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 a4 %.
To solve this problem, Carl needs to use the z-score formula and the standard normal distribution table.
The z-score formula is given by:
z = (x - μ) / σ
where:
- x is the given value (weight in this case),
- μ is the mean weight of the sample (180 oz), and
- σ is the standard deviation of the sample (8 oz).
1. Calculating the z-score for 169 oz:
z1 = (169 - 180) / 8
z1 ≈ -1.375 (rounded to the nearest tenth)
2. Using the standard normal distribution table, Carl needs to find the percentage associated with z1 = -1.375. This represents the percentage of the population that falls below 169 oz. From the table, he finds the corresponding percentage to be 0.0847 (or 8.47% as a percentage).
3. Calculating the z-score for 191 oz:
z2 = (191 - 180) / 8
z2 ≈ 1.375 (rounded to the nearest tenth)
4. Similar to the previous step, Carl needs to find the percentage associated with z2 = 1.375. This represents the percentage of the population that falls below 191 oz. From the table, he finds the corresponding percentage to be 0.9153 (or 91.53% as a percentage).
5. To find the percentage between 169 oz and 191 oz, he needs to subtract the percentage associated with the lower z-score (8.47%) from the percentage associated with the higher z-score (91.53%):
Percentage between 169 oz and 191 oz = 91.53% - 8.47%
Percentage between 169 oz and 191 oz ≈ 83.06% (rounded to the nearest hundredth)
Therefore, Carl concludes that approximately 83.06% of the vegetable boxes he has grouped for sale will have a weight between 169 oz and 191 oz.