To solve these problems, we will use the properties of the normal distribution and technology (like Python, R, or a statistical calculator) to compute the probabilities.
We are given:
- Mean (\(\mu\)) = 33 liters
- Standard deviation (\(\sigma\)) = 4.9 liters
Part A: Probability that daily production is less than 21.8 liters
We need to calculate \( P(X < 21.8) \).
Using the formula for the z-score:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \(X = 21.8\)
- \(\mu = 33\)
- \(\sigma = 4.9\)
First, we find the z-score:
\[ z = \frac{21.8 - 33}{4.9} \approx \frac{-11.2}{4.9} \approx -2.2857 \]
Next, we use technology to find the probability \( P(Z < -2.2857) \). In Python, for instance, we can use:
from scipy.stats import norm
mean = 33
std_dev = 4.9
value = 21.8
# Calculate probability
probability_A = norm.cdf(value, mean, std_dev)
print(round(probability_A, 4))
The result from the above calculation gives us \( P(X < 21.8) \) approx equal to 0.0110.
Part B: Probability that daily production is more than 38.7 liters
Now, we need to calculate \( P(X > 38.7) \).
First, we calculate the z-score:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \(X = 38.7\)
Calculating the z-score:
\[ z = \frac{38.7 - 33}{4.9} \approx \frac{5.7}{4.9} \approx 1.1633 \]
Next, we find the probability:
Using the formula \( P(X > 38.7) = 1 - P(X < 38.7) \).
Using technology:
value_B = 38.7
probability_B = 1 - norm.cdf(value_B, mean, std_dev)
print(round(probability_B, 4))
The result from this calculation gives us \( P(X > 38.7) \) approx equal to 0.1224.
Final Answers:
A) The probability that daily production is less than 21.8 liters is approximately 0.0110.
B) The probability that daily production is more than 38.7 liters is approximately 0.1224.