The heights of adult men in America are normally distributed, with a mean of 69.1 inches and a standard deviation of 2.62 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.1 inches and a standard deviation of 2.59 inches.
a) If a man is 6 feet 3 inches tall, what is his z-score?
z =
b) What percentage of men are SHORTER than 6 feet 3 inches?
c) If a woman is 5 feet 6 inches tall, what is her z-score?
z =
d) What percentage of women are TALLER than 5 feet 6 inches?
To answer these questions, we need to calculate the z-score for each height and then use the standard normal distribution table to determine the corresponding percentage.
a) The z-score represents the number of standard deviations above or below the mean a particular observation is. To calculate the z-score for a man who is 6 feet 3 inches tall (or 75 inches), we can use the formula:
z = (x - mean) / standard deviation
Where x is the observed value (75 inches), mean is the mean height of men (69.1 inches), and the standard deviation is 2.62 inches. Plugging in the values, we get:
z = (75 - 69.1) / 2.62 = 2.25
Therefore, the z-score for a man who is 6 feet 3 inches tall is 2.25.
b) To find the percentage of men who are shorter than 6 feet 3 inches, we need to calculate the area under the normal distribution curve to the left of the z-score. We can use the standard normal distribution table or a statistical software to find this percentage. In this case, the area to the left of a z-score of 2.25 is approximately 0.9878. This means that approximately 98.78% of men are shorter than 6 feet 3 inches.
c) Similarly, to calculate the z-score for a woman who is 5 feet 6 inches tall (or 66 inches), we use the formula:
z = (x - mean) / standard deviation
Substituting the values, we get:
z = (66 - 64.1) / 2.59 = 0.73
Therefore, the z-score for a woman who is 5 feet 6 inches tall is 0.73.
d) To find the percentage of women who are taller than 5 feet 6 inches, we need to calculate the area under the normal distribution curve to the right of the z-score. Using the standard normal distribution table or a statistical software, the area to the left of a z-score of 0.73 is approximately 0.7669. Therefore, the percentage of women who are taller than 5 feet 6 inches is approximately 1 - 0.7669 = 0.2331, or 23.31%.