Use the Empirical Rule to answer each question.
During 1 week an overnight delivery company found that the weights of its parcels were normally distributed, with a mean of 16 ounces and a standard deviation of 4 ounces.
(a) What percent of the parcels weighed between 8 ounces and 20 ounces? (Round your answer to one decimal place.)
(b) What percent of the parcels weighed more than 28 ounces? (Round your answer to two decimal places.)
A. 13.5% + 34% + 34%
= 81.5%
B. (100% - 99.7%) / 2
=0.15%
To use the Empirical Rule, we first need to convert the given weights to z-scores. The z-score formula is:
z = (x - μ) / σ
where:
- z is the z-score
- x is the observed value
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
(a) To find the percent of parcels weighing between 8 ounces and 20 ounces, we need to find the area under the normal distribution curve between the corresponding z-scores.
For 8 ounces:
z1 = (8 - 16) / 4 = -2
For 20 ounces:
z2 = (20 - 16) / 4 = 1
The area between z1 and z2 represents the percentage of parcels weighing between 8 and 20 ounces. Using the Empirical Rule, we know that approximately 68% of observations fall within one standard deviation of the mean. Therefore, the percentage between z1 and z2 is approximately 68%.
(b) To find the percent of parcels weighing more than 28 ounces, we calculate the z-score for 28 ounces:
z = (28 - 16) / 4 = 3
Since we are interested in the area to the right of z, we find the complement of the area to the left. Using the Empirical Rule, we know that approximately 95% of observations fall within two standard deviations of the mean. Therefore, the percentage of parcels weighing more than 28 ounces is approximately 1 - 95% = 5%.
Therefore:
(a) The percent of parcels weighing between 8 ounces and 20 ounces is approximately 68%.
(b) The percent of parcels weighing more than 28 ounces is approximately 5%.
To answer both of these questions using the Empirical Rule, we need to understand the concept of standard deviations and how they relate to the normal distribution.
The Empirical Rule provides an approximation for the spread of data in a normal distribution. It states that:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
Now let's apply this rule to the given scenario.
(a) To find the percent of parcels that weigh between 8 ounces and 20 ounces, we need to determine how many standard deviations away these weights are from the mean.
Step 1: Find the z-scores for the weights of 8 ounces and 20 ounces.
The z-score formula is:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For 8 ounces:
z = (8 - 16) / 4 = -2
For 20 ounces:
z = (20 - 16) / 4 = 1
Step 2: Determine the area under the normal curve between these two z-scores.
Using a standard normal distribution table or a calculator, we can find that the area between -2 and 1 is approximately 0.8186.
Step 3: Convert the area to a percentage.
Since the Empirical Rule states that approximately 68% of the data falls within one standard deviation of the mean, we know that 68% - 50% = 18% falls between one and two standard deviations.
Therefore, the percentage of parcels that weigh between 8 ounces and 20 ounces is approximately 50% + 18% + (0.8186 * 68%) = 68.186%, rounded to one decimal place.
(b) To find the percent of parcels that weigh more than 28 ounces, we need to determine how many standard deviations away this weight is from the mean.
Step 1: Find the z-score for the weight of 28 ounces.
z = (28 - 16) / 4 = 3
Step 2: Determine the area under the normal curve to the right of this z-score.
Using a standard normal distribution table or a calculator, we can find that the area to the right of 3 is approximately 0.0013.
Step 3: Convert the area to a percentage.
Since the Empirical Rule states that approximately 99.7% of the data falls within three standard deviations of the mean, we know that 99.7% - 95% = 4.7% falls between two and three standard deviations.
Therefore, the percentage of parcels that weigh more than 28 ounces is approximately 0.0013 + 4.7% = 4.7013%, rounded to two decimal places.
8 = µ-2σ
20 = µ+1σ
So what does your Z table say?