You test 8 automobiles and find that their mileage ratings are 28, 28, 29, 30, 33, 34, 35, and 340 mpg, arranged in order from smallest to largest. The average of these numbers is 557 ÷ 8, or 70 mpg. However, the last data value (340 mpg) seems impossibly large, and you decide that it must be due to an error in the data gathering process.
Rather than doing all your measurements all over again, at considerable expense, you decide to simply ignore the "bad" value of 340 mpg. (When they do this, scientists say they are "rejecting the outliers," by ignoring unreasonable values that are nowhere close to the main cluster of data points. This procedure has some risks.) What is the mean of the remaining mileage numbers?
sorry all one question
what i did was (557-340)/7
and got 31
that´s correct
\[ \frac{28+28+ 29+ 30+ 33+ 34+ 35}{7}\]
donkey work, i would cheat http://www.wolframalpha.com/input/?i=mean++28%2C+28%2C+29%2C+30%2C+33%2C+34%2C+35
but then you wouldent learn anything ;p
what do you learn by punching buttons on a calculator? or moving beads on an abacus?
beads on an abacus of course haha
i know you added up in your head, and then divided by 7 ok but my slide rule is rusty, so i used stones
what @summertimesadness proposed as solution is no donkey work and much more reasonable than yours @satellite73
ohh that sounds so much more efficient @satellite73 haha
the sum and the outlier are given data, so no need to add the remaining seven
ohh yeah huh
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