Respuesta :
z = (x - mean) / SD = (79 - 70) / 3 = 3
P (Z > 3)? = 1 - F (z) = 1 - F (3) = 0.00135
P (Z > 3)? = 1 - F (z) = 1 - F (3) = 0.00135
Answer:
A. 0.135%
Step-by-step explanation:
We have been given that adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketball player is 79 inches tall.
We need to find the area of normal curve above the raw score 79.
First of all let us find the z-score corresponding to our given raw score.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Raw-score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Upon substituting our given values in z-score formula we will get,
[tex]z=\frac{79-70}{3}[/tex]
[tex]z=\frac{9}{3}[/tex]
[tex]z=3[/tex]
Now we will find the P(z>3) using formula:
[tex]P(z>a)=1-P(z<a)[/tex]
[tex]P(z>3)=1-P(z<3)[/tex]
Using normal distribution table we will get,
[tex]P(z>3)=1-0.99865 [/tex]
[tex]P(z>3)=0.00135[/tex]
Let us convert our answer into percentage by multiplying 0.00135 by 100.
[tex]0.00135\times 100=0.135%[/tex]
Therefore, approximately 0.135% of the adult male population is taller than the average basketball player and option A is the correct choice.
