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.
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Identify the parameters:
- Mean (\(\mu\)) = 30 liters
- Standard deviation (\(\sigma\)) = 5.7 liters
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Standardize the values: We need to convert the raw scores (21.5 and 40.3) into Z-scores using the formula: \[ Z = \frac{(X - \mu)}{\sigma} \]
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For \(X = 21.5\): \[ Z_{21.5} = \frac{(21.5 - 30)}{5.7} = \frac{-8.5}{5.7} \approx -1.49123 \]
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For \(X = 40.3\): \[ Z_{40.3} = \frac{(40.3 - 30)}{5.7} = \frac{10.3}{5.7} \approx 1.80351 \]
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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.
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Let \(P(Z < -1.49123)\) and \(P(Z < 1.80351)\) be the probabilities for our Z-scores:
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Calculating \(P(Z < -1.49123)\): \[ P(Z < -1.49123) \approx 0.0689 \quad \text{(using software/calculator)} \]
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Calculating \(P(Z < 1.80351)\): \[ P(Z < 1.80351) \approx 0.9642 \quad \text{(using software/calculator)} \]
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Calculate the probability of being between 21.5 and 40.3 liters: \[ P(21.5 < X < 40.3) = P(Z < 1.80351) - P(Z < -1.49123) \] \[ P(21.5 < X < 40.3) = 0.9642 - 0.0689 = 0.8953 \]
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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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