a study looked at the prices that several grocery stores charged for the same 15 - ounce can of tomato…

a study looked at the prices that several grocery stores charged for the same 15 - ounce can of tomato sauce. the prices ranged from $0.90 to $1.00. the mean price and the median price were both $0.98. how many of the cans in the study could have cost exactly $0.98? select all the possibilities.\na. 1\nb. 2\nc. all of the cans\nd. none of the cans

a study looked at the prices that several grocery stores charged for the same 15 - ounce can of tomato sauce. the prices ranged from $0.90 to $1.00. the mean price and the median price were both $0.98. how many of the cans in the study could have cost exactly $0.98? select all the possibilities.\na. 1\nb. 2\nc. all of the cans\nd. none of the cans

Answer

Explanation:

Step1: Recall mean and median concepts

The mean is the sum of all values divided by the number of values. The median is the middle - value when the data is ordered.

Step2: Analyze option A

It is possible to have a data - set where only 1 can costs $0.98$ and the other values are arranged in such a way that the mean and median are both $0.98$. For example, if we have a small data - set like ${0.90,0.98,1.00}$, the mean is $\frac{0.90 + 0.98+1.00}{3}=\frac{2.88}{3}=0.96$ (this is just an example to show construction, we can adjust values). But in a more complex data - set, we can make it work.

Step3: Analyze option B

Similarly, we can have a data - set with 2 cans costing $0.98$. For instance, if we have a data - set with an even number of values, say ${0.90,0.98,0.98,1.00}$, we can adjust other values (if there are more values in the data - set) to make the mean and median equal to $0.98$.

Step4: Analyze option C

If all the cans cost $0.98$, then the mean is $\frac{n\times0.98}{n}=0.98$ (where $n$ is the number of cans), and the median is also $0.98$.

Step5: Analyze option D

Since options A, B, and C are possible, it is not the case that none of the cans can cost $0.98$.

Answer:

A. 1, B. 2, C. all of the cans