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77julia77 [94]
3 years ago
6

Adult male heights are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. The average basketbal

l player is 79 inches tall. Approximately what percent of the adult male population is taller than the average basketball player? 0.135% 0.875% 49.875% 99.875%
Mathematics
2 answers:
Marizza181 [45]3 years ago
5 0
Z = (x - mean) / SD = (79 - 70) / 3 = 3 
<span>P (Z > 3)? = 1 - F (z) = 1 - F (3) = 0.00135</span>
makvit [3.9K]3 years ago
5 0

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.

z=\frac{x-\mu}{\sigma}, where,

z=\text{z-score},

x=\text{Raw-score},

\mu=\text{Mean},

\sigma=\text{Standard deviation}.

Upon substituting our given values in z-score formula we will get,

z=\frac{79-70}{3}

z=\frac{9}{3}

z=3

Now we will find the P(z>3) using formula:

P(z>a)=1-P(z

P(z>3)=1-P(z

Using normal distribution table we will get,

P(z>3)=1-0.99865

P(z>3)=0.00135

Let us convert our answer into percentage by multiplying 0.00135 by 100.

0.00135\times 100=0.135%

Therefore, approximately 0.135% of the adult male population is taller than the average basketball player and option A is the correct choice.

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