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The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 30 liters, and standard deviat...Question
                The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 30 liters, and standard deviation of 5.7 liters.
A) What is the probability that daily production is between 21.5 and 40.3 liters? Do not round until you get your your final answer.
Answer=
(Round your answer to 4 decimal places.)
Warning: Do not use the Z Normal Tables...they may not be accurate enough since WAMAP may look for more accuracy than comes from the table.
The answer is not 0.8960 or 0.8955
            
        A) What is the probability that daily production is between 21.5 and 40.3 liters? Do not round until you get your your final answer.
Answer=
(Round your answer to 4 decimal places.)
Warning: Do not use the Z Normal Tables...they may not be accurate enough since WAMAP may look for more accuracy than comes from the table.
The answer is not 0.8960 or 0.8955
Answers
                                                    There are no human answers yet.
                                            
                
                    Answered by
            GPT-4o mini
            
    To calculate the probability that the daily production of milk from cows is between 21.5 and 40.3 liters, we'll be using the properties of the normal distribution. 
1. **Identify the parameters**:
- Mean (\(\mu\)) = 30 liters
- Standard deviation (\(\sigma\)) = 5.7 liters
2. **Standardize the values**:
We need to convert the raw scores (21.5 and 40.3) into Z-scores using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
- For \(X = 21.5\):
\[
Z_{21.5} = \frac{(21.5 - 30)}{5.7} = \frac{-8.5}{5.7} \approx -1.49123
\]
- For \(X = 40.3\):
\[
Z_{40.3} = \frac{(40.3 - 30)}{5.7} = \frac{10.3}{5.7} \approx 1.80351
\]
3. **Use the cumulative distribution function (CDF)**:
We need to find the probabilities corresponding to these Z-scores. Instead of using the Z-table, it's recommended to use a calculator or software that provides the CDF for the standard normal distribution.
- Let \(P(Z < -1.49123)\) and \(P(Z < 1.80351)\) be the probabilities for our Z-scores:
- Calculating \(P(Z < -1.49123)\):
\[
P(Z < -1.49123) \approx 0.0689 \quad \text{(using software/calculator)}
\]
- Calculating \(P(Z < 1.80351)\):
\[
P(Z < 1.80351) \approx 0.9642 \quad \text{(using software/calculator)}
\]
4. **Calculate the probability of being between 21.5 and 40.3 liters**:
\[
P(21.5 < X < 40.3) = P(Z < 1.80351) - P(Z < -1.49123)
\]
\[
P(21.5 < X < 40.3) = 0.9642 - 0.0689 = 0.8953
\]
5. **Final answer**:
Rounding to 4 decimal places, the probability that daily production is between 21.5 and 40.3 liters is:
\[
\boxed{0.8953}
\]
    
1. **Identify the parameters**:
- Mean (\(\mu\)) = 30 liters
- Standard deviation (\(\sigma\)) = 5.7 liters
2. **Standardize the values**:
We need to convert the raw scores (21.5 and 40.3) into Z-scores using the formula:
\[
Z = \frac{(X - \mu)}{\sigma}
\]
- For \(X = 21.5\):
\[
Z_{21.5} = \frac{(21.5 - 30)}{5.7} = \frac{-8.5}{5.7} \approx -1.49123
\]
- For \(X = 40.3\):
\[
Z_{40.3} = \frac{(40.3 - 30)}{5.7} = \frac{10.3}{5.7} \approx 1.80351
\]
3. **Use the cumulative distribution function (CDF)**:
We need to find the probabilities corresponding to these Z-scores. Instead of using the Z-table, it's recommended to use a calculator or software that provides the CDF for the standard normal distribution.
- Let \(P(Z < -1.49123)\) and \(P(Z < 1.80351)\) be the probabilities for our Z-scores:
- Calculating \(P(Z < -1.49123)\):
\[
P(Z < -1.49123) \approx 0.0689 \quad \text{(using software/calculator)}
\]
- Calculating \(P(Z < 1.80351)\):
\[
P(Z < 1.80351) \approx 0.9642 \quad \text{(using software/calculator)}
\]
4. **Calculate the probability of being between 21.5 and 40.3 liters**:
\[
P(21.5 < X < 40.3) = P(Z < 1.80351) - P(Z < -1.49123)
\]
\[
P(21.5 < X < 40.3) = 0.9642 - 0.0689 = 0.8953
\]
5. **Final answer**:
Rounding to 4 decimal places, the probability that daily production is between 21.5 and 40.3 liters is:
\[
\boxed{0.8953}
\]
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