7. According to a National Health Survey

Question

7. According to a National Health Survey, American men’s heights are normally distributed with a mean given by 

μ=69.7

 inches and a standard deviation given by σ=2.8

 inches.   

 

a). If a man is randomly selected, find the probability that his height is more 72 inches. (10 pints) 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

b). If a man is randomly selected, find the probability that his height is between 68 and 72 inches. (10 points) 

 

Answer 1

Normal distribution (bell curve) with mean of 69.7" and a standard deviation of 2.8".

(a) P(h > 72)

1 SD is 72.5" so 72" is slightly less, so my guesstimate is that 25% will be taller (because 34% within 1 SD of mean).

Exact value requires Z-Score and Z-Score Table.

Z-Score = (value - mean) / SD with positive values to right of mean, negative to left

or Z = ( X - μ ) / σ

In this case, Z = (72 - 69.7) / 2.8 = 0.82

From table 0.82 = 0.2061 (actually from -0.82 as curve is symmetrical)

Hence P(h > 72) = 20.61%

(b) P(68 < h < 72)

Again, I prefer to make a guesstimate. We already know 72" is slightly within 1 SD. 68 is also less than 1 SD (to left of mean). Again, based on knowing that 1 SD of mean = 34% of bell curve, I'd guess about 20% to left and 25% to right, or around 45%.

This time we need two Z-scores, one of which we already worked out.

From (a) we know 20.61% is above 72".

For 68", Z = ( 68 - 69.7 ) / 2.8 = -0.61

From Z table -0.61 = 0.2709 or 27.09% (which is probabiliy below 68")

Hence P(68 < h < 72) = 100 - 27.09 - 20.61 = 52.3%