in part d, we have a 'choice' about which bean to take as the median bean.
Usually, when n (the number of items in our sample) is greater than 50, then the median item is the (n/2)th item.
When n is less than 50, the median item is the (n+1)/2 th item.
Hence the MS allowing both
if we choose the 24th bean to be the median, we need to know the class interval in which that lies, and further, what is the position of that item within the interval.
In the question, there are 9 + 12 = 21 beans in the first two classes. So the 24th bean will be the third item in the next class.
Now we use linear interpolation to find its mass.
We make an assumption : that the beans in this class are spread evenly. the spread of masses is 0.2g, and there are 11 of them, so the difference in mass btwn successive beans in this class will be 0.2 / 11.
we want the third bean in this class, so this will be 3 * 0.2/11 more than the lower limit of the class. eg 1.3 + 0.6/11
You should now see how the alternate response in the MS has been calculated (for using the 24.5th bean as our median bean)
Can you see how much explanation is needed here to explain interpolation fully & properly?
Now you understand the reason for my first post to you.
Go and learn the basics. Look up youtube videos of how to use these types of calculations BEFORE you start asking such open questions on here.