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Answer :
Let x be a random variable representing the heights of adult American men. Since it is normally distributed and the population mean and standard deviation are known, we would apply the formula,
z = (x - mean)/Standard deviation
From the information given,
mean = 68
standard deviation = 2.5
The probability that the height of a selected adult is between 63 and 73 is expressed as
[tex]P(68\leq x\leq73)[/tex]For x = 63,
z = (63 - 68)/2.5 = -2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
For x = 73,
z = (73 - 68)/2.5 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.97725
Therefore,
[tex]P(68\text{ }\leq x\leq73)\text{ = 0.97725 - 0.0}2275\text{ = 0}.9545[/tex]Thus, the percentage of men are between 63 and 73 is
0.9545 * 100 = 95.45%
Rounding up to the nearest percentage, the answer is 95%
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