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 more than 196 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 196 oz. ____
Using the table (column .00), Carl sees the area associated with this z-score is 0. _____
Carl rounds this value to the nearest thousandth or 0. ______
Now, Carl subtracts 0.50 - 0. = %.
To answer these questions, Carl needs to perform the following steps:
1. Calculate the z-score corresponding to the weight 196 oz:
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = 196 oz (Weight of the box)
μ = Mean weight of the boxes (180 oz)
σ = Standard deviation (8 oz)
Plugging in the values,
z = (196 - 180) / 8
z = 2
So in this case, the z-score is 2.
2. Using the table (column .00), Carl needs to find the area associated with this z-score:
Carl should look up the z-score of 2 in the z-table. The columns of the z-table indicate the first digit of the z-score, while the rows indicate the second digit. As Carl needs to find the area for a z-score of 2, he should check the row for 2.0 and the column for 0.00.
After referring to the table, Carl finds that the area associated with a z-score of 2.0 (rounded to two decimal places) is 0.9772.
3. Carl needs to round this value to the nearest thousandth or 0:
The value from step 2, 0.9772, should be rounded to the nearest thousandth. This means Carl needs to keep three decimal places. Therefore, the rounded value is 0.977.
4. Now, Carl subtracts 0.50 - 0 to find the percentage:
Since Carl wants to find the percentage of vegetable boxes with a weight more than 196 oz (which is represented by the right tail of the distribution), he subtracts the area to the left of the z-score (0.977) from 0.5 (which represents 50% of the distribution).
0.50 - 0.977 = -0.477
However, since percentages are always positive, Carl takes the absolute value of -0.477.
| -0.477 | = 0.477
So, the percentage of vegetable boxes with a weight more than 196 oz is 0.477% (rounded to three decimal places).
To calculate the z-score corresponding to the weight 196 oz, Carl can use the formula:
z = (x - μ) / σ
Where:
x is the value (196 oz),
μ is the mean (180 oz), and
σ is the standard deviation (8 oz).
Substituting the values into the formula:
z = (196 - 180) / 8
z = 16 / 8
z = 2
The z-score corresponding to the weight 196 oz is 2.
Using the z-table, Carl can find the area associated with the z-score of 2 in the column .00. The closest value in the table to 2 in the .00 column is 0.9772.
Rounding this value to the nearest thousandth or 0 gives us 0.977.
Now, Carl subtracts 0.50 - 0 which gives us 0.50.
Therefore, Carl calculates the percentage of vegetable boxes that weigh more than 196 oz to be 50%.
I suggest you follow the instructions in the steps outlined for you.
Looks very straight-forward to me
look in your notes or textbook to see how you can find the z-score.
As for tables, they are really not used anymore, there are enough on-line charts or applets to find the above values
e.g. http://davidmlane.com/normal.html
is one of the best.