The percentage of all men between 66.5 inches and 71.5 inches is 99.99%
The mean of the heights of adult men in inches = 69
The standard deviation of the adult men in inches = 2.5
The percentage of all men between the heights of 66.5 inches and 71.5 inches can be calculated using the normal distribution.
P(66.5 < x < 71.5)
First, let us find the standard form for these random samples using the formula,
Z = (X - μ) / σ
where Z is the standard form
X is the random sample
μ is the mean
σ is the standard deviation.
For x = 66.5,
z = (66.5 - 69) / 2.5
= -2.5 / 2.5
= -1
For x = 71.5,
z = (71.5 - 69) / 2.5
= 2.5 / 2.5
= 1
The standardized form of those random samples are
P(66.5 < x < 71.5) = P( -1 < z < 1)
which can be written as
= P(-1 < z < 0) + P(0 < z < 1)
By using the z-table we can find the value of z
P(-1 < z < 1) = 0.15866 + 0.8413
= 0.99996
Then, the percentage of heights between 66.5 and 71.5 is
= 0.99996 x 100
= 99.99 %
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