Respuesta :
Answer:
a) 0.1138 = 11.38% probability that 14 of them were very confident their major would lead to a good job
b) 0.0483 = 4.83% probability that 10 of them are NOT confident that their major would lead to a good job
Step-by-step explanation:
Binomial probability distribution
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}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations 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 p is the probability of X happening.
a. A 2017 poll found that 53% of college students were very confident that their major will lead to a good job. If 30 college students are chosen at random, what's the probability that 14 of them were very confident their major would lead to a good job?
Here, we have that [tex]p = 0.53, n = 30[/tex], and we want to find [tex]P(X = 14)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 14) = C_{30,14}.(0.53)^{14}.(0.47)^{16} = 0.1138[/tex]
0.1138 = 11.38% probability that 14 of them were very confident their major would lead to a good job.
b. A 2017 poll found that 53% of college students were very confident that their major will lead to a good job. If 30 college students are chosen at random, what's the probability that 10 of them are NOT confident that their major would lead to a good job?
10 not confident, so 30 - 10 = 20 confident. This is P(X = 20).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 20) = C_{30,20}.(0.53)^{20}.(0.47)^{10} = 0.0483[/tex]
0.0483 = 4.83% probability that 10 of them are NOT confident that their major would lead to a good job
