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anonymous
 one year ago
WILL FAN AND MEDAL!
The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:
anonymous
 one year ago
WILL FAN AND MEDAL! The table below shows data from a survey about the amount of time students spend doing homework each week. The students were either in college or in high school:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Which of the choices below best describes how to measure the spread of this data? (Hint: Use the minimum and maximum values to check for outliers.) a. Both spreads are best described with the IQR. b. Both spreads are best described with the standard deviation. c. The college spread is best described by the IQR. The high school spread is best described by the standard deviation. d. The college spread is best described by the standard deviation. The high school spread is best described by the IQR.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it's A? I know it's not B.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it may be D, actually. Cause I saw a couple of people say that they got A and C wrong. I put B before and got it wrong, so it must be D? I don't know.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it says use the min and max data to check for outliers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I've already graphed it. I'll attach the box plots.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't they both have an outlier?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the mean > median generally an outlier exists

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The median is more accurate than the mean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm just having a hard time understanding when to use standard deviation and when to use IQR. I know my textbook says that if the distribution of data is symmetrical, then the best measurement to use is standard deviation. And if the data is asymmetrical, the best measurement is IQR. I don't know if I'm making any sense here, so I'm sorry. I really suck at math.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0beuase it takes things in order, so the mean is shifgted upwards or downwards depending on the data

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0standard deviation is a relative measure (i.e. relative to the mean)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know, and IQR is relative to the median.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry it took me so long to type, I was looking at my textbook.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0IQR , think about it, "quartile" means "quarter" divide into 4 quarters

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the lower quartile is the value one way into the data the upper quartile is the value 3 quarters into the data the upper quartile minus the lower quartile is the IQR

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I keep thinking it's A because the data isn't evenly distributed, and it has outliers. But I remember somebody saying they got that wrong, so I'm just completely stuck.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0think about the mean.,. now what is going to upset the mean, only outliers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0outliers dont upset the median as much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I didn't read what you typed above, it didn't show because my computer froze for a couple minutes. I'm gonna read it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is becuase the median will shift as the data extends

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the IQR for the high school would be 5 and the IQR for the college would be 7.5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so the outliers upset the mean, not the median. So it would be IQR, not standard deviation, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0take simple examples: use 1, 2, 3, 4 5 what is the mean?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct, now what is the median?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct also, now chose 1, 2, 3, 4 & 13 what is the mean and what is the median

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0see what happens to the mean when you have an outlier?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, like you said before, it messes up the mean.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you are gettin it, cool!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well thanks for helping so much :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now you can look at your table and guess your answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0550 average should be about 25

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0however, the mean is 13.8 which means there must be an outlier somewhere

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it would be A? Because there's also an outlier in the high school plot.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cause outliers would cause the shape of the data to be asymmetric, and IQR is the best measure of spread for asymmetrical data.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which shows how to best measure the spread of the data, well you cant use measure of central tendency (mean, median etc) to show spread, you have to use measures of spread (IQR or standard deviation etc)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Once again you need to be able to "see" the data, i.e relative to the mean, to be able to judge which one is best for judging this, this is the whole point of the exercise
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