Question

NASA excludes anyone under 5'2" in height and anyone over 6'3" from being an astronaut pilot...

NASA excludes anyone under 5'2" in height and anyone over 6'3" from being an astronaut pilot (NASA 2004). In metric units, these values for the lower and upper height restrictions are 157.5cm and 190.5cm, respectively.

The distribution of American adult heights within a sex and age group is reasonably well approximated by normal distributions as follows:

20-29 year old males:  

20-29 year old females:

We want to answer the question: What fraction of the young male adults is eligible to be an astronaut pilot by these height constraints?

Restated, we want to find Pr(males){157.5cm  height  190.5cm}, (put the greater than and/or less than symbols in correct places).

The standard normal deviate for the lower bound =  (round to 2 decimals)

The standard normal deviate for the upper bound =  (round to 2 decimals)

One way to find the answer is 1-Pr(Z < lower bound<lower]) + Pr(Z > upper bound)

Using the Statistical Table B: The standard normal (Z) distribution,

Pr(Z<157.5) =  (report exactly the number from the table, do not round)

Pr(Z>190.5) =  (report exactly the number from the table, do not round)

Thus, the percentage of young male adults that are eligible to be an astronaut pilot based on height alone is  (report answer as a percent, rounded to one decimal place)

Homework Answers

Answer #1

Given that

* NASA Includes anyone under 5'2'' in height and anyone over 6'3" from being an astronaut pilot.

* In metric units, these values for the lower and upper height restrictions are 157.5 cm and 190.5 cm respectively.

1. pr(males)=157.5 cm<height<190.5 cm.

2. Standard normal deviate for the lower bound

=(157.5-177.6)/9.7

=-2.07

* Standard normal deviate for the upper bound

(190.5-177.6)/9.7

=1.33

pr(x<157.5)=p(z<-2.07)=0.0192

pr(x<190.5)=p(z>1.33)=0.0918

Hence, the % of young is 1-(0.0192+0.0918)

=0.8890 (or) 88.9%

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