To find the probability that the daily production of the cows is between 42 and 68.7 liters, we will use the properties of the normal distribution.
Given:
- Mean (\(\mu\)) = 39 liters
- Standard Deviation (\(\sigma\)) = 11.6 liters
Step 1: Calculate the Z-scores for the values 42 and 68.7.
The Z-score formula is:
\[ Z = \frac{(X - \mu)}{\sigma} \]
For \(X = 42\):
\[ Z_{42} = \frac{(42 - 39)}{11.6} = \frac{3}{11.6} \approx 0.2586 \]
For \(X = 68.7\):
\[ Z_{68.7} = \frac{(68.7 - 39)}{11.6} = \frac{29.7}{11.6} \approx 2.5593 \]
Step 2: Use the Z-scores to find the probabilities.
You can use a calculator or statistical software to find the probabilities corresponding to these Z-scores.
-
Let \(P(Z \leq Z_{42})\) be the cumulative probability for \(Z_{42}\): \[ P(Z \leq 0.2586) \approx 0.6016 \]
-
Let \(P(Z \leq Z_{68.7})\) be the cumulative probability for \(Z_{68.7}\): \[ P(Z \leq 2.5593) \approx 0.9947 \]
Step 3: Calculate the probability that production is between 42 and 68.7 liters.
\[ P(42 < X < 68.7) = P(Z \leq 2.5593) - P(Z \leq 0.2586) \]
\[ P(42 < X < 68.7) \approx 0.9947 - 0.6016 = 0.3931 \]
Step 4: Round the answer to four decimal places.
The probability that daily production is between 42 and 68.7 liters is:
\[ \text{Answer} = 0.3931 \]
1 answer