MEAN (AVERAGE):
average of the data set
ALL you do is ADD the number in the data set and DIVIDE by the amount of number in the set.
((-10)+(-7)+2+4+14+15) / 6
=18/6
= 3
mean (average) = 3
MEDIAN:
middle of the data set
Put the number in order (smallest to largest)
if there are an ODD amount of the number, the MIDDLE is the MEDIAN.
if you have an EVEN amount of number, the MEDIAN is the two MIDDLE number average
-10, -7, 2, 4, 14, 15,
the two middle number are
2, 4
(2+4)/2
= 3
median=3
2007-07-30 14:18:27
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answer #1
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answered by Anonymous
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Mean=sum of the terms / total number of the terms
add all numbers and divide by 6
-10-7+2+4+14+15=18/6=3
median: Arrange the numbers in order
-10,-7,2,4,14,15
mid value of the above is 2,4
average the two mid-values 2,4
=(2+4)/2
=6/2
=3
therefore mean=3 and median=3
2007-07-29 13:24:34
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answer #2
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answered by sss 2
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Adding this we get
-10+-7+2+4+14+15=18
Since the frequency of no is 6
therefore divide by 6
18/6=3
2007-07-29 12:14:04
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answer #3
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answered by Nikhil B 2
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Mean : add all numbers and divide by 6
-10-7+2+4+14+15=18/6=3
median: Arrange the numbers in order
-10,-7,2,4,14,15
average the two mid-values namely 2,4
(2+4)/2=6/2=3
2007-07-29 12:03:41
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answer #4
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answered by cidyah 7
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Meanaverage=sunos/no.onos. = (-10-7+2+4+14+15)/6
so Mean =18/6=3
Median is the most central valuve,
If the nos are odd ,it is one no.,
But if the nos are even,we have to take tke average of the two nos.,like in this case,so Median=(2+4)/2=6/2=3
2007-07-29 12:19:15
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answer #5
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answered by MAHAANIM07 4
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Sum = - 17 + 35 = 18
mean = 18 / 6 = 3
median = (2 + 4) / 2 = 3
2007-07-29 11:58:41
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answer #6
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answered by Como 7
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mean
add them all up and divide by the number of bits of data
-10 + -7 + 2 + 4 + 14 + 15
18 / 6 = 3
median is the middle number whihc lies between the middle pair ie. between 2 and 4 so 3
2007-07-29 11:44:36
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answer #7
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answered by JAMES C 2
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its not good to try and have people do your work for you, try and work through the problem, if you still dont get it ask your teacher tomorrow
2007-07-29 11:46:57
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answer #8
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answered by Anonymous
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