Respuesta :
In a certain college, 33% of the physics majors belong to ethnic minorities. 10 students are selected at random from physics majors.
Let x be the number of students selected from physics majors belongs to ethnic minorities.
Let p be the probability that selected student belongs to ethnic minorities. Let n be the total number of students selected from physics majors.
Given: n=10 and p=0.33
X follows Binomial distribution with n=10 and p=0.33
The probability that no more than 6 belongs to ethnic majority
P(x ≤ 6) = P(x=6) + p(x=5)+P(x=4) + p(x=3) + P(x=2) + p(x=1) + p(x=0)
Using Binomial distribution to find the probabilities
P(x=k) = [tex] (10Ck) 0.33^{k} (1-0.33)^{10-k} [/tex]
P(x=6) = [tex] (10C6) 0.33^{6} (1-0.33)^{10-6} [/tex]
P(x=6) = 0.0546
P(x=5) = [tex] (10C5) 0.33^{5} (1-0.33)^{10-5} [/tex]
P(x=5) = 0.1332
P(x=4) = [tex] (10C4) 0.33^{4} (1-0.33)^{10-4} [/tex]
P(x=4) = 0.2253
P(x=3) = [tex] (10C3) 0.33^{3} (1-0.33)^{10-3} [/tex]
P(x=3) = 0.2613
P(x=2) = [tex] (10C2) 0.33^{2} (1-0.33)^{10-2} [/tex]
P(x=2) = 0.1989
P(x=1) = [tex] (10C1) 0.33^{1} (1-0.33)^{10-1} [/tex]
P(x=1) = 0.0897
P(x=0) =[tex] (10C0) 0.33^{0} (1-0.33)^{10-0} [/tex]
P(x=0) = 0.0182
P(x ≤ 6) = 0.0546 + 0.1332+ 0.2253 +0.2613+ 0.1989 + 0.0897 + 0.0182
P(x ≤ 6) = 0.9821 ~ 0.982
The probability that no more than 6 belong to an ethnic minority is 0.982
Using the binomial distribution, it is found that there is a 0.982 = 98.2% probability that no more than 6 belong to an ethnic minority
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For each student, there are only two possible outcomes. Either they belong to an ethnic minority, or they do not. The probability of a student belonging to an ethnic minority is independent of any other student, which means that the binomial probability distribution is used to solve this question.
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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 a success on a single trial.
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- 33% are ethnic minorities, thus, [tex]p = 0.33[/tex]
- 10 students are chosen, thus, [tex]n = 10[/tex].
- The probability that no more than 6 are ethnic minorities is:
[tex]P(X \leq 6) = 1 - P(X > 6)[/tex]
In which
[tex]P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)[/tex]
Thus
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 7) = C_{10,7}.(0.33)^{7}.(0.67)^{3} = 0.0154[/tex]
[tex]P(X = 8) = C_{10,8}.(0.33)^{8}.(0.67)^{2} = 0.0028[/tex]
[tex]P(X = 9) = C_{10,9}.(0.33)^{9}.(0.67)^{1} = 0.0003[/tex]
[tex]P(X = 10) = C_{10,9}.(0.33)^{10}.(0.67)^{0} \approx 0[/tex]
Then
[tex]P(X > 6) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.0154 + 0.0028 + 0.0003 + 0 = 0.0185[/tex]
[tex]P(X \leq 6) = 1 - P(X > 6) = 1 - 0.0185 = 0.9815[/tex]
0.982 = 98.2% probability that no more than 6 belong to an ethnic minority
A similar problem is given at https://brainly.com/question/11404081
