Solution 2
The sum of L and R is equal to 1. If L and R are given as percentages, then L + R = 100%.
Practice 3
Suppose we are told that Thermington is exactly at the 81st percentile point in the distribution. What does this mean in terms of the areas of the regions L and R?
Solution 3
It means that L represents 81% of the area under the curve, and therefore that R represents 100% – 81%, or 19%, of the area under the curve.
Practice 4
Suppose we are told that Thermington is exactly at the 8th decile point in the distribution. What does this mean in terms of the areas of the regions L and R?
Solution 4
It means that L represents 8/10 of the area under the curve, and therefore that R represents 1 – 8/10, or 2/10, of the area under the curve.
Practice 5
Suppose we are told that Thermington is exactly at the 3rd quartile point in the distribution. What does this mean in terms of the areas of the regions L and R?
Solution 5
It means that L represents 3/4 of the area under the curve, and therefore that R represents 1–3/4, or 1/4, of the area under the curve.
Practice 6
Suppose we are told that Thermington is among the onequarter of towns in the experiment that saw the greatest temperature increase between last year and 100 years ago. What does this mean in terms of the areas L and R?
Solution 6
The statement of this problem is ambiguous. It could mean either of two things:
 We are considering all the towns in the experiment.
 We are considering only those towns in the experiment that witnessed increases in temperature.
If we mean (1) above, then the above specification means that L represents more than 3/4 of the area under the curve, and therefore that R represents less than 1/4 of the area under the curve. If we mean (2) above, we can say that R represents less than 1/4, or 25%, of the area under the curve to the right of the vertical axis (the axis in the center of the graph showing the number of locations experiencing a given temperature change); but we can't say anything about L unless we know more about the distribution.
It looks like the curve in Fig. 810 is symmetrical around the vertical axis. In fact, it's tempting to think that the curve is a normal distribution. But we haven't been told that this is the case. We mustn't assume it without proof. Suppose we run the data through a computer and determine that the curve is a normal distribution. Then if (2) above is true, L represents more than 7/8 of the total area under the curve, and R represents less than 1/8 of the total area under the curve.
Practice 7
Suppose that the curve in Fig. 810 represents a normal distribution, and that Thermington happens to lie exactly one standard deviation to the right of the vertical axis. What can be said about the areas L and R in this case?
Solution 7
Imagine the "sister city" of Thermington, a town called Frigidopolis. Suppose Frigidopolis was cooler last year, in comparison to 100 years ago, by exactly the same amount that Thermington was warmer. This situation is shown graphically in Fig. 811. Because Thermington corresponds to a point (or dashed vertical line) exactly one standard deviation to the right of the vertical axis, Frigidopolis can be represented by a point (or dashed vertical line) exactly one standard deviation to the left of the vertical axis.
Fig. 811. Illustration for Solution 7.
The mean (μ) in Fig. 811 happens to coincide with the vertical axis, or the point where the temperature change is 0. Recall the empirical rule concerning standard deviation (σ) and normal distributions. We learned about this in Chapter 3. The empirical rule applies directly to this problem. The vertical dashed line on the left, representing Frigidopolis, is – σ from μ. The vertical dashed line on the right, representing Thermington, is +σ from μ. The empirical rule tells us that the proportion of the area under the curve between these two dashed lines is 68% of the total area under the curve, because these two dashed vertical lines represent that portion of the area within ± of μ. The proportion of the area under the curve between the vertical axis and either dashed line is half this, or 34%.
The fact that Fig. 811 represents a normal distribution also tells us that exactly 50% of the total area under the curve lies to the left of μ, and 50% of the area lies to the right of μ, if we consider the areas extending indefinitely to the left or the right. This is true because a normal distribution is always symmetrical with respect to the mean.
Knowing all of the above, we can determine that L = 50% + 34% = 84%. That means R = 100% – 84% = 16%.
More practice problems for these concepts can be found at:
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