Mean: 113
Median: 108-116
Mode: 108-116
The mean was the hard one, I actually took the middle score from each range (eg 90-98=94) and multipled it by the number of times it occured (eg 94x6=564) and added all 5 (564+2266+4816+3388+1170= 12204). Then divided it by the frequency (6+22+43+28+9=108). Total (12204/108= 113). It's impossible to find a mean of a set of ranges, you need exact figures.
The mode is the most commonly occuring range, the median is the central range. These figures can't be accurate without the full actual IQ scores. But based on the figures provided, the mean, median and mode are as I stated above. If you provide the exact figures I'll work out the exact mean, mode and median.
2006-11-30 08:03:17
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answer #1
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answered by Velouria 6
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Peta is right, and her maths is correct. However, for a range like this, you should also quote limits. Since all ranges are the same distance apart (8), the mean should be quoted as 113 +/- 4.
To clarify, the median is the middle number. So if there are 3 numbers, e.g. 2,3 and 16, the median is 3. If there are 4 numbers, e.g. 2,3,11, and 16, the median is the average of the 2 middle numbers, i.e. (3+11) / 2, i.e. 7.
In this case, the 54th and 55th number both lie in the range 108 - 116, so the median is correctly stated as 108 - 116.
2006-12-01 01:00:43
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answer #2
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answered by Trevor P 2
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All you may desire to do is upload the frequencies of the classification limits that are above 107. 108-116 40 3 117-a hundred twenty five 28 126-134 9 40 3+28+9=80 80 pupils do no longer could desire to sign up in this technique.
2016-12-29 17:31:37
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answer #3
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answered by Anonymous
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Mean -- average -- can't really figure it out without the real numbers.
Median -- middle number -- again, since we don't have the actual numbers we don't know, but it falls within 108-116.
Mode -- most common number -- no idea without the actual results.
Thus, for all 3, we need the actual results, not the number that fall with a 9 or 10 point range
2006-11-30 07:59:24
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answer #4
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answered by Justin B 2
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She(Peta) is actually right. Assuming there's no mistake in calculations.
2006-11-30 08:30:21
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answer #5
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answered by najahian85 2
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