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secondary 4 | E Maths
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Date Posted: 4 years ago
Views: 505
The use of x is actually ridiculous for the question. The variable x has already been used for the data values, so it should not be used again for the frequency. Hence, there is a fault with the question. Nevertheless, you can just take it that the “x” in the frequency is different from the “x” in the data values.
Suppose the modal class is 30 to 35. Then, the median must lie somewhere within the 13th and the 20th people.
Supposing the median is the 13th person. Then, there must be a total of (13 x 2 - 1) = 25 people, leading to x = 1.
Supposing the median is the 20th person. Then, there must be a total of (20 x 2 - 1) = 39 people, leading to x = 15.
The problem with x = 15 is that this will lead to the mode of the distribution being different.
As such, 1 <= x <= 7.
There is a second consideration for this. What happens if, say, the median lies somewhere in there class containing 40 <= x <= 45 if the number of people in this class is large enough?
Well, could it be possible?
Now, there are a total of 22 people in the group below the group 40 <= x <= 45.
It is also possible that the 23rd person, who falls in the group 40 <= x <= 45, is the median person. For this to happen, there must be a total of (23 x 2 - 1) = 45 people.
In this scenario, x = 21.
In fact, for any x > 21, the median will fall in the range 40 <= x <= 45.
Therefore, our solutions are 1 <= x <= 7, x >= 21.
Suppose the modal class is 30 to 35. Then, the median must lie somewhere within the 13th and the 20th people.
Supposing the median is the 13th person. Then, there must be a total of (13 x 2 - 1) = 25 people, leading to x = 1.
Supposing the median is the 20th person. Then, there must be a total of (20 x 2 - 1) = 39 people, leading to x = 15.
The problem with x = 15 is that this will lead to the mode of the distribution being different.
As such, 1 <= x <= 7.
There is a second consideration for this. What happens if, say, the median lies somewhere in there class containing 40 <= x <= 45 if the number of people in this class is large enough?
Well, could it be possible?
Now, there are a total of 22 people in the group below the group 40 <= x <= 45.
It is also possible that the 23rd person, who falls in the group 40 <= x <= 45, is the median person. For this to happen, there must be a total of (23 x 2 - 1) = 45 people.
In this scenario, x = 21.
In fact, for any x > 21, the median will fall in the range 40 <= x <= 45.
Therefore, our solutions are 1 <= x <= 7, x >= 21.
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Hope this helps.
Date Posted:
4 years ago
im sorry but this not the answer
It’s 1 to 7. Because if x exceeds 7, then it would become the modal class instead.
Thank you for pointing out the mistake.
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I am really sorry for making the careless mistake. I apologise for providing the incorrect answer.
I failed to realised that the mode will change with my original answer.
As such, the maximum value for x is 7 so that the mode doesn't change and the distribution remains valid.
Really sorry.
I failed to realised that the mode will change with my original answer.
As such, the maximum value for x is 7 so that the mode doesn't change and the distribution remains valid.
Really sorry.
Date Posted:
4 years ago
Oh ya, forgot to mention that x >= 21 is also possible. I have discussed it in the main question chat box a few hours ago.
That's true! Making the unknown (22 or more) sufficiently big to make it both the median and mode.
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