Question

he heights of young women aged 20 to 29 follow approximately the N(64, 2.7) distribution. Young men the same age have heights distributed as N(69.3, 2.8).Height is measured in inches.A young woman in that age group is randomly selected; independently, a young man in that age group is randomly selected. What is the probability that the young man is taller than the young woman by more than 5 inches

Answer 1

Answer:

0.5319 = 53.19% probability that the young man is taller than the young woman by more than 5 inches

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

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 p-value, we get the probability that the value of the measure is greater than X.

Subtraction between normal variables:

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

Distribution of the difference in height between men and woman:

Applying the concept above:

$\mu = 69.3 - 64 = 5.3$

$\sigma = \sqrt{2.7^2 + 2.8^2} = 3.89$

What is the probability that the young man is taller than the young woman by more than 5 inches?

Probability of the difference being more than 5 inches, that is, 1 subtracted by the pvalue of Z when X = 5.

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

$Z = \frac{5 - 5.3}{3.89}$

$Z = -0.08$

$Z = -0.08$ has a pvalue of 0.4681

1 - 0.4681 = 0.5319

0.5319 = 53.19% probability that the young man is taller than the young woman by more than 5 inches