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

Answer:

$P(z> \frac{(69.3-64)-0}{\sqrt{2.74^2 +2.7^2}}) =P(z>1.38)$

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:

$Y-X \sim N (\mu_Y -\mu_X , \sqrt{\sigma^2_Y +\sigma^2_X})$

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

$z =\frac{z -\mu_z}{\sigma_z}$

And using the z score formula we got:

$P(z> \frac{(69.3-64)-0}{\sqrt{2.74^2 +2.7^2}}) =P(z>1.38)$

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