The distribution of heights of adult American men is approximately normal with mean 69 inches and standard deviation 2.5 inches. Use the 68-95-99.7 rule to answer the following questions:
(d) A height of 71.5 inches corresponds to what percentile of adult male American heights?
==> I already know the answer is the 84th percentile, but how do you get this answer?
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.
So since its asking you to use the empirical rule. if you notice 71.5 is 2.5 inches higher than 69. So 71.5 is exactly 1 standard deviation above the mean which we know is 68%. 50% + (68/2) = 84%
To find the percentile corresponding to a height of 71.5 inches, we need to calculate the z-score first. The z-score is a measure of how many standard deviations away from the mean a particular value is.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = the value
μ = mean
σ = standard deviation
In this case:
x = 71.5 inches
μ = 69 inches
σ = 2.5 inches
Substituting the values into the formula, we have:
z = (71.5 - 69) / 2.5
z = 2.5 / 2.5
z = 1
The z-score of 1 indicates that the height of 71.5 inches is one standard deviation above the mean.
Now, using the 68-95-99.7 rule, we know that approximately 68% of the data falls within one standard deviation of the mean. Since the normal distribution is symmetric, we can infer that 34% lies below the mean and 34% lies above the mean within one standard deviation.
Therefore, to calculate the percentile of the height of 71.5 inches, we need to find the area under the normal curve to the left of the z-score of 1. This area represents the percentage of data below 71.5 inches.
Consulting a standard normal distribution table or using a statistical calculator, we find that the area to the left of the z-score of 1 is approximately 0.8413. This means that approximately 84.13% of the data falls below 71.5 inches.
Therefore, a height of 71.5 inches corresponds to the 84th percentile of adult male American heights.
To find the percentile that corresponds to a specific height in a normal distribution, you can use the Z-score formula. The Z-score measures the number of standard deviations a particular value is away from the mean. Here's how you can calculate the percentile for a height of 71.5 inches:
1. Calculate the Z-score:
Z = (X - μ) / σ
where X is the given height (71.5 inches), μ is the mean (69 inches), and σ is the standard deviation (2.5 inches).
Z = (71.5 - 69) / 2.5
= 2.5 / 2.5
= 1
2. Look up the Z-score in the Z-score table or use a calculator to find the area under the normal curve to the left of the Z-score. In this case, the Z-score of 1 corresponds to a left-tailed area of 0.8413.
3. Convert the left-tailed area to a percentile by multiplying by 100:
Percentile = 0.8413 * 100
= 84.13 (rounded to the nearest whole number)
Therefore, a height of 71.5 inches corresponds to approximately the 84th percentile of adult male American heights according to the given normal distribution.