A quality control department recorded the processing time, in days, for a batch of request… — GRE General Test

A quality control department recorded the processing time, in days, for a batch of requests. The table below displays the frequency distribution of the processing times:

Processing Time (Days)	Number of Requests
1	5
2	8
3	$k$
4	6
5	4

If $k$ is a positive integer and the median processing time for all requests in the batch is equal to 3 days, what is the minimum possible value of $k$ ?

A
2
B
3
4Answer
D
5
E
6

Answer

The correct answer is 4. There are

5 + 8 = 13

values less than

3

. For

3

to be the median of the dataset, the median position must be greater than

13

. With

k = 4

, the total number of items is

N = 23 + 4 = 27

. For an odd dataset of size

27

, the median is the

\frac{27 + 1}{2} = 14

th item. Since the first

13

items are less than

3

, the

14

th item is

3

, making

3

the median. Any value of

k \le 3

results in a median less than

3

Step-by-Step Solution

Calculate the cumulative frequency of values below 3 days.

Number of requests with processing times of 1 or 2 days is

5 + 8 = 13

To determine the median position, we first count how many data points lie strictly below 3 days.

Express the total number of requests

N

in terms of

k

N = 5 + 8 + k + 6 + 4 = 23 + k

The total number of requests determines whether

N

is odd or even and where the median position lies.

Determine the condition for 3 days to be the median.

For 3 days to be the median, the median position must be greater than 13 so that it falls into the category of 3 days.

Since 13 values are strictly less than 3, the median rank must be at least 14.

Test minimum integer values for

k

k = 3

N = 26

. The median is the average of the

13

th value (

2

) and the

14

th value (

3

), which is

2.5

. If

k = 4

N = 27

. The median is the

14

th value, which is

3

Testing

k = 4

yields

N = 27

where the

14

th value is

3

, satisfying the median requirement.

Key Concept

Median of a Frequency Distribution Table

Estimated Time:1m 30s

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