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.