Answer:
a) 20<h≤30.
b) 26.17 hrs
Step-by-step explanation:
The missing table is shown in attachment.
Part a)
We need to find the class interval that contains the median.
The total frequency is

The median class corresponds to half

That is the 15th value.
We start adding the frequency from the top obtain the least cumulative frequency greater or equal to 15.
2+8+9=19
This corresponds to the class interval 20<h≤30.
Adding from the bottom also gives the same result.
Therefore the median class is 20<h≤30.
b) Since this is a grouped data we use the midpoint to represent the class.
The median is given by :



