Final answer:
To determine the percentages of average Americans who spend a certain amount of time shopping online, we need to use the properties of a normal distribution. For each scenario, we can calculate the z-scores and then use a standard normal distribution table or a calculator to find the corresponding probabilities. The percentage of average Americans who spend less than 15 hours per month shopping online is approximately 53.07
Step-by-step explanation:
To determine the percentages of average Americans who spend a certain amount of time shopping online, we need to use the properties of a normal distribution. For each scenario, we can calculate the z-scores and then use a standard normal distribution table or a calculator to find the corresponding probabilities.
(a) To determine the percentage of average Americans who spend less than 15 hours per month shopping online, we can calculate the z-score:
z = (15 - 14) / 15 = 1/15 = 0.067
Using a standard normal distribution table or calculator, we find that the probability corresponding to a z-score of 0.067 is approximately 0.5307. This means that approximately 53.07% of average Americans spend less than 15 hours per month shopping online.
(b) To determine the percentage of average Americans who spend at least 8.5 hours per month shopping online, we can calculate the z-score:
z = (8.5 - 14) / 15 = -5.5/15 = -0.367
Using a standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -0.367 is approximately 0.3573. To find the percentage of average Americans who spend at least 8.5 hours, we subtract this probability from 1: 1 - 0.3573 = 0.6427. This means that approximately 64.27% of average Americans spend at least 8.5 hours per month shopping online.
(c) To determine the percentage of average Americans who spend either less than 12 hours or more than 15.5 hours shopping online, we need to calculate the z-scores for both values:
z1 = (12 - 14) / 15 = -0.133
z2 = (15.5 - 14) / 15 = 1.5/15 = 0.1
Using a standard normal distribution table or calculator, we find that the probability corresponding to a z-score of -0.133 is approximately 0.45, and the probability corresponding to a z-score of 0.1 is approximately 0.5400. To find the percentage of average Americans who spend either less than 12 hours or more than 15.5 hours, we add these probabilities: 0.45 + 0.54 = 0.99. This means that approximately 99% of average Americans spend either less than 12 hours or more than 15.5 hours shopping online.