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. Consider the data from the Anthropology 105 class. The mean height for women in theclass is 64.33 in and the standard deviation is 2.64 in.1. The average height of men in the US is approximately 5ft 10in. What proportion of womenrepresented here are shorter than the average man? (enter the answer as a percent rounded to thenearest hundredth as needed)2. Assume professional volleyball players who are women are at least 6 foot tall. What proportion ofthe population could even hope to be a professional volleyball player by using these standards? Hint:Find the proportion of women who are at least 6 foot tall. (enter the answer as a percent rounded tothe nearest hundredth as needed)

Respuesta :

Given:

The mean height for women in the class = μ = 64.33

The standard deviation = σ = 2.64 in

We will use the formula of the z-score

[tex]z=\frac{x-\mu}{\sigma}[/tex]

We will find the following:

1. The average height of men in the US is approximately 5ft 10in. What proportion of women represented here are shorter than the average man?

1 foot = 12 inches

So, x = 5ft 10in = 5*12 + 10 = 70 in

[tex]z=\frac{70-64.33}{2.64}=2.1477[/tex]

From the normal distribution curve we will find P(z < 2.1477)

So, the percentage will be = 0.9841

rounding to the nearest hundredth,

So, the answer will be 98.41%

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2. Assume professional volleyball players who are women are at least 6 feet tall. What proportion of the population could even hope to be a professional volleyball players by using these standards?

So, x = 6 foot = 72 in

[tex]z=\frac{72-64.33}{2.64}=2.9053[/tex]

From the normal distribution curve, we will find P( z > 2.9053)

So, the answer will be 0.0018 = 0.18%

So, the answer will be 0.18%

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