Question 1

Here is the distribution of workers a restaurant keeps on staff during a 1-hour shift. 
$\begin{array}{|l|l|l|l|l|}
\hline   \text {probability} & 0.10 & 0.25 & 0.55 & 0.10 \
\hline  \text { \# of workers} & 3 & 4 & 5 & 6 \
\hline
\end{array}$

If the workers are paid $10/hour, what is the expected long-term cost of a 1-hour shift?
A) none of these
B) $180
C) $46.50
D) $4.65
E) $45

Accepted Answer

To find the expected long-term cost, we must multiply the number of workers in each category by their respective hourly wage and then sum those products.

Category 1: 4 workers x $10/hour = $40/hour
Category 2: 10 workers x $10/hour = $100/hour
Category 3: 6 workers x $10/hour = $60/hour

Total cost = $40 + $100 + $60 = $200

However, we are looking for the expected long-term cost, which means we need to account for the probability of each category occurring during a shift.

Category 1 occurs with probability 0.1, category 2 occurs with probability 0.7, and category 3 occurs with probability 0.2.

Expected cost = (0.1 x $40) + (0.7 x $100) + (0.2 x $60) = $46.

Therefore, the expected long-term cost of a 1-hour shift is $46.50. Answer choice C is correct.

Question 2

A supermarket claims that their checkout scanners correctly price 99.8% of the items sold. How many items would you expect to buy, on average, to find one that scans incorrectly?
A) 200
B) 998
C) 99.8
D) 2
E) 500

Accepted Answer

To find the average number of items you need to buy to get one that scans incorrectly, you take the inverse of the error rate. The error rate is 0.2% (100% - 99.8%), which is 0.002 when expressed as a decimal. The inverse of 0.002 is 1/0.002 = 500.

Question 3

An ice cream stand reports that 12% of the cones they sell are "jumbo" size. You want to see what a "jumbo" cone looks like, so you stand and watch the sales for a while. What is the probability that the first jumbo cone is the fourth cone you see them sell?
A) 33%
B) 60%
C) 40%
D) 8%
E) 93%

Accepted Answer

The probability of seeing a jumbo cone on any given sale is 0.12, and the probability of not seeing a jumbo cone is 0.88. To calculate the probability of seeing the first jumbo cone on the fourth sale, we need to multiply the probability of not seeing a jumbo cone on the first three sales (0.88 x 0.88 x 0.88) by the probability of seeing a jumbo cone on the fourth sale (0.12). This gives us 0.088, or 8% chance of the first jumbo cone being the fourth sale we see.

Question 4

According to the National Telecommunication and Information Administration, 50.5% of U.S. households had Internet access in 2001. What is the probability that four randomly selected U.S. households all had Internet access in 2001?
A) 12.6%
B) 50.5%
C) 49.5%
D) 93.5%
E) 6.5%

Accepted Answer

To find the probability of four randomly selected U.S. households all having internet access in 2001, we need to multiply the probability of one household having internet access by itself four times, since the households are being selected randomly and independently.

The probability of one household having internet access is 50.5%, or 0.505. Therefore, the probability of four households all having internet access is:
0.505 x 0.505 x 0.505 x 0.505 = 0.065, or 6.5%

Therefore, the best choice is E, with explanation as above.

Quiz 4: Displaying and Comparing Qualitative Data

A supermarket claims that their checkout scanners correctly price 99.8% of the items sold. How many items would you expect to buy, on average, to find one that scans incorrectly?

An ice cream stand reports that 12% of the cones they sell are "jumbo" size. You want to see what a "jumbo" cone looks like, so you stand and watch the sales for a while. What is the probability that the first jumbo cone is the fourth cone you see them sell?

According to the National Telecommunication and Information Administration, 50.5% of U.S. households had Internet access in 2001. What is the probability that four randomly selected U.S. households all had Internet access in 2001?