To find the probability of a household generating between 27 and 31 pounds of newspaper per month, we will use the properties of the normal distribution. Given:
- Mean (μ) = 28 pounds
- Standard deviation (σ) = 2 pounds
We want to find \( P(27 \leq X \leq 31) \).
First, we will standardize the values using the z-score formula:
\[ z = \frac{X - \mu}{\sigma} \]
Calculating the z-scores for 27 pounds and 31 pounds:
-
For \( X = 27 \): \[ z_{27} = \frac{27 - 28}{2} = \frac{-1}{2} = -0.5 \]
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For \( X = 31 \): \[ z_{31} = \frac{31 - 28}{2} = \frac{3}{2} = 1.5 \]
Next, we will look up these z-scores in the standard normal distribution table (or use a calculator to find the probabilities):
- For \( z = -0.5 \), the probability \( P(Z < -0.5) \) is approximately 0.3085.
- For \( z = 1.5 \), the probability \( P(Z < 1.5) \) is approximately 0.9332.
Now we can find the probability of \( P(27 \leq X \leq 31) \) by calculating:
\[ P(27 \leq X \leq 31) = P(Z < 1.5) - P(Z < -0.5) \] \[ P(27 \leq X \leq 31) = 0.9332 - 0.3085 = 0.6247 \]
To express this as a percentage, we multiply by 100:
\[ P(27 \leq X \leq 31) \approx 62.47% \]
Therefore, the answer is:
A) 62.47%