If 28% of students in College are near-sighted, the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is closest to

(A) 17%
(B) 18%
(C) 12%
(D) 16%

Answer :

JeanaShupp

Answer: (D) 16%

Step-by-step explanation:

Binomial probability formula :-

[tex]P(x)=^nC_xp^x(1-p)^n-x[/tex], where n is the sample size , p is population proportion and P(x) is the probability of getting success in x trial.

Given : The proportion of students in College are near-sighted : p= 0.28

Sample size : n= 20

Then, the the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is given by :_

[tex]P(x=4)=^{20}C_4(0.28)^4(1-0.28)^{20-4}\\\\=\dfrac{20!}{4!16!}(0.28)^4(0.72)^{16}\\\\=0.155326604912\approx0.16\%[/tex]

Hence, the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is closest to 16%.

We will see that the probability is closest to 16%, so the correct option is D.

How to get the probability?

We know that 28% of the students are near-sighted, then the other 72% arent.

This means that:

  • There is a probability of 0.28 that a random student is near-sighted.
  • There is a probability of 0.72 that a random student is not near-sighted.

Then the probability that out of 20, there are 4 near-sighted students is:

[tex]P = (0.72)^{16}*(0.28)^4*C(20, 4)[/tex]

Where C(20, 4) is the number of different possible groups of 4 that we can make out from the 20 students, it is given by:

[tex]C(N, K) = \frac{N!}{(N - K)!*K!}[/tex]

Then we have:

[tex]P = (0.72)^{16}*(0.28)^4*\frac{20!}{16!*4!}\\ \\P = (0.72)^{16}*(0.28)^4*\frac{20*19*18*17}{4*3*2} = 0.155[/tex]

Written in percentage form, and rounding to the nearest whole number, we have:

P = (0.155)*100% = 15.5% ≈ 16%

So the correct option is D.

If you want to learn more about probability, you can read:

https://brainly.com/question/251701

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