[tex]P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873[/tex]
[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838[/tex]
[tex]P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162[/tex]
There is a 71.62% probability that more than two students use credit cards because of the rewards program.
(c) between two and five inclusive
This is:
[tex]P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873[/tex]
[tex]P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573[/tex]
[tex]P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320[/tex]
[tex]P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434[/tex]
[tex]P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82[/tex]
There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.
