The problem involves two independent normal random variables, X and Y. We are interested in the probability that X Y, or equivalently, that X - Y 0.

The difference of two independent normal random variables is also normally distributed. The mean of the difference is the difference of the means, and the variance of the difference is the sum of the variances.

So, we have a new random variable Z = X - Y, which is normally distributed with mean μ_Z = μ_X - μ_Y = 60 - 70 = -10, and variance σ_Z^2 = σ_X^2 + σ_Y^2 = 20^2 + 10^2 = 400 + 100 = 500. Therefore, the standard deviation σ_Z = sqrt(500) = 22.36.

We want to find Pr(Z 0), which is equivalent to finding the probability that a standard normal variable is greater than (0 - (-10))/22.36 = 0.447.

Looking up 0.447 in a standard normal table, we find that the probability that a standard normal variable is less than 0.447 is approximately 0.673.

Because the standard normal distribution is symmetric about 0, the probability that a standard normal variable is greater than 0.447 is 1 - 0.673 = 0.327.

So, the probability that a randomly chosen male is longer than a randomly chosen female is approximately 0.327. Therefore, the correct answer is 0.327.

Question

The problem involves two independent normal random variables, X and Y. We are interested in the probability that X > Y, or equivalently, that X - Y > 0.

The difference of two independent normal random variables is also normally distributed. The mean of the difference is the difference of the means, and the variance of the difference is the sum of the variances.

So, we have a new random variable Z = X - Y, which is normally distributed with mean μ_Z = μ_X - μ_Y = 60 - 70 = -10, and variance σ_Z^2 = σ_X^2 + σ_Y^2 = 20^2 + 10^2 = 400 + 100 = 500. Therefore, the standard deviation σ_Z = sqrt(500) = 22.36.

We want to find Pr(Z > 0), which is equivalent to finding the probability that a standard normal variable is greater than (0 - (-10))/22.36 = 0.447.

Looking up 0.447 in a standard normal table, we find that the probability that a standard normal variable is less than 0.447 is approximately 0.673.

Because the standard normal distribution is symmetric about 0, the probability that a standard normal variable is greater than 0.447 is 1 - 0.673 = 0.327.

So, the probability that a randomly chosen male is longer than a randomly chosen female is approximately 0.327. Therefore, the correct answer is 0.327.

Knowee AI · Accepted Answer

The problem involves two independent normal random variables, X and Y. We are interested in the probability that X > Y, or equivalently, that X - Y > 0.

The difference of two independent normal random variables is also normally distributed. The mean of the difference is the difference of the means, and the variance of the difference is the sum of the variances.

So, we have a new random variable Z = X - Y, which is normally distributed with mean μ_Z = μ_X - μ_Y = 60 - 70 = -10, and variance σ_Z^2 = σ_X^2 + σ_Y^2 = 20^2 + 10^2 = 400 + 100 = 500. Therefore, the standard deviation σ_Z = sqrt(500) = 22.36.

We want to find Pr(Z > 0), which is equivalent to finding the probability that a standard normal variable is greater than (0 - (-10))/22.36 = 0.447.

Looking up 0.447 in a standard normal table, we find that the probability that a standard normal variable is less than 0.447 is approximately 0.673.

Because the standard normal distribution is symmetric about 0, the probability that a standard normal variable is greater than 0.447 is 1 - 0.673 = 0.327.

So, the probability that a randomly chosen male is longer than a randomly chosen female is approximately 0.327. Therefore, the correct answer is 0.327.

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