Question

this list shows the ages of children who have appointments at a local pediatrician’s office.
16, 12, 13, 15, 17, 11, 14, 14, 14, 5
how does the data set change when the outlier is removed?
a. the mean age increases.
b. the median age increases.
c. the standard deviation increases.
d. the interquartile range increases.

Explanation:

Step1: Identify the outlier

In the data set \(16, 12, 13, 15, 17, 11, 14, 14, 14, 5\), the value \(5\) is much smaller than the other values, so it is the outlier.

Step2: Calculate mean with outlier

First, find the sum of the data set with outlier: \(16 + 12 + 13 + 15 + 17 + 11 + 14 + 14 + 14 + 5=\sum_{i = 1}^{10}x_i=121\). The mean is \(\frac{121}{10} = 12.1\).

Step3: Calculate mean without outlier

After removing \(5\), the sum is \(121 - 5=116\) and the number of data points is \(9\). The new mean is \(\frac{116}{9}\approx12.89\), which is greater than the previous mean.

Step4: Analyze median

With outlier (10 data points), the median is the average of the 5th and 6th values. Sorting the data: \(5, 11, 12, 13, 14, 14, 14, 15, 16, 17\). The 5th value is \(14\) and the 6th value is \(14\), so median is \(\frac{14 + 14}{2}=14\). Without outlier (9 data points), the median is the 5th value. Sorting the data without \(5\): \(11, 12, 13, 14, 14, 14, 15, 16, 17\). The 5th value is \(14\), so the median remains the same.

Step5: Analyze standard deviation

Standard deviation measures the spread of data. Removing the outlier (which is far from the other data points) will decrease the spread, so the standard deviation will decrease, not increase.

Step6: Analyze interquartile range (IQR)

IQR is the difference between the third quartile (\(Q_3\)) and the first quartile (\(Q_1\)). With outlier:

• For \(n = 10\), \(Q_1\) is the median of the first 5 values: \(5, 11, 12, 13, 14\), so \(Q_1 = 12\).
• \(Q_3\) is the median of the last 5 values: \(14, 14, 15, 16, 17\), so \(Q_3 = 15\).
• \(IQR=15 - 12 = 3\).

Without outlier (\(n = 9\)):

• \(Q_1\) is the median of the first 4 values (since \(n = 9\), the median is the 5th value, and \(Q_1\) is the median of the first 4 values? Wait, for \(n = 9\), the positions are: \(Q_1\) is at \(\frac{n + 1}{4}=\frac{10}{4}=2.5\)th value, so \(Q_1=\frac{12+13}{2}=12.5\)? Wait, no, more accurately, for \(n = 9\), the data is sorted as \(11, 12, 13, 14, 14, 14, 15, 16, 17\). The first quartile (\(Q_1\)) is the median of the first 4 values? No, the formula for quartiles: for \(n\) data points, \(Q_1\) is the value at the \(\frac{n + 1}{4}\)th position. For \(n = 9\), \(\frac{9+1}{4}=2.5\), so \(Q_1 = 12+\frac{13 - 12}{2}=12.5\). \(Q_3\) is at \(\frac{3(n + 1)}{4}=\frac{3\times10}{4}=7.5\)th position, so \(Q_3=15+\frac{16 - 15}{2}=15.5\). Then \(IQR = 15.5 - 12.5 = 3\)? Wait, maybe my initial calculation was wrong. Alternatively, using the method of dividing the data into four equal parts. Anyway, the key point is that removing an outlier that is not in the middle 50% of the data (the IQR range) will not increase the IQR. In fact, the IQR is relatively stable with respect to outliers, and in this case, it either stays the same or changes slightly, but it does not increase.

Answer:

A. The mean age increases.