Each day for 28 days, the owner of a store kept track of the number of

Question

https://gmatclub.com/forum/download/file.php?mode=view&id=73862 Each day for 28 days, the owner of a store kept track of the number of customers who came into her store between 6 a.m. and 8 a.m. She displayed the results in the bar graph above. For example, on 2 of the days, 1 customer came into the store between 6 a.m. and 8 a.m., and on 7 of the days, 2 customers came into the store between 6 a.m. and 8 a.m. What was the median of the numbers of customers who came into the store between 6 a.m. and 8 a.m. during the 28 days the owner kept track? A) 5 B) 4.5 C) 4 D) 3.5 E) 3 Screenshot 2023-12-19 204413.png

Bunuel · Accepted Answer

The median of an even count of values (28) is the average of the two middle values, specifically the 14th and 15th values. According to the graph, both of these values are 5, making the median also equal to 5.

{1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 5,   5  ,   5  , 5, 6, 6, 6, 6, 6, 6, 7, 8, 8, 8, 8, 8}

Answer: A.

MartyMurray · Answer

You can just treat the graph like a table, with the numbers of customers as the values and the numbers of days as the frequencies.

Reading the graph, we see the equivalent of what appears in the following table:

Number of Customers   1  2  3  4  5  6  7  8      Frequency in Days   2  7  1  2  4  6  1  5

To find the median of 28 values, we must find the mean of the 14th and 15th values.

Using the second row of the table, we see that the 14th and 15th values are both in the 5 column.

So, the median is 5 customers.

Correct answer:  A

ashdank94 · Answer

I know how to calculate median using values and frequencies, however I get very confused when there are 2 types of numbers. Is there an easy way to calculate this using a table?  MartyMurray

bumpbot · Answer

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Each day for 28 days, the owner of a store kept track of the number of

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