Interquartile range, Median of Cumulative Frequency.

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JimCrown

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Class width (N/mm2)FrequencyCumulative Frequency
0 - 4.900
5 - 9.911 (= 0 + 1)
10 - 14.945 (=0 +1 +4)
15 - 19.9611 (=0 + 1 + 4 + 6)
20 - 24.9516 (=0 + 1 + 4 + 6 + 5)
25 - 29.9319 (= 0 + 1 + 4 + 6 + 5 + 3)
30 - 34.9120 (=0 + 1 + 4 + 6 + 5 + 3 + 1)
35 - 39.9020 (=0 + 1 + 4 + 6 + 5 + 3 + 1 + 0)

What would be:
i) The Median
ii) The Inter-quartile range
iii) How many bricks had a compressive strengthmore than 22 N/mm2?

I believe it to be:
i) There are 20 sizes of brick. The median is the 10thvalue. By using cumulative frequency (0+1+4=6) gives us the 10thvalue that is in the 15-19.9 class width. So, the median is 15-19.9.
ii) The interquartile range is 24 (upper quartile) – 15(lower quartile) = 9
iii) 20 – 13 = 7. 7 bricks have a compressive strength morethan 22 N/mm2

But I am not sure.
 
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JimCrown said:
Class width (N/mm2)FrequencyCumulative Frequency
0 - 4.900
5 - 9.911 (= 0 + 1)
10 - 14.945 (=0 +1 +4)
15 - 19.9611 (=0 + 1 + 4 + 6)
20 - 24.9516 (=0 + 1 + 4 + 6 + 5)
25 - 29.9319 (= 0 + 1 + 4 + 6 + 5 + 3)
30 - 34.9120 (=0 + 1 + 4 + 6 + 5 + 3 + 1)
35 - 39.9020 (=0 + 1 + 4 + 6 + 5 + 3 + 1 + 0)

What would be:
i) The Median

I believe it to be:
i) There are 20 sizes of brick. The median is the 10thvalue. By using cumulative frequency (0+1+4=6) gives us the 10thvalue that is in the 15-19.9 class width. So, the median is 15-19.9.
Click to expand...

No. To learn how to find the median from grouped data, try here.

JimCrown said:
ii) The Inter-quartile range
I believe it to be:
ii) The interquartile range is 24 (upper quartile) – 15(lower quartile) = 9
Click to expand...
How did you arrive at these values?

JimCrown said:
iii) How many bricks had a compressive strength [of] more than 22 N/mm2?
I believe it to be:
iii) 20 – 13 = 7. 7 bricks have a compressive strength morethan 22 N/mm2

But I am not sure.
Click to expand...
What relation were you given between the posted information and compressive strength? By what reasoning did you obtain your result?

Please be complete. Thank you! :wink:
 
bY using cumulative frequency graph I got the upper and lower quartile. So, would the median be 19 or the 10th term in the cumulative frequency. I AM Completely confused at the moment.
 
JimCrown said:
bY using cumulative frequency graph I got the upper and lower quartile.
Click to expand...
"By using" it how? What were your steps? What was your reasoning?

JimCrown said:
So, would the median be 19 or the 10th term in the cumulative frequency.
Click to expand...
What did they explain and demonstrate at the link I gave you? How did you use that information? By what steps did you arrive at your answer?

Please be complete, as we cannot help resolve issues that we cannot see. Thank you! ;)
 
JimCrown said:
ii) The Inter-quartile range
Click to expand...

Any set of data can be described by its five-number summary. These five numbers, which give you the information you need to find patterns and outliers, consist of (in ascending order). The minimum or lowest value of the dataset The first quartile Q1, which represents a quarter of the way through the list of all data The median of the data set, which represents the midpoint of the whole list of data The third quartile Q3, which represents three-quarters of the way through the list of all data The maximum or highest value of the data set.

IQR = Q3 - Q1. The interquartile range shows how the data is spread about the median. It is less susceptible than the range to outliers and can, therefore, be more helpful.
 
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