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Part III: Distributions and the Central Limit Theorem
But exactly how long is that fish, in inches? In Step 3, you change the z-value
back to an x-value (fish length in inches) using the z-formula solved for x; you
get x = 16 + –1.28 ∗ 4 = 10.88 inches. So 10.88 inches marks the lowest 10 per-
cent of fish lengths. Ten percent of the fish are shorter than that.
Working with a higher percentile
Now suppose you want to find the length that marks the top 25 percent of all
the fish in the pond. This problem calls for Step 1b (in “Finding a percentile
for a normal distribution”) because being in the top part of the distribution
means you’re dealing with a greater-than probability. The number you are
looking for is somewhere in the right tail (upper area) of the X-distribution,
with p = 25 percent of the probability to its right and 1 – p = 75 percent to its
left. Thinking in terms of the Z-table and how it only uses less-than probabili-
ties, you need to find the 75th percentile for Z, then change it to an x-value.
Step 2: The 75th percentile of Z is the z-value where p(Z < z) = 0.75. Using the
Z-table (in the appendix), you find the probability closest to 0.7500 is 0.7486,
and its corresponding z-value is in the row for 0.6 and column for 0.07. Put
these together and you get a z-value of 0.67. This is the 75th percentile for Z.
In Step 3, change the z-value back to an x-value (length in inches) using the
z-formula solved for x to get x = 16 + 0.67 ∗ 4 = 18.68 inches. So, 75% of the
fish are shorter than 18.68 inches. And to answer the original question, the
top 25% of the fish in the pond are longer than 18.68 inches.
Translating tricky wording
in percentile problems
Some percentile problems are especially challenging to translate. For example,
suppose the amount of time for a racehorse to run around a track in a quali-
fying round has a normal distribution with mean 120 seconds and standard
deviation 5 seconds. The best 10 percent of the times qualify; the rest don’t.
What’s the cutoff time for qualifying?
Because “best times” mean “lowest times” in this case, the percentage of
times that lie below the cutoff must be 10, and the percentage above the cutoff
must be 90. (It’s an easy mistake to think it’s the other way around.) The per-
centile of interest is therefore the 10th, which is down on the left tail of the
distribution. You now work this problem the same way I worked Problem 1
regarding fish lengths (see the section, “Finding Probabilities for a Normal
Distribution”). The standard score for the 10th percentile is z = –1.28 look-
ing at the Z-table (in the appendix). Converting back to original units, you get
seconds. So the cutoff time needed for a
racehorse to qualify (that is, to be among the fastest 10%) is 113.6 seconds.
(Notice this number is less than the average time of 120 seconds, which makes
sense; a negative z-value is what makes this happen.)

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