Question

Each 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 generation between 27 and 31 pounds per month. Assume the variable is normally distributed. Express your answer as a percentage. (do not include % sign in your answer). Round your answer to the nearest hundredth.

Answer 1

To find the probability of a household generating between 27 and 31 pounds of newspaper per month, we can use the properties of the normal distribution.

1. **Identify the parameters of the normal distribution:**
- Mean (\(\mu\)): 28 pounds
- Standard deviation (\(\sigma\)): 2 pounds

2. **Convert the raw scores (27 and 31) to z-scores:**

The formula for the z-score is:
\[
z = \frac{(X - \mu)}{\sigma}
\]

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

For \(X = 31\):
\[
z_{31} = \frac{(31 - 28)}{2} = \frac{3}{2} = 1.5
\]

3. **Use the standard normal distribution table or calculator to find probabilities:**

Now we find the probabilities corresponding to \(z_{27} = -0.5\) and \(z_{31} = 1.5\).

- The probability for \(z = -0.5\):
\[
P(Z < -0.5) \approx 0.3085
\]

- The probability for \(z = 1.5\):
\[
P(Z < 1.5) \approx 0.9332
\]

4. **Calculate the probability between the two z-scores:**
\[
P(27 < X < 31) = P(Z < 1.5) - P(Z < -0.5)
\]
\[
P(27 < X < 31) = 0.9332 - 0.3085 = 0.6247
\]

5. **Convert the probability to a percentage:**
\[
0.6247 \times 100 \approx 62.47
\]

Finally, rounding to the nearest hundredth, the probability that a household generates between 27 and 31 pounds of newspaper per month is **62.47**.