Respuesta :
Answer:
14.19% probability that from a group of 200 people picked at random from this population, at least four people will be over 6 feet 3 inches tall
Step-by-step explanation:
For each person, there are only two possible outcomes. Either they are over 6 feet 3 inches tall, or they are not. The probability of a person being over this height is independent of any other person. So we use the binomial probability distribution to solve this question.
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.
1% of people in a large population are over 6 feet 3 inches tall.
This means that [tex]p = 0.01[/tex]
200 people
This means that [tex]n = 200[/tex]
At least four people will be over 6 feet 3 inches tall
Either there is less than four people over this height, or there is more than 4 people. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 4) + P(X \geq 4) = 1[/tex]
We want [tex]P(X \geq 4)[/tex]. So
[tex]P(X \geq 4) = 1 - P(X < 4)[/tex]
In which
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{200,0}.(0.01)^{0}.(0.99)^{200} = 0.1340[/tex]
[tex]P(X = 1) = C_{200,1}.(0.01)^{1}.(0.99)^{199} = 0.2707[/tex]
[tex]P(X = 2) = C_{200,2}.(0.01)^{2}.(0.99)^{198} = 0.272[/tex]
[tex]P(X = 3) = C_{200,3}.(0.01)^{3}.(0.99)^{197} = 0.1814[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1340 + 0.2707 + 0.272 + 0.1814 = 0.8581[/tex]
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.8581 = 0.1419[/tex]
14.19% probability that from a group of 200 people picked at random from this population, at least four people will be over 6 feet 3 inches tall
