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.
Z = (x - μ)/SD
You have values for μ and SD, but need to determine Z to find x. Look in the back of your stat text for a table labeled something like "areas under the normal distribution" with the proportions (.10, ±.375 and .05 respectively) to find the corresponding Z scores to solve for x in the above equation.
I hope this helps. Thanks for asking.
Sure, I'd be happy to help!
To answer these questions, we need to use the concept of z-scores and the standard normal distribution. A z-score measures how many standard deviations an observation is from the mean. It allows us to compare values in a normal distribution.
a) The shortest 10% of women refers to the lower tail of the distribution. To find the height corresponding to the shortest 10%, we need to find the z-score that corresponds to that percentile.
We can use the z-score formula: z = (x - μ) / σ
Where:
- z is the z-score
- x is the observed value
- μ is the mean
- σ is the standard deviation
To find the z-score corresponding to the 10th percentile, we can use a standard z-table or a statistical calculator. The z-score corresponding to the 10th percentile is approximately -1.28.
Now we can solve for x:
-1.28 = (x - 64) / 3
Rearranging the equation:
-3.84 = x - 64
x = 64 - 3.84
x ≈ 60.16
Therefore, the shortest 10% of women are shorter than approximately 60.16 inches in height.
b) The middle 75% of all women refers to the interquartile range (IQR), which spans from the 25th percentile to the 75th percentile.
To find the heights within the middle 75%, we need to calculate the z-score for the 25th and 75th percentiles.
The z-score corresponding to the 25th percentile is approximately -0.67, and the z-score corresponding to the 75th percentile is approximately 0.67.
Now we can solve for x:
For the lower height:
-0.67 = (x - 64) / 3
x - 64 = -0.67 * 3
x - 64 ≈ -2.01
x ≈ 61.99
For the upper height:
0.67 = (x - 64) / 3
x - 64 = 0.67 * 3
x - 64 ≈ 2.01
x ≈ 66.01
Therefore, the middle 75% of women are between approximately 61.99 inches and 66.01 inches in height.
c) The tallest 5% of women refers to the upper tail of the distribution. To find the height corresponding to the tallest 5%, we need to find the z-score that corresponds to that percentile.
Using the z-table or a statistical calculator, we find that the z-score corresponding to the 95th percentile is approximately 1.645.
Now we can solve for x:
1.645 = (x - 64) / 3
Rearranging the equation:
4.935 = x - 64
x = 64 + 4.935
x ≈ 68.94
Therefore, the tallest 5% of women are taller than approximately 68.94 inches in height.
I hope this helps! Let me know if you have any further questions.