Question

The heights of women aged 20 to 29 follow approximately the N(64, 2.74) distribution. Men the same age have heights distributed as N(69.3, 2.7). What percent of young men are taller than the mean height of young women

Answer 1

To find the percentage of young men who are taller than the mean height of young women, we need to calculate the z-score for the mean height of young women and then find the area to the right of that z-score in the standard normal distribution curve. Approximately 97.5% of young men are taller than the mean height of young women.

Step-by-step explanation:

In order to find the percentage of young men who are taller than the mean height of young women, we need to calculate the z-score for the mean height of young women and then find the area to the right of that z-score in the standard normal distribution curve.

The z-score formula is:

z = (X - µ) / σ

Where X is the value you want to convert to a z-score, µ is the mean, and σ is the standard deviation.

In this case, the mean height of young women is 64 inches and the standard deviation is 2.74 inches. The mean height of young men is 69.3 inches with a standard deviation of 2.7 inches.

Using the z-score formula for the mean height of young women:

z = (64 - 69.3) / 2.7

z = -1.96

Now we need to find the area to the right of -1.96 in the standard normal distribution curve, which represents the percentage of young men who are taller than the mean height of young women.

Using a standard normal distribution table or a calculator, we find that the area to the right of -1.96 is approximately 0.975.

Therefore, approximately 97.5% of young men are taller than the mean height of young women.

Answer 2

`P(z> ((69.3-64)-0)/(√(2.74^2 +2.7^2))) =P(z>1.38)`

and we can find this probability using the complement rule and we got:

`P(z>1.38)=1-P(z<1.38) = 1-0.9162= 0.0838`

So then we can conclude that approximately 8.38% of the men are taller than women

Step-by-step explanation:

Let X the random variable who represent the heights of women aged 20 to 29 and the distribution is given by:

`X \sim N(64, 2.74)`

And let Y the heights of men aged 20 to 29 and the distribution for Y is given by:

`Y \sim N(69.3, 2.7)`

And for this case we want to find the following probability:

`P(Y>X) = P(Y-X >0)`

The distribution for Y-X is given by:

`Y-X \sim N (\mu_Y -\mu_X , √(\sigma^2_Y +\sigma^2_X))`

We can define the random variable Z= Y-X and the we can use the z score formula given by:

`z =(z -\mu_z)/(\sigma_z)`

And using the z score formula we got:

`P(z> ((69.3-64)-0)/(√(2.74^2 +2.7^2))) =P(z>1.38)`

and we can find this probability using the complement rule and we got:

`P(z>1.38)=1-P(z<1.38) = 1-0.9162= 0.0838`

So then we can conclude that approximately 8.38% of the men are taller than women