Question 1171482: The height of adult males in Japan is approximately normally
distributive with mean 68.5 inches and standard deviation of 1.7 inches.
(These numbers are fictious.)
(a) What percent of Japanese adult males have a height between 68
inches and 69.2 inches?
Found 2 solutions by CubeyThePenguin, greenestamps:
Answer by CubeyThePenguin(3113)   (Show Source):
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
Unfortunately, new tutor @CubeyThe Penguin often doesn't take time to answer the given question; and he rarely provides any explanation of HOW to get an answer.
In this problem, it is a complete mystery how he converts the given information to 0.5 standard deviations.
His response can be ignored, because it does nothing to help YOU understand how to get the answer. And you can ignore his answer, since there is no indication where it came from.
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The mean is 68.5 and the standard deviation is 1.7. We want to know the percentage of Japanese adult males with heights between 68 and 69.2 inches.
68 inches is 0.5 inches below the mean; that is 0.5/1.7 = 0.294 standard deviations below the mean. A z score table shows that the fraction of measurements less than 0.294 standard deviations below the mean is about 0.3844.
69.2 inches is 0.7 inches above the mean; that is 0.7/1.7 = 0.412 standard deviations above the mean. A z score table shows that the fraction of measurements less than 0.412 standard deviations above the mean is about 0.6598.
So the fraction of measurements between 0.5 inches below the mean and 0.7 inches above the mean is about 0.6598-0.3844 = 0.2754, or about 27.54%.
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If you are working problems like this, you should know how to use a z score table to get the answer, as described above.
However, if you have and know how to use a TI-83 or similar calculator, you can get this answer as follows:
2nd-VARS (gets you to DISTR)
normalcdf(68,69.2,68.5,1.7)
Certainly if you were in a job where you are performing calculations like this frequently, you would want to use a calculator to solve problems like this, instead of going to z score tables.
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