A manufacturer of cereal finds that the masses of cereal in a company's 200-g package are normally distributed with a mean of 200g and a standard deviation of 16.3g
a) What proportion of these boxes contain between 183.7g and 216.3g of cereal?
b) What is the probability that a box selected at random contains more than 216.3g of cereal?
a manufacturer of cereal finds that the masses of cereal in the company's 200-g packages are normally distributed with a mean of 200 g and a standard deviation of 16.3 g
To solve this problem, we will use the properties of the normal distribution.
a) To find the proportion of boxes containing between 183.7g and 216.3g of cereal, we need to calculate the z-scores for the lower and upper limits and then find the area between them.
First, we calculate the z-score for the lower limit (183.7g):
z₁ = (x - μ) / σ
= (183.7 - 200) / 16.3
≈ -1.0
Next, we calculate the z-score for the upper limit (216.3g):
z₂ = (x - μ) / σ
= (216.3 - 200) / 16.3
≈ 1.0
Using a standard normal distribution table or a calculator, we find the area to the left of z = -1.0 is approximately 0.1587 and the area to the left of z = 1.0 is also approximately 0.8413.
To find the proportion of boxes containing cereal between 183.7g and 216.3g, we subtract the area to the left of z = -1.0 from the area to the left of z = 1.0:
Proportion = Area(z < 1.0) - Area(z < -1.0)
≈ 0.8413 - 0.1587
≈ 0.6826
Therefore, approximately 68.26% of the boxes contain between 183.7g and 216.3g of cereal.
b) To find the probability that a randomly selected box contains more than 216.3g of cereal, we need to find the area to the right of z = 1.0.
Using a standard normal distribution table or a calculator, we find the area to the left of z = 1.0 is approximately 0.8413.
To find the probability of a box containing more than 216.3g of cereal, we subtract the area to the left of z = 1.0 from 1 (the total area under the curve):
Probability = 1 - Area(z < 1.0)
= 1 - 0.8413
≈ 0.1587
Therefore, there is approximately a 15.87% chance that a randomly selected box contains more than 216.3g of cereal.
To solve these questions, we will use the standard normal distribution, also known as the Z-distribution, which is a mathematical model used for analyzing normally distributed data.
The standard normal distribution has a mean of 0 and a standard deviation of 1. To work with different means and standard deviations, we need to standardize the values using the formula:
Z = (X - μ) / σ
Where:
- Z is the standardized value (also called the Z-score)
- X is the value we want to standardize
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
Now, let's solve each part of the question:
a) What proportion of these boxes contain between 183.7g and 216.3g of cereal?
First, we need to standardize the lower value, 183.7g:
Z1 = (183.7 - 200) / 16.3
Similarly, we standardize the upper value, 216.3g:
Z2 = (216.3 - 200) / 16.3
Now, we need to find the area under the standard normal distribution curve between Z1 and Z2. We can do this by using a Z-table or a calculator with a built-in Z-table function. Let's assume we have access to a Z-table:
Look up the Z-values in the table, and find the corresponding probabilities for each Z-value. Subtract the probability of Z1 from the probability of Z2 to find the proportion of boxes that contain between 183.7g and 216.3g of cereal.
b) What is the probability that a box selected at random contains more than 216.3g of cereal?
In this case, we only need to find the probability of the upper value, 216.3g. We standardize it to get the Z-score using the same formula:
Z = (216.3 - 200) / 16.3
Now, we find the probability corresponding to this Z-score from the Z-table. However, instead of calculating the area between two values, we need to find the area to the right of the Z-score (since we are interested in values greater than 216.3g).
This will give us the probability that a randomly selected box contains more than 216.3g of cereal.
Note: If you have access to statistical software like Excel or programming languages like Python, you can also calculate the probabilities using the cumulative distribution function (CDF) of the standard normal distribution.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.