As.1997
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Can someone please explain part a) and b) of the question below. Thanks.

The range of 99 different integers is 98, and the median is 350.

a) What is the lower quartile, Q1? Answer: 325

b) What is the interquartile range? Answer: 50
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Equation_Addict
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Maths is a social construct don't listen to the teachers there turning you into a puppet for the state!!!!!!
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As.1997
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I guess there is some truth in what you're saying. Nonetheless, I would still like to know how to get the answer.
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haxhacks
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you really went and did the etika
(Original post by Equation_Addict)
Maths is a social construct don't listen to the teachers there turning you into a puppet for the state!!!!!!
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dextrous63
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(Original post by As.1997)
Can someone please explain part a) and b) of the question below. Thanks.

The range of 99 different integers is 98, and the median is 350.

a) What is the lower quartile, Q1? Answer: 325

b) What is the interquartile range? Answer: 50
You should ask yourself, what is the position of the median, which has 49 different integers either side?
Since the range of these 99 is 350 and 98 is the range, what must this list of these integers actually be?

As for Equation_Addict's input, one fears that he's on the vino again
Last edited by dextrous63; 1 year ago
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As.1997
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The position of the median: 99+1= 50th position?
I think you meant 350 is the median. Range = biggest number (B) - smallest number (S)
Therefore, 98=B-S.

Need some more clues bud.
(Original post by dextrous63)
You should ask yourself, what is the position of the median, which has 49 different integers either side?
Since the range of these 99 is 98 and 350 is the range, what must this list of these integers actually be?

As for Equation_Addict's input, one fears that he's on the vino again
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dextrous63
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(Original post by As.1997)
The position of the median: 99+1= 50th position?
I think you meant 350 is the median. Range = biggest number (B) - smallest number (S)
Therefore, 98=B-S.

Need some more clues bud.
Oop, yep, sorry for typo
So, you know that the 50th number into the list is 350. If the range is 98 and 350 is bang centre, how much bigger than 350 must the biggest number be. This will also be the answer for how much smaller must the smallest number be?

The important bit of the question is that all of these numbers are integers, and different from each other.
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As.1997
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Would be better if you illustrate what you're saying with some numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9
Based on these 9 numbers the range is 9-1 = 8 and the median is (9+1)/2= 5th value = 5


(Original post by dextrous63)
Oop, yep, sorry for typo
So, you know that the 50th number into the list is 350. If the range is 98 and 350 is bang centre, how much bigger than 350 must the biggest number be. This will also be the answer for how much smaller must the smallest number be?

The important bit of the question is that all of these numbers are integers, and different from each other.
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dextrous63
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(Original post by As.1997)
Would be better if you illustrate what you're saying with some numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9
Based on these 9 numbers the range is 9-1 = 8 and the median is (9+1)/2= 5th value = 5
You might be right. Either way, you now know the answer and what I was banging on about

So, what's the list of numbers in your original question?
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mnot
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(Original post by As.1997)
Can someone please explain part a) and b) of the question below. Thanks.

The range of 99 different integers is 98, and the median is 350.

a) What is the lower quartile, Q1? Answer: 325

b) What is the interquartile range? Answer: 50
If u were to order the data from lowest to highest, the lower quartile would be the value at the 25th% ie if the data consisted of 100 integers 1-100 it would be 25 (1-200, it would be 50)...

Interquartile range IQR is the range between the lower and upper quartile ie 75th % to 25th %, so if the data consisted of 100 integers 1-100 it would be 50 (75-25) & again similarly (1-200, would be 100 as 150-50 = 100)

So apply this theory to your boundary conditions (the numbers in your question)

FYI: it is worth noting this can be applied to discrete or continuous data sets, your data is discrete.
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As.1997
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Guys, I need more hints.
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dextrous63
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(Original post by As.1997)
Guys, I need more hints.
Not sure how to hint any more. What is 350+49, since that's gotta be the biggest number. Given this fact, what are all the other numbers as they're all different whole numbers?

Once you know this, then "cover up the median" and find the median of the upper list, this will give you the upper quartile.
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As.1997
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I got it. Prior to the last hint.

301 lowest middle: 350 399 highest

301+350/2 = 325.
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mnot
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(Original post by As.1997)
Guys, I need more hints.
ok well median is 350 thats the 50th percentile

u need to go from 50th to 25th percentile
and you know the range of 99 integers is 98
find 25% of the range, work down from the 50th percentile and move to the integer answer...
350- (98/4)=325 (as your answer has to be an integer), same principle for IQR as 375-325 = 50
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dextrous63
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(Original post by As.1997)
I got it. Prior to the last hint.

301 lowest middle: 350 399 highest

301+350/2 = 325.
Well done. Nice feeling when the penny drops
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As.1997
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Thanks bud. This is what I did. ^^
(Original post by mnot)
ok well median is 350 thats the 50th percentile

u need to go from 50th to 25th percentile
and you know the range of 99 integers is 98
find 25% of the range, work down from the 50th percentile and move to the integer answer...
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Muttley79
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(Original post by As.1997)
Can someone please explain part a) and b) of the question below. Thanks.

The range of 99 different integers is 98, and the median is 350.

a) What is the lower quartile, Q1? Answer: 325

b) What is the interquartile range? Answer: 50
Are you sure this is the complete question?
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As.1997
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Yup, I copied it word for word.
(Original post by Muttley79)
Are you sure this is the complete question?
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Muttley79
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(Original post by As.1997)
Yup, I copied it word for word.
So you should be able to list the integers from smallest to largest - they must be consecutive
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