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cestrela7 [59]
4 years ago
13

The table 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:
High Low Q1 Q3 IQR Median Mean σ
College 50 5 7.5 15 7.5 11 13.8 6.4
High School16 0 9.5 14.5 5 13 10.7 5.3

Which 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.)
Here are the answer choices:
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.
Mathematics
1 answer:
Fynjy0 [20]4 years ago
4 0
 in table 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:High Low Q1 Q3 IQR Median Mean σ
College 50 5 7.5 15 7.5 11 13.8 6.4
<span>High School16 0 9.5 14.5 5 13 10.7 5.3 </span>Both spreads are best described with the standard deviation. 
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Answer:

A.The mean would increase.

Step-by-step explanation:

Outliers are numerical values in a data set that are very different from the other values. These values are either too large or too small compared to the others.

Presence of outliers effect the measures of central tendency.

The measures of central tendency are mean, median and mode.

The mean of a data set is a a single numerical value that describes the data set. The median is a numerical values that is the mid-value of the data set. The mode of a data set is the value with the highest frequency.

Effect of outliers on mean, median and mode:

  • Mean: If the outlier is a very large value then the mean of the data increases and if it is a small value then the mean decreases.
  • Median: The presence of outliers in a data set has a very mild effect on the median of the data.
  • Mode: The presence of outliers does not have any effect on the mode.

The mean of the test scores without the outlier is:

   \bar{x}=\frac{Total of the observations-Outlier value}{n-1} \\=\frac{(86*16)-72}{15} \\=\frac{1304}{15}\\ =86.9333

*Here <em>n</em> is the number of observations.

So, with the outlier the mean is 86 and without the outlier the mean is 86.9333.

The mean increased.

Since the median cannot be computed without the actual data, no conclusion can be drawn about the median.

Conclusion:

After removing the outlier value of 72 the mean of the test scores increased from 86 to 86.9333.

Thus, the the truer statement will be that when the outlier is removed the mean of the data set increases.

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3 years ago
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