The average U.S. yearly capita consumption of citrus fruit is 26.8 pounds. Suppose that the distribution of fruit amounts consumed is bell shaped with a standard deviation equal to 4.2 pounds.
a. What percentage of Americans would you expect to consume more than 31 pounds of citrus fruit per year?
b. What percentage of Americans would you expect to consume less than 18.4 pounds of citrus fruit per year?
you can play around with Z table stuff here:
http://davidmlane.com/hyperstat/z_table.html
To solve this problem, we will use the concept of standard deviation and the normal distribution. The normal distribution is a bell-shaped curve that is often used to model random variables like the consumption of citrus fruit.
a. To find the percentage of Americans who consume more than 31 pounds of citrus fruit per year, we will use the standard deviation. We need to calculate the z-score for 31 pounds using the formula:
z = (x - μ) / σ
where:
x = 31 pounds (the value we are interested in)
μ = 26.8 pounds (the mean consumption)
σ = 4.2 pounds (the standard deviation)
z = (31 - 26.8) / 4.2
z ≈ 1.00
Now, we can use a standard normal distribution table or a calculator to find the percentage associated with a z-score of 1.00. Looking up the z-score of 1.00 in the table, we find that approximately 0.8413 (or 84.13%) of Americans consume less than 31 pounds of citrus fruit per year.
Therefore, the percentage of Americans who consume more than 31 pounds of citrus fruit per year is approximately 100% - 84.13% = 15.87% (or 15.87%).
b. To find the percentage of Americans who consume less than 18.4 pounds of citrus fruit per year, we will again use the standard deviation and the z-score formula.
z = (x - μ) / σ
where:
x = 18.4 pounds (the value we are interested in)
μ = 26.8 pounds (the mean consumption)
σ = 4.2 pounds (the standard deviation)
z = (18.4 - 26.8) / 4.2
z ≈ -2.00
Using a standard normal distribution table or a calculator, we find that the percentage associated with a z-score of -2.00 is approximately 0.0228 (or 2.28%).
Therefore, the percentage of Americans who consume less than 18.4 pounds of citrus fruit per year is approximately 2.28%.
To determine the percentages requested, we will use z-scores and the standard normal distribution table. By converting the given values to z-scores, we can then find the corresponding areas under the normal curve which represent the percentages.
Let's calculate the z-scores first:
a. To find the z-score for consuming more than 31 pounds of citrus fruit per year, we use the formula:
z = (x - μ) / σ
where:
x = 31 pounds (value we want to convert)
μ = 26.8 pounds (mean)
σ = 4.2 pounds (standard deviation)
Plugging in these values, we get:
z = (31 - 26.8) / 4.2
z ≈ 1.0
b. To find the z-score for consuming less than 18.4 pounds of citrus fruit per year, we use the same formula:
z = (x - μ) / σ
where:
x = 18.4 pounds (value we want to convert)
μ = 26.8 pounds (mean)
σ = 4.2 pounds (standard deviation)
Plugging in these values, we get:
z = (18.4 - 26.8) / 4.2
z ≈ -2.0
Now that we have the z-scores, we can use the standard normal distribution table to find the corresponding areas.
a. To find the percentage of Americans who would consume more than 31 pounds of citrus fruit per year, we need to find the area to the right of the z-score of 1.0 in the standard normal distribution table. This area represents the percentage of the population that falls above that z-score.
Looking up the z-score of 1.0 in the table, we find that the area is approximately 0.8413. To convert this to a percentage, we multiply by 100:
0.8413 * 100 ≈ 84.13%
Therefore, we would expect approximately 84.13% of Americans to consume more than 31 pounds of citrus fruit per year.
b. To find the percentage of Americans who would consume less than 18.4 pounds of citrus fruit per year, we need to find the area to the left of the z-score of -2.0 in the standard normal distribution table. This area represents the percentage of the population that falls below that z-score.
Looking up the z-score of -2.0 in the table, we find that the area is approximately 0.0228. To convert this to a percentage, we multiply by 100:
0.0228 * 100 ≈ 2.28%
Therefore, we would expect approximately 2.28% of Americans to consume less than 18.4 pounds of citrus fruit per year.