Respuesta :
Answer:
a) 99.9996% probability that X is at least four.
b) 0.0034% probability that X is exactly four.
Step-by-step explanation:
For each resident, there are only two possible outcomes. Either they are at risk of a heart disease, or they are not. The individuals are chosen at random, which means that the probability of a person being at risk is independent from other people. So we use the binomial probability distribution to solve this problem.
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.
75% of all residents of Wellsburg, West Virginia, are at risk of heart disease.
This means that [tex]p = 0.75[/tex]
Sixteen of these residents
This means that [tex]n = 16[/tex]
Find the probability that X is
(a) at least four
Either X is lower than four, or it is at least four. The sum of the probabilities of these events is 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_{16,0}.(0.75)^{0}.(0.25)^{16} \cong 0[/tex]
[tex]P(X = 1) = C_{16,1}.(0.75)^{1}.(0.25)^{15} \cong 0[/tex]
[tex]P(X = 2) = C_{16,2}.(0.75)^{2}.(0.25)^{14} \cong 0[/tex]
[tex]P(X = 3) = C_{16,3}.(0.75)^{3}.(0.25)^{13} = 0.000004[/tex]
[tex]P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.000004[/tex]
So
[tex]P(X \geq 4) = 1 - P(X < 4) = 1 - 0.000004 = 0.999996[/tex]
99.9996% probability that X is at least four.
(b) exactly four.
This is P(X = 4).
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{16,4}.(0.75)^{4}.(0.25)^{12} = 0.000034[/tex]
0.0034% probability that X is exactly four.
Answer:
Step-by-step explanation:
Hello!
Given the discrete variable:
X: number, out of 16, of residents who are at risk for heart disease.
Binomial criteria:
1. The number of observation of the trial is fixed n=16
2. Each observation in the trial is independent, this means that none of the trials will affect the probability of the next trial. There are two possible outcomes, that the resident is at risk of heart disease, which will be the "success" of this experiment, or that the resident isn't at risk of heart disease, this will be the "failure" of the experiment.
3. The probability of success in the same from one trial to another The probability of the residents to be at heart risk is p=0.75
So X≈ Bi (n;ρ)
To calculate these probabilities I'll use a table of binomial distribution which gives the information of cumulative probabilities P(X≤r) where "r" is a possible value of the variable.
a) P(X≤4) This value can be obtained directly from the binomial table with n=16;p=0.75; r=4
P(X≤4)= 0.000038
b) P(X=4)
Using the cumulative probabilities, to reach the value you have to subtract to the probability accumulated until for the value of the probability of the previous integer:
P(X=4)= P(X≤4) - P(X≤3)= 0.000038 - 0.0000037= 0.0000343.
I hope it helps!
