The height of adult women in the U.S. is normally distributed, with a mean height of 64 inches and a standard deviation of 3 inches. With this in mind, answer the following:
a) The shortest 10% of women are shorter than what height?
b) The middle 75% of all women are between what two heights?
c) The tallest 5% of women are taller than what height?
I'm so confused. Please help.
Try using this website for making calculations that require the normal distribution function.
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
For part a), I get
60.15 inches.
For art (c), I get 68.94 inches
I had to guess the limits and see what the fractions fell within those limits, using the computation tool provided. Some iteration was required.
To answer these questions, we can utilize the concept of z-scores, which is a measure of how many standard deviations an observation is away from the mean.
a) To find the height for the shortest 10% of women, we need to determine the z-score corresponding to the 10th percentile. We can do this by using a standard normal distribution table, or by making use of the inverse cumulative distribution function (also known as the inverse normal function) with a probability of 0.10.
The formula to calculate the z-score is:
z = (x - mean) / standard deviation
Where:
x = the particular value we want to find
mean = mean height of women (64 inches)
standard deviation = standard deviation of women's heights (3 inches)
Using the inverse normal function or a z-score table, we find that the z-score corresponding to the 10th percentile is approximately -1.28.
Then, we rearrange the formula to solve for x:
-1.28 = (x - 64) / 3
Solving for x, we get:
x - 64 = -1.28 * 3
x - 64 = -3.84
x = 64 - 3.84
x ≈ 60.16
Therefore, the shortest 10% of women are shorter than approximately 60.16 inches.
b) To find the range for the middle 75% of women's heights, we need to determine the z-scores corresponding to the 25th and 75th percentiles (which divide the distribution into the lower and upper quartiles).
The z-scores for the 25th and 75th percentiles are approximately -0.674 and 0.674, respectively.
Using the z-score formula, we can solve for the corresponding values of x:
For the lower quartile:
-0.674 = (x - 64) / 3
x - 64 = -0.674 * 3
x - 64 = -2.022
x ≈ 61.978
For the upper quartile:
0.674 = (x - 64) / 3
x - 64 = 0.674 * 3
x - 64 = 2.022
x ≈ 66.022
Therefore, the middle 75% of women's heights fall between approximately 61.978 inches and 66.022 inches.
c) To find the height for the tallest 5% of women, we need to determine the z-score corresponding to the 95th percentile.
The z-score for the 95th percentile is approximately 1.645.
Using the z-score formula and solving for x:
1.645 = (x - 64) / 3
x - 64 = 1.645 * 3
x - 64 = 4.935
x ≈ 68.935
Therefore, the tallest 5% of women are taller than approximately 68.935 inches.