Statistics

Suppose heights of men are normally distributed with a mean 69.0 inches and standard deviation 2.8 inches. What percent of men are over 6 feet tall? Round to the nearest tenth of a percent.

I understanf finding the z score, but I'm not sure how to find the percent.

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  1. Find the z-score

    72-69 divided by 2.8 = 1.07 for a z-score.

    What percent are over 6 feet tall, so we want the right tail.

    You can use a z-table to find the value. Unfortunately, not all z-tables are constructed the same way. If you have a TI-83 or 84 calculator, you can use the DISTR function found above the VARS key.
    2nd distr Normalcdf
    you enter (1.07, 1000000000000) use a very large number to take you all the way out to the right tail. I got .1423 that means 14.23% would be taller than 6 feet.

    If you are using a z-table, compare your answer with what the calculator gives.

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