Option Seminar 2
5.19) According to data from the U.S. Energy information Administration, the 60.0 million U.S. households with personal computers were distributed as shown here with regard to geographic location and access to the Internet. The entries represent millions of households. What is the probability that a randomly selected computer household would be in the category described by (South or Midwest or Yes)? In the category (West and No)?
Access Region
To Internet? Northeast Midwest South West Total
Yes 9.7 11.8 16.9 12.2 50.6
No 1.2 2.3 3.8 2.1 9.6
Total 10.9 14.1 20.7 14.3 60.0
We don ot have the data that is "shown here."
However, for "West and No," find that cell and divide by the grand total.
For the "South or Midwest or Yes," find each of those totals and divide each by the grand total. For "either-or" probabilities, you have to add these individual probabilities.
I hope this helps. Thanks for asking.
To find the probability of a randomly selected computer household being in the category described by (South or Midwest or Yes), we need to sum the entries for South, Midwest, and Yes, and then divide it by the total number of computer households.
South: 16.9 million households
Midwest: 11.8 million households
Yes: 50.6 million households
Total: 60.0 million households
Probability = (16.9 + 11.8 + 50.6) / 60.0 = 79.3 / 60.0
So, the probability that a randomly selected computer household would be in the category described by (South or Midwest or Yes) is 1.32 (rounded to two decimal places).
To find the probability of a randomly selected computer household being in the category described by (West and No), we need to consider the intersection of West and No.
West and No: 2.1 million households
Probability = 2.1 / 60.0
So, the probability that a randomly selected computer household would be in the category described by (West and No) is 0.035 (rounded to three decimal places).
To find the probability that a randomly selected computer household would be in the category described by (South or Midwest or Yes), we need to add up the number of households in those categories and divide by the total number of households.
According to the table, there are 16.9 million households in the South, 11.8 million households in the Midwest, and 50.6 million households with access to the Internet. To avoid double counting, we need to subtract the number of households that are both in the South and the Midwest.
So, the number of households in (South or Midwest or Yes) would be:
(16.9 + 11.8 + 50.6) - (11.8) = 67.5 million households
The total number of households is given as 60.0 million.
Therefore, the probability that a randomly selected computer household would be in the category described by (South or Midwest or Yes) is:
67.5 million households / 60.0 million households = 1.125
Since probability is a value between 0 and 1, we can conclude that there is a 112.5% chance, which is not possible. It is likely that there is an error in the data provided or in the calculations. You may want to double-check the values or consult the original source of the data.
To find the probability that a randomly selected computer household would be in the category described by (West and No), we can refer to the table.
According to the table, there are 2.1 million households in the West and with no access to the Internet.
The total number of households is given as 60.0 million.
Therefore, the probability that a randomly selected computer household would be in the category described by (West and No) is:
2.1 million households / 60.0 million households = 0.035
So, the probability is 0.035 or 3.5%.