Questions: A website reports that 50% of their users are from outside the United States and that 40% of their users log on to their website every day. Suppose that 15% of their users are United States users who log on every day. Complete parts a through e below. a) What percentage of the website's users are from the United States? The percentage of the website's users that are from the United States is 50%. b) What type of probability is the 15% mentioned above? - marginal probability - empirical probability - personal probability - joint probability c) Construct a contingency table showing all the joint and marginal probabilities. Assign user from the United States and users from outside the United States to the columns. Assign users who log on every day and users who do not log on every day to the rows. Log on every day Do not log on every day Total (Do not round.)

Solution Steps

The problem states that 50% of the website's users are from the United States.

The 15% mentioned refers to the probability that a user is from the United States and logs on every day. This is a joint probability.

To construct the contingency table, we need to use the given probabilities:

• 50% of users are from the United States.
• 50% of users are from outside the United States.
• 15% of users are from the United States and log on every day.
• 40% of users log on every day.

Let's denote:

• \( P(A) \) = Probability of being from the United States = 0.50
• \( P(B) \) = Probability of logging on every day = 0.40
• \( P(A \cap B) \) = Joint probability of being from the United States and logging on every day = 0.15

We can find the remaining probabilities:

• \( P(A \cap B^c) \) = Probability of being from the United States and not logging on every day = \( P(A) - P(A \cap B) = 0.50 - 0.15 = 0.35 \)
• \( P(A^c \cap B) \) = Probability of being from outside the United States and logging on every day = \( P(B) - P(A \cap B) = 0.40 - 0.15 = 0.25 \)
• \( P(A^c \cap B^c) \) = Probability of being from outside the United States and not logging on every day = \( 1 - P(A \cap B) - P(A \cap B^c) - P(A^c \cap B) = 1 - 0.15 - 0.35 - 0.25 = 0.25 \)

Now, we can fill in the contingency table:

| | United States | Not United States | Total | |--------------------------|---------------|-------------------|-------| | Log on every day | 0.15 | 0.25 | 0.40 | | Do not log on every day | 0.35 | 0.25 | 0.60 | | Total | 0.50 | 0.50 | 1.00 |

Final Answer