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Color-blindness is any abnormality of the color vision system that causes a person to see colors differently than most people, or to have difficulty distinguishing among certain colors (www.visionrx.com).Color-blindness is gender-based, with the majority of sufferers being males.Roughly 8% of white males have some form of colorblindness, while the incidence among white females is only 1%.A random sample of 20 white males and 40 white females was chosen.LetXbe the number of males (out of the 20) who are color-blind.LetYbe the number of females (out of the 40) who are color-blind.LetZbe the total number of color-blind individuals in the sample (males and females together).Which of the following is true about the random variables X, Y, and Z?X is binomial with n = 20 and p = .08.Y is binomial with n = 40 and p = .01.Z is not binomial.All of the above are true.Only (A) and (B) are true.What is the probability that exactly 2 of the 20 males are color-blind? (Note: Some answers are rounded.).08.2711.0143.5422.0159

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Answer:

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Step-by-step explanation:

User RobertoT
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Final Answer:

Only (A) and (B) are true.

Step-by-step explanation:

Variable distributions:

X: Since we are sampling a specific number (20) of individuals with a known probability of color blindness (8%), X follows a binomial distribution with n = 20 and p = 0.08.

Y: Similarly, Y follows a binomial distribution with n = 40 and p = 0.01.

Z: Z is not a simple binomial because it combines two independent binomial variables (X and Y) with different parameters. Therefore, Z's distribution is not directly binomial.

Probability of 2 color-blind males:

Using the binomial probability formula for X, the probability of exactly 2 color-blind males (out of 20) is:

P(X = 2) = 20C2 * 0.08^2 * (1 - 0.08)^18 ≈ 0.2711

Therefore, only statements (A) and (B) are true:

(A) True: X is binomial with n = 20 and p = 0.08.

(B) True: Y is binomial with n = 40 and p = 0.01.

Statements (C), (D), and the answer choices for the probability of 2 color-blind males are incorrect.

User Elcan
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