Question

In the United States, the mean and standard deviation of adult men's heights are 70 inches (5 feet 10 inches) and 4 inches, respectively. Suppose the American adult men's heights have a normal distribution Whe probability that a randomly chosen American man is taller than 6 feet (72 inches) is equal to:___________. (round off to fourth decimal place, use the given table)
a. 0.6853
b. 0.0062
c. 0.3085
d.0.6915
e. None of these

Answer 1

Answer:

$P(X>72)=P(\frac{X-\mu}{\sigma}>\frac{72-\mu}{\sigma})=P(Z>\frac{72-70}{4})=P(z>0.5)$

And we can find this probability using the complement rule and the normal standard table:

$P(z>0.5)=1-P(z$

And the best solution would be:

c. 0.3085

Step-by-step explanation:

For this case we can convert all the values to inches in order to standardize the solution:

$5ft * \frac{12 in}{1ft}= 60 in$

$6ft * \frac{12 in}{1ft}= 72 in$

Let X the random variable that represent the heights of US mens, and for this case we know the distribution for X is given by:

$X \sim N(70,4)$  

Where $\mu=70$ and $\sigma=4$

We are interested on this probability

$P(X>72)$

We can use the z score formula given by:

$z=\frac{x-\mu}{\sigma}$

Using this formula we got:

$P(X>72)=P(\frac{X-\mu}{\sigma}>\frac{72-\mu}{\sigma})=P(Z>\frac{72-70}{4})=P(z>0.5)$

And we can find this probability using the complement rule and the normal standard table:

$P(z>0.5)=1-P(z$

And the best solution would be:

c. 0.3085