anonymous
  • anonymous
Means and Medians. It makes sense to count the negatives as positve when taking the mean, but I'm not sure.
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
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anonymous
  • anonymous
no, you shouldnt count the negatives as positive when calculating mean.
anonymous
  • anonymous
for example, if you made a profit of 1000 dollars in a month and made a loss of 1000 dollars the next month, the average profit is 1000-1000/2 = 0 in two months, not 1000+1000/2 = 1000 in two months. it doesnt make any sense.

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anonymous
  • anonymous
Right.
anonymous
  • anonymous
But this is about predicting temperature. If you are off by -5 on one day and off by +5 another day, you are still off by an average of 5 over both days.
anonymous
  • anonymous
if I kept one negative then it would average to 0?
anonymous
  • anonymous
oh. i didnt see that part. it said difference between actual high and predicted high. In that case, it should be the absolute difference, or the absolute value of the difference between what was predicted and what actually happened
anonymous
  • anonymous
So when I take the mean I should treat all the numbers as positive? It makes sense to do that but I haven't encountered a problem like this before.
anonymous
  • anonymous
the error is absolute, whether it was made above or below the actual temperature.
anonymous
  • anonymous
you are talking about the magnitude of the error here.
anonymous
  • anonymous
if you predicted the temperature was going to be 50 and it turned out to be 100, then the magnitude of error in your prediction was 50. conversely if you predicted it was going to be 100 and it turned out to be 50, the magnitude of the error is still 50. so on average, you make an error of 50 degrees in your predictions.
anonymous
  • anonymous
Yet they include the negative in the chart given with the problem?
anonymous
  • anonymous
the lesser the error, the more the accuracy. yes, because the chart given indicates predicted temperature - actual temperature. so if they predicted 90 degree and it turned out to be 93, then the predicted-actual temperature is 90-93 = -3. But the magnitude of the error they made was 3 degrees
anonymous
  • anonymous
So when I calculate the mean I should or shouldn't include the negatives?
anonymous
  • anonymous
When I include the negatives the mean of the 5 day predictions ends up being lower than the 1 day predictions, making it sort of confusing when I have to answer the last question.
anonymous
  • anonymous
it depends. if you are talking about accuracy, then you should calculate using the magnitude of the error. if you are talking about mean difference in temperature between actual and predicted, then you should take the values as posted.(without changing the negatives to positives)
anonymous
  • anonymous
The last question talks about accuracy. I'll leave them negative and see how it comes out. Thanks :)
anonymous
  • anonymous
for example, if you predicted 100, and it was 50 for 2 days in a row, then your accuracy is off by 50 degrees. but the average difference in actual temperature to predicted is -50 degrees.
anonymous
  • anonymous
That makes sense. But like in the problem, in the 5 day predictions, one is off by 6 and another by -6, that comes out as 0 as if the being off in the opposite direction should compensate.. I'm just not sure exactly what they are after, I'll find out when I submit it :)
anonymous
  • anonymous
yes, one is off by 6 and the other is off by -6. so the accuracy is off by 6 degrees. accuracy deals with how much error you made in the prediction. on average, you are accurate up to within 6 degrees of actual temperature. but on average, the difference between actual and predicted temperature is 0 degrees. the difference between actual and predicted temperature doesn't deal with the accuracy of the prediction. it just tells you what the difference between the two was. On average, it is 0 in your case.
anonymous
  • anonymous
Sounds right, I'll answer that way. At first glance it made sense to disregard the negatives so the mean would say something useful about the error, but that's not what the problem is looking for. Thanks.
anonymous
  • anonymous
you are welcome.

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