Question

2Each month, an American household generates an average of 28 pounds of newspaper for
garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at
random, find the probability of its generating between 27 and 31 pounds per month. Assume
the variable is approximately normally distributed.
A) 62.47%
B) 67.62%
C) 75.25%
D) 80. 56%

Answer 1

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:

1. For \( X = 27 \):
\[
z_{27} = \frac{27 - 28}{2} = \frac{-1}{2} = -0.5
\]

2. 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%**