Assuming that the heights of college women are normally distributed with mean 66 inches and standard deviation 2 inches, what percentage of women are shorter than 72 inches?
Try z-scores:
z = (x - mean)/sd
x = 72
mean = 66
sd = 2
Calculate the z-score, then check a z-table for your probability (remember the question is asking "what percentage of women are shorter than 72 inches" when looking at the table).
To calculate the percentage of women shorter than 72 inches, we need to find the area under the normal distribution curve to the left of 72 inches.
Step 1: Calculate the z-score.
The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the given value, μ is the mean, and σ is the standard deviation.
In this case:
x = 72 inches
μ = 66 inches
σ = 2 inches
z = (72 - 66) / 2
z = 3
Step 2: Find the area under the standard normal distribution curve to the left of the z-score.
To find the area under the curve, we can use a z-table or a calculator with a built-in function.
Using a standard normal distribution table, we find that the area to the left of z = 3 is approximately 0.9987.
Step 3: Convert the z-score back to the original units.
Since the area calculated in Step 2 represents the percentage of the distribution, we can convert it back to a percentage.
Percent = Area * 100
Percent = 0.9987 * 100
Percent ≈ 99.87%
Therefore, approximately 99.87% of women will be shorter than 72 inches.
To find the percentage of women who are shorter than 72 inches, we need to calculate the area under the normal distribution curve up to the value of 72 inches. This can be done by finding the z-score corresponding to 72 inches and then looking up the probability from the standard normal distribution table.
First, we need to calculate the z-score using the formula:
z = (x - μ) / σ
Where:
x = 72 (the value we want to find the percentage for)
μ = 66 (mean)
σ = 2 (standard deviation)
z = (72 - 66) / 2
z = 3
Now, we need to find the probability associated with the z-score of 3 from the standard normal distribution table.
Looking up the z-score of 3, we find that the value is approximately 0.9987. This represents the area under the curve up to the z-score of 3.
However, since we are interested in the percentage of women shorter than 72 inches, we need to find the area to the left of the z-score. This can be obtained by subtracting the value from 1.
Probability = 1 - 0.9987
Probability ≈ 0.0013
So, approximately 0.13% of college women are shorter than 72 inches.