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alexandr402 [8]
3 years ago
12

With a height of 68 ​in, Nelson was the shortest president of a particular club in the past century. The club presidents of the

past century have a mean height of 70.7 in and a standard deviation of 2.3 in. a. What is the positive difference between Nelson​'s height and the​ mean? b. How many standard deviations is that​ [the difference found in part​ (a)]? c. Convert Nelson​'s height to a z score. 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?
Mathematics
1 answer:
Ivahew [28]3 years ago
7 0

Answer:

a. The positive difference between Nelson's height and the population mean is: \\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

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: \\ z = -1.1739 \approx -1.174 (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 \\ -2 < -1.174 < 2.

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:

\\ z = \frac{x - \mu}{\sigma} [1]

Where

\\ z is the <em>z-score</em>.

\\ \mu is the <em>population mean</em>.

\\ \sigma 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 \\ x = 68 in.
  • \\ \mu = 70.7in.
  • \\ \sigma = 2.3in.

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):

\\ \lvert 68-70.7 \rvert = \lvert 70.7-68 \rvert\;in = 2.7\;in.

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:

\\ \frac{2.7\;in}{2.3\;in} \approx 1.1739 \approx 1.174

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

\\ z = \frac{x - \mu}{\sigma}

\\ z = \frac{68\;in - 70.7\;in}{2.3\;in}

\\ z = \frac{-2.7\;in}{2.3\;in}

\\ z = -1.1739 \approx -1.174

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:

\\ -2 < z_{Nelson} < 2

\\ -2 < -1.174 < 2

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

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