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Bezzdna [24]
4 years ago
12

Use z scores to compare the given values. The tallest living man at one time had a height of 252 cm. The shortest living man at

that time had a height of 79.2 cm. Heights of men at that time had a mean of 176.74 cm and a standard deviation of 8.06 cm. Which of these two men had the height that was moreâ extreme?
Mathematics
1 answer:
Sever21 [200]4 years ago
6 0

Answer:

The more extreme height was the case for the <em>shortest living man</em> at that time (12.1017 standard deviation units below the population's mean) compare with the tallest living man (at that time) that was 9.3374 standard deviation units above the population's mean.

Step-by-step explanation:

To answer this question, we need to use standardized values, and we can obtain them using the formula:

\\ z = \frac{x - \mu}{\sigma} [1]  

Where

  • x is the <em>raw score</em> we want to standardize.
  • \\ \mu is the population's mean.
  • \\ \sigma is the population standard deviation.

A z-score "tells us" the distance from \\ \mu in <em>standard deviation units</em>, and a <em>positive value</em> indicates that the raw score is <em>above</em> the mean and a <em>negative</em> that the raw score is <em>below</em> the mean.

In a normal distribution, the more extreme values are those with <em>bigger z-scores</em>, above and below the mean. We also need to remember that the normal distribution is symmetrical.

Heights of men at that time had:

  • \\ \mu = 176.74 cm.
  • \\ \sigma = 8.06 cm

Let us see the z-score for each case:

Case 1: The tallest living man at that time

The tallest man had a height of 252 cm.

Using [1], we have (without using units):

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

\\ z = \frac{252 - 176.74}{8.06}

\\ z = \frac{75.26}{8.06}

\\ z = 9.3374  

That is, the tallest living man was 9.3374 standard deviation units <u><em>above</em></u> the population's mean.

Case 2: The shortest living man at that time

The shortest man had a height of 79.2 cm.

Following the same procedure as before, we have:

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

\\ z = \frac{79.2 - 176.74}{8.06}

\\ z = \frac{-97.54}{8.06}

\\ z = -12.1017

That is, the shortest living man was 12.1017 standard deviation units <u><em>below</em></u> the population's mean (because of the negative value for the standardized value.)

The normal distribution is symmetrical (as we previously told). The height for the shortest man was at <em>the other extreme of the normal distribution</em> in \\ 12.1017 - 9.3374 = 2.7643 standard deviation units more than the tallest man.

Then, the more extreme height was the case for the shortest living man (12.1017 standard deviation units below the population's mean) compare with the tallest man that was 9.3374 standard deviation units above the population's mean.

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