It is well known that the heights of individual American men are normally distributed with mean 70 inches and standard deviation 2.8 inches.
The Central Limit Theorem states that if n men are randomly chosen, then their average height will also be normally distributed with mean 70 inches (so the mean is unchanged), but the standard deviation will not be 2.8 inches--it will be 2.8 divided by the square root of n (the standard deviation is smaller for groups than for individuals). This means that there is less variation among group averages than there is between individuals--I hope that makes intuitive sense.
a) How likely is it that a randomly chosen man would be more than six feet tall (i.e., what percentage of men are over six feet)?
b) How likely is it that a randomly chosen group of ten men would have an average height exceeding six feet?
Remember: The answer is never sufficient:
What's of interest is always the explanation!
a) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
b) Standard Error of the mean, SEm = SD/√(n-1)
Z = (score-mean)/SEm
Use same table.
To find the probability of a randomly chosen man being more than six feet tall, we can use the normal distribution and convert the height to a z-score. A z-score represents the number of standard deviations a data point is from the mean.
First, we convert six feet (72 inches) to a z-score. The formula to calculate the z-score is:
z = (x - μ) / σ
where:
x = the data point (72 inches in this case)
μ = the mean (70 inches)
σ = the standard deviation (2.8 inches)
Plugging in the values, we get:
z = (72 - 70) / 2.8
z ≈ 0.714
Next, we need to find the probability associated with this z-score. We can look up the z-score in a standard normal distribution table or use a calculator or software that provides the cumulative distribution function (CDF) for the standard normal distribution.
From the standard normal distribution table or software, we find that the probability corresponding to a z-score of 0.714 is approximately 0.7623. This means that about 76.23% of men have a height less than or equal to six feet. To find the percentage of men over six feet, we subtract this value from 1 (or multiply by 100%):
Percentage of men over six feet = 1 - 0.7623 ≈ 0.2377 or 23.77%
Therefore, the probability of randomly choosing a man who is more than six feet tall is approximately 23.77%.
Moving on to the second question, we want to find the likelihood of a randomly chosen group of ten men having an average height exceeding six feet.
To solve this, we can use the Central Limit Theorem as mentioned earlier. According to the theorem, the mean of the group will still be 70 inches, but the standard deviation will be the standard deviation of individual heights divided by the square root of the group size (10).
Standard deviation for the group = σ / sqrt(n)
= 2.8 / sqrt(10)
≈ 0.883
Now, we can convert the six feet (72 inches) average height to a z-score using the same formula:
z = (x - μ) / σ
= (72 - 70) / 0.883
≈ 2.26
Next, we find the probability corresponding to this z-score. From the standard normal distribution table or using software, we determine that the probability of a z-score of 2.26 or higher is approximately 0.0113.
Therefore, the likelihood of a randomly chosen group of ten men having an average height exceeding six feet is approximately 0.0113 or 1.13%.