suppose X and Y are independent random variables, both with normal distributions. If X has a mean of 45 with a standard deviation of 4 and Y has a mean of 35 with a standard deviation of 3, what is the probability that a randomly generated value of X is greater than a randomly generated value of Y

Answer:

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

[tex]\mu_X = 45, \sigma_X = 4, \mu_Y = 35, \sigma_Y = 3[/tex]

What is the probability that a randomly generated value of X is greater than a randomly generated value of Y

This means that the subtraction of X by Y has to be positive.

When we subtract two normal variables, the mean is the subtraction of their means, and the standard deviation is the square root of the sum of their variances. So

[tex]\mu = \mu_X - \mu_Y = 45 - 35 = 0[/tex]

[tex]\sigma = \sqrt{\sigma_X^2+\sigma_Y^2} = \sqrt{25} = 5[/tex]

We want to find P(X > 0), that is, 1 subtracted by the pvalue of Z when X = 0. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{0 - 10}{5}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228

1 - 0.0228 = 0.9772

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

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