For the given heights of U.S. men, calculate the 𝑧 z -score, and comment on whether the height would be unusual for a U.S. man.

Question

Strata	Mean Height (inches)	Standard Deviation Height (inches)	Mean Weight (pounds)	Standard Deviation Weight (pounds)
U.S. Men	69.3	2.8	191	28
U.S. Women	64	2.8	145	32
NFL Quarterbacks	76.5	1.8	245	25
Top Female Models	70	2.2	115

For the given heights of U.S. men, calculate the 𝑧z-score, and comment on whether the height would be unusual for a U.S. man.

(a) 76.7 

𝑧z-score:  

Unusual

Not unusual

(b) 75.1 

𝑧z-score:  

Unusual

Not unusual

(c) 65.5 

𝑧z-score:  

Unusual

Not unusual

(d) 68.4 

𝑧z-score:  

Unusual

Not unusual

Answer 1

To get the z-score, subtract the mean from the height and divide the result by the standard deviation.'

For question "a." that would be (76.7 - 69.3) / 2.8 = 2.64

Doing the same procedure for the other 3 yield the following z-scores:

b. (75.1 - 69.3) / 2.8 = 2.07

c. (65.5 - 69.3) / 2.8 = -1.38

d. (68.4 - 69.3) / 2.8 = -0.32

"Usual / unusual" is not technical, but is normally interpreted in terms of the z-score. If -1.96 ≤ z ≤ +1.96, the measurement is in the central 95% and is considered usual. Otherwise, the measurement is either in the low 2.5% or the high 2.5% and would be labeled "unusual." Using this standard, a and b are unusual, c and d are usual.

Answer 2

Briana

I will model (a) then you can try the others.

For (a), Z = (76.7 - 69.3) / 2.8 = 2.64

Using an area table for a standard normal area to the left of Z = 2.64 is 0.9959, so the chance of a height 76.7 or higher is 1 - 0.9959 = 0.0041 = 0.41 %

So such a height would be unusual for a US man, as this probability is very low

Mike