Question

Each month, an American household generates an average of 28 pounds of newspaper for garbage and/or recycling. Assume this is approximately normally distributed and that the standard deviation is 2 pounds. If a household is selected at random, find the probability of it generating between 27–31 pounds per month.

Answer 1

68.26% probability that a randomly selected household will generate between 27 and 31 pounds of newspaper per month.

Step-by-step explanation:

The probability of a randomly selected household generating between 27 and 31 pounds of newspaper per month can be found using the normal distribution.

First, we need to standardize the random variable by subtracting the mean (28 pounds) and dividing by the standard deviation (2 pounds):

(27 - 28) / 2 = -0.5

(31 - 28) / 2 = 1.5

These standardized values correspond to the z-scores for the lower and upper bounds of the range. We can use a standard normal table or a calculator to find the probability that a random household will generate a value within this range.

The probability of a value falling between -0.5 and 1.5 on the standard normal distribution is approximately 0.6826, or 68.26%. This means that there is a 68.26% probability that a randomly selected household will generate between 27 and 31 pounds of newspaper per month.

It's important to note that this is an approximate value, as the normal distribution is a continuous distribution and the probability of a specific value occurring is technically zero. However, the normal distribution can be used to approximate the probability of a value falling within a given range, as long as the range is relatively small compared to the standard deviation.