Final answer:
The height at the 95th percentile for the specified normal distribution of women's heights is approximately 69.34 inches, calculated using the mean of 64.7 inches and a standard deviation of 2.82 inches along with the z-score for the 95th percentile.
Step-by-step explanation:
To find the height that represents the 95th percentile for the specified normal distribution of women's heights, you would use a z-score table or calculator, considering that the sample follows a normal distribution with a mean (μ) of 64.7 inches and a standard deviation (σ) of 2.82 inches. The z-score for the 95th percentile in a standard normal distribution is approximately 1.645. To convert this z-score to the corresponding value in the distribution of women's heights, the following formula is used:
X = μ + (z * σ), where X is the height at the 95th percentile.
Substituting the values in:
X = 64.7 inches + (1.645 * 2.82 inches)
X ≈ 64.7 inches + 4.6439 inches
X ≈ 69.3439 inches
So the height that represents the 95th percentile for women in this age category in that country is approximately 69.34 inches.