# Maths question

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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

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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#2

Maths is a social construct don't listen to the teachers there turning you into a puppet for the state!!!!!!

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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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#4

you really went and did the etika

(Original post by

Maths is a social construct don't listen to the teachers there turning you into a puppet for the state!!!!!!

**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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#5

(Original post by

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

**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

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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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.

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

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

**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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#7

(Original post by

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.

**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.

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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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

Based on these 9 numbers the range is 9-1 = 8 and the median is (9+1)/2= 5th value = 5

(Original post by

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.

**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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#9

(Original post by

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

**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

So, what's the list of numbers in your original question?

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#10

**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

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.

Last edited by mnot; 1 year ago

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#12

(Original post by

Guys, I need more hints.

**As.1997**)Guys, I need more hints.

Once you know this, then "cover up the median" and find the median of the upper list, this will give you the upper quartile.

Last edited by dextrous63; 1 year ago

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I got it. Prior to the last hint.

301 lowest middle: 350 399 highest

301+350/2 = 325.

301 lowest middle: 350 399 highest

301+350/2 = 325.

Last edited by As.1997; 1 year ago

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#14

(Original post by

Guys, I need more hints.

**As.1997**)Guys, I need more hints.

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

Last edited by mnot; 1 year ago

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#15

(Original post by

I got it. Prior to the last hint.

301 lowest middle: 350 399 highest

301+350/2 = 325.

**As.1997**)I got it. Prior to the last hint.

301 lowest middle: 350 399 highest

301+350/2 = 325.

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Thanks bud. This is what I did. ^^

(Original post by

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...

**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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#17

**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

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Yup, I copied it word for word.

(Original post by

Are you sure this is the complete question?

**Muttley79**)Are you sure this is the complete question?

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#19

(Original post by

Yup, I copied it word for word.

**As.1997**)Yup, I copied it word for word.

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