The heights of adult men in America are normally distributed, with a mean of 69.7 inches and a standard deviation of 2.68 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.6 inches and a standard deviation of 2.59 inches.
a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)?
z =
b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.
%
c) If a woman is 5 feet 11 inches tall, what is her z-score (to two decimal places)?
z =
d) What percentage of women are TALLER than 5 feet 11 inches? Round to nearest tenth of a percent.
%
e) Who is relatively taller: a 6'3" American man or a 5'11" American woman? Defend your choice in a meaningful sentence.
a) To find the z-score, we divide the deviation of the man's height from the mean by the standard deviation.
Height of man = 6 feet 3 inches = 75 inches
Mean height of men = 69.7 inches
Standard deviation of men's heights = 2.68 inches
z = (75 - 69.7) / 2.68 = 1.97
Therefore, the z-score for a man who is 6 feet 3 inches tall is 1.97.
b) To find the percentage of men shorter than 6 feet 3 inches, we need to find the area to the left of the z-score calculated in part a on the standard normal distribution table.
Using the z-table, we find that the area to the left of 1.97 is approximately 0.9764. To convert this into a percentage, we multiply by 100.
Percentage of men shorter than 6 feet 3 inches ≈ 0.9764 * 100 ≈ 97.6%
Therefore, approximately 97.6% of men are shorter than 6 feet 3 inches.
c) To find the z-score, we divide the deviation of the woman's height from the mean by the standard deviation.
Height of woman = 5 feet 11 inches = 71 inches
Mean height of women = 64.6 inches
Standard deviation of women's heights = 2.59 inches
z = (71 - 64.6) / 2.59 ≈ 2.49
Therefore, the z-score for a woman who is 5 feet 11 inches tall is 2.49.
d) To find the percentage of women taller than 5 feet 11 inches, we need to find the area to the right of the z-score calculated in part c on the standard normal distribution table.
Using the z-table, we find that the area to the right of 2.49 is approximately 0.0063. To convert this into a percentage, we multiply by 100.
Percentage of women taller than 5 feet 11 inches ≈ 0.0063 * 100 ≈ 0.63%
Therefore, approximately 0.63% of women are taller than 5 feet 11 inches.
e) To determine who is relatively taller, we compare the z-scores calculated for the two individuals. The higher the z-score, the taller the individual is compared to their respective gender.
The z-score for the 6'3" American man is 1.97, while the z-score for the 5'11" American woman is 2.49.
Therefore, the 5'11" American woman has a higher z-score, indicating that she is relatively taller compared to her gender than the 6'3" American man.