Weights of the vegetables in a field are normally distributed. From a sample Carl Cornfield determines the mean weight of a box of vegetables to be 180 oz. with a standard deviation of 8 oz. He wonders what percent of the vegetable boxes he has grouped for sale will have a weight between 169 oz. and 191 oz. Carl decides to answer the following questions about the population of vegetables from these sample statistics.
Carl calculates the z-score corresponding to the weight 169 oz. (to the nearest tenth). -
Using the table, Carl sees the percentage associated with this z-score is %.
Carl calculates the z-score corresponding to the weight 191 oz. (to the nearest tenth).
Using the table below, Carl sees the percentage associated with this z-score is %
Adding these together, Carl sees the percentage between 169 oz. and 191 oz. is %.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z scores. Multiply probability by 100 to get percentage.
To calculate the z-score corresponding to a specific weight in a normally distributed population, you can use the formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the desired weight
- μ is the mean weight of the population
- σ is the standard deviation of the population
Let's calculate the z-score for the weight of 169 oz. using the given values:
μ = 180 oz. (mean weight)
σ = 8 oz. (standard deviation)
x = 169 oz.
Substituting these values into the formula:
z = (169 - 180) / 8
z = -11 / 8
z ≈ -1.4
Now, let's find the percentage associated with this z-score using a z-table or standard normal distribution table. The z-table provides the percentage of data below a given z-score.
Looking up the z-score of -1.4 in the table, we find that the corresponding percentage is approximately 0.0808. So, Carl sees that the percentage associated with the z-score of -1.4 to be approximately 0.0808 or 8.08%.
For the weight of 191 oz., we follow the same steps:
μ = 180 oz. (mean weight)
σ = 8 oz. (standard deviation)
x = 191 oz.
z = (191 - 180) / 8
z = 11 / 8
z ≈ 1.4
Using the z-table, we find that the percentage associated with the z-score of 1.4 is also approximately 0.0808 or 8.08%.
To calculate the percentage between 169 oz. and 191 oz., we add the percentages associated with the z-scores.
0.0808 + 0.0808 = 0.1616
Therefore, Carl sees that the percentage between 169 oz. and 191 oz. is approximately 0.1616 or 16.16%.