Answer:
a. The positive difference between Nelson's height and the population mean is:
.
b. The difference found in part (a) is 1.174 standard deviations from the mean (without taking into account if the height is above or below the mean).
c. Nelson's z-score:
(Nelson's height is <em>below</em> the population's mean 1.174 standard deviations units).
d. Nelson's height is <em>usual</em> since
.
Step-by-step explanation:
The key concept to answer this question is the z-score. A <em>z-score</em> "tells us" the distance from the population's mean of a raw score in <em>standard deviation</em> units. A <em>positive value</em> for a z-score indicates that the raw score is <em>above</em> the population mean, whereas a <em>negative value</em> tells us that the raw score is <em>below</em> the population mean. The formula to obtain this <em>z-score</em> is as follows:
[1]
Where
is the <em>z-score</em>.
is the <em>population mean</em>.
is the <em>population standard deviation</em>.
From the question, we have that:
- Nelson's height is 68 in. In this case, the raw score is 68 in
in.
in.
in.
With all this information, we are ready to answer the next questions:
a. What is the positive difference between Nelson's height and the mean?
The positive difference between Nelson's height and the population mean is (taking the absolute value for this difference):
.
That is, <em>the positive difference is 2.7 in</em>.
b. How many standard deviations is that [the difference found in part (a)]?
To find how many <em>standard deviations</em> is that, we need to divide that difference by the <em>population standard deviation</em>. That is:

In words, the difference found in part (a) is 1.174 <em>standard deviations</em> from the mean. Notice that we are not taking into account here if the raw score, <em>x,</em> is <em>below</em> or <em>above</em> the mean.
c. Convert Nelson's height to a z score.
Using formula [1], we have




This z-score "tells us" that Nelson's height is <em>1.174 standard deviations</em> <em>below</em> the population mean (notice the negative symbol in the above result), i.e., Nelson's height is <em>below</em> the mean for heights in the club presidents of the past century 1.174 standard deviations units.
d. If we consider "usual" heights to be those that convert to z scores between minus2 and 2, is Nelson's height usual or unusual?
Carefully looking at Nelson's height, we notice that it is between those z-scores, because:


Then, Nelson's height is <em>usual</em> according to that statement.