Dataset X consists of seven numerical values: \{12, 16, 20, 24, 28, 32, 36\}. Dataset Y is… — GRE General Test

Dataset $X$ consists of seven numerical values: $\{12, 16, 20, 24, 28, 32, 36\}$ . Dataset $Y$ is formed by replacing the value $36$ in Dataset $X$ with $56$ , while keeping all other six values the same. Which of the following statements correctly compares the interquartile range (IQR) and standard deviation of Dataset $Y$ to those of Dataset $X$ ?

The IQR remains unchanged, while the standard deviation increases.Answer
B
Both the IQR and the standard deviation increase.
C
The IQR increases, while the standard deviation remains unchanged.
D
Both the IQR and the standard deviation remain unchanged.
E
The IQR decreases, while the standard deviation increases.

Answer

The IQR remains unchanged, while the standard deviation increases.

The statement specifying that the IQR remains unchanged while the standard deviation increases is correct because the quartiles Q1 (16) and Q3 (32) are unaffected by changing the largest value from 36 to 56, maintaining an IQR of 16. In contrast, standard deviation measures spread relative to the mean, so pulling the maximum value further out increases overall deviation.

Step-by-Step Solution

Calculate the IQR of Dataset X

For Dataset X = {12, 16, 20, 24, 28, 32, 36}, the median is 24. The lower half is {12, 16, 20} with Q1 = 16. The upper half is {28, 32, 36} with Q3 = 32. Thus, IQR_X = 32 - 16 = 16.

IQR is calculated as the difference between the third quartile (Q3) and the first quartile (Q1).

Calculate the IQR of Dataset Y

For Dataset Y = {12, 16, 20, 24, 28, 32, 56}, the median is still 24. The lower half remains {12, 16, 20} (Q1 = 16). The upper half becomes {28, 32, 56}, so Q3 is still 32. Thus, IQR_Y = 32 - 16 = 16.

The third quartile is the middle number of the upper half, which remains 32 regardless of replacing 36 with 56.

Compare the standard deviation of Dataset X and Dataset Y

Replacing 36 with a significantly higher value 56 increases the distance of the maximum data point from the mean, increasing overall variance and thus increasing the standard deviation.

Standard deviation measures the average distance of data points from the mean and is highly sensitive to extreme values/outliers.

Key Concept

Sensitivity of Measures of Dispersion to Outliers

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