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%

1 answer

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%