Question

Use 2 scores to compare the given valuesThe talest living man at one time had a height of 229 cm. The shortest living man at that time had a height of 77.2 em Heights of men at that time had a mean of172.16 cm and a standard deviation of 5.95 em. Which of these two men had height that was more extreme?and the score for the shortest man iszatheman had the height that was more extremeSince the score for the tallest man is z(Round to be decimal places)

Answer 1

As usual, let us make a drawing of the distribution:

In general, a z-score says how far from the mean a (data) point is. If we have a data point x, its corresponding z-score can be calculated by

$z_x=\frac{x-\mu}{\sigma}$

where "mu" represents the mean, and "sigma" represents the standard deviation. Let's calculate the z-scores for 229cm and 77.2cm:

$z_{229}=\frac{229\operatorname{cm}-172.16\operatorname{cm}}{5.95\operatorname{cm}}\approx9.55$$z_{77.2}=\frac{77.2\operatorname{cm}-172.16\operatorname{cm}}{5.95\operatorname{cm}}\approx-15.96$

Now, note that

$15.96=|z_{77.2}|>|z_{229}|=9.55$

When this is the case, we say that 77.2cm is more extreme (or is further away from the mean) than 229cm. Namely, the shortest living man had a more extreme height than the tallest one.

Comment: (Be careful!) Note that in the inequality, I consider the absolute value of the z-scores. The sign just says whether x is on the left or on the right of the mean; if z_x is negative, x is on the left, and if z_x is positive, x is on the right of the mean.