The table shows data from a survey about the number of times families eat at restaurants during a week. the families are either from rome, italy or new york, new york:

high low q1 q3 iqr median mean σ
rome 18 1 3 7 4 6.5 6.4 4.3
ny 14 1 4.5 8.5 4 5.5 6.1 3.2

which of the choices below best describes how to measure the center of this data?

- both centers are best described with the mean.

-both centers are best described with the median.

- the rome data center is best described by the mean. the new york data center is best described by the median.

-the rome data center is best described by the median. the new york data center is best described by the mean.

To solve this problem, let us first define what is mean and median. Mean is the average of all the numbers in the data set while the median is the number in the middle of the data set in ascending order.

If we create a box plot for the data of Rome and New York, we can see that there is an outlier in the data for New York. Since New York has an outlier, so the mean is not a good representation on the central tendency of the data. The mean is skewed (distorted) by the outlier. So in this case it is better to use the median.

While the Rome data is nice and symmetrical, it does not seem to have an outlier,  so we can use the mean for this data set.

Therefore the answer is:

The Rome data center is best described by the mean. The New York data center is best described by the median

Mean and median are used to find the central tendency of the data. For Rome, the mean could be used and for NY, the median could be used to find the center of the data. Option C is correct.

Given:

The given data set of families from Rome and NewYork is:

          High   Low   Q1     Q3    IQR   Median    Mean        σ

Rome   18         1       3       7       4         6.5         6.4          4.3

NY        14         1      4.5    8.5    4         5.5         6.1           3.2

Now, talking about Rome. The distribution is not varying very much and it doesn't show very much deviation from the mean value. It means there is not any outlier formed in the data set of Rome families.

So, for Rome families, mean could be a good fit for measuring the central tendency.

Now, for NY families, the data shows a clear outlier at Q3. Here, the difference is very high, and the mean could not be a suitable fir for measuring the central tendency.

So, the median will be used to measure the central tendency of the data set for NY.

Therefore, for Rome, the mean could be used and for NY, the median could be used to find the center of the data.

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