Question

Suppose X and Y are independent random variables, both with normal distribution. If X has mean 30 with standard deviation 6,

Answer 1

Using the normal distribution, it is found that there is a 0.7549 = 75.49% probability that a randomly generated value of X is greater than a  randomly generated value of Y.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.  

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

• When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variables.

In this problem:

• For variable X, $\mu_X = 30, \sigma_X = 6$.

• For variable Y, $\mu_Y = 25, \sigma_Y = 4$.

The the distribution of X - Y, we have that:

$\mu = \mu_X - \mu_Y = 30 - 25 = 5$

$\sigma = \sqrt{\sigma_X^2 + \sigma_Y^2} = \sqrt{6^2 + 4^2} = 7.2111$

The probability that a randomly generated value of X is greater than a  randomly generated value of Y is P(X - Y) > 0, which is 1 subtracted by the p-value of Z when X = 0, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0 - 5}{7.2111}$

$Z = -0.69$

$Z = -0.69$ has a p-value of 0.2451.

1 - 0.2451 = 0.7549.

0.7549 = 75.49% probability that a randomly generated value of X is greater than a  randomly generated value of Y.

To learn more about the normal distribution, you can take a look at brainly.com/question/24663213