I have no idea how to calculate z scores
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 a
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 a
Adding these together, Carl sees the percentage between 169 oz. and 191 oz. is a
Z = (score-mean)/SD
Use your table or find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities of Z scores to the mean. Add.
To calculate the z-scores and find the corresponding percentages, you can use the formula:
z = (x - μ) / σ
where:
- z is the z-score
- x is the value you are interested in
- μ is the mean
- σ is the standard deviation
Let's calculate the z-scores and find the percentages:
1. Calculate the z-score for the weight 169 oz.:
z = (169 - 180) / 8
z = -11 / 8
z ≈ -1.4 (to the nearest tenth)
2. Look up the percentage associated with -1.4 in the z-table. From the table, you will find that the percentage is approximately 0.0808 or 8.08%.
3. Calculate the z-score for the weight 191 oz.:
z = (191 - 180) / 8
z = 11 / 8
z ≈ 1.4 (to the nearest tenth)
4. Look up the percentage associated with 1.4 in the z-table. From the table, you will find that the percentage is approximately 0.9192 or 91.92%.
5. To find the percentage between 169 oz. and 191 oz., add the percentages you obtained in steps 2 and 4:
Percentage = 8.08% + 91.92%
Percentage = 100%
Therefore, Carl Cornfield can expect that 100% of the vegetable boxes he has grouped for sale will have a weight between 169 oz. and 191 oz.
To calculate z-scores, you can use the formula:
z = (x - μ) / σ
Where:
x represents the observed value,
μ represents the mean of the distribution, and
σ represents the standard deviation of the distribution.
In this case, Carl has determined the mean weight of a box of vegetables to be 180 oz, with a standard deviation of 8 oz.
To calculate the first z-score for the weight of 169 oz:
z1 = (169 - 180) / 8
Calculating this, you would get a z1 value.
To find the percentage associated with this z-score, Carl can refer to a standard normal distribution table. The table provides the cumulative probabilities for different z-scores.
Carl then finds the z1 value in the table and determines the percentage associated with it. This will give him the percentage of boxes of vegetables with a weight less than 169 oz.
To calculate the second z-score for the weight of 191 oz:
z2 = (191 - 180) / 8
Calculating this, you would get a z2 value.
Carl again refers to the standard normal distribution table and finds the z2 value, determining the percentage associated with it. This will give him the percentage of boxes of vegetables with a weight less than 191 oz.
To find the percentage between 169 oz and 191 oz, Carl adds the percentages associated with z1 and z2 together.
Explanation:
The z-score represents the number of standard deviations an observation is away from the mean. By calculating the z-score, you can standardize the measurement and compare it to a standard normal distribution.
The standard normal distribution is a specific case of the normal distribution with a mean of 0 and a standard deviation of 1. The standard normal distribution table provides the cumulative probabilities (percentages) for different z-scores. By looking up the z-score in the table, you can find the percentage associated with it.