Answer: 96.2%
Step-by-step explanation:
Assume that the heights of American men are normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = heights of American men.
µ = mean height
σ = standard deviation
From the information given,
µ = 69.0 inches
σ = 2.8 inches
the probability of men that have heights between 64 and 78 inches is expressed as
P(64 ≤ x ≤ 78)
For x = 64,
z = (64 - 69)/2.8 = - 1.79
Looking at the normal distribution table, the probability corresponding to the z score is 0.037
For x = 78,
z = (78 - 69)/2.8 = 3.2
Looking at the normal distribution table, the probability corresponding to the z score is 0.999
Therefore,
P(64 ≤ x ≤ 78) = 0.999 - 0.037 = 0.962
Therefore, the percent of men meeting these height requirements is
0.962 × 100 = 96.2%
