Dataset X consists of 80 numerical observations with a standard deviation of s (s > 0) and… — GRE General Test

Dataset $X$ consists of $80$ numerical observations with a standard deviation of $s$ ( $s > 0$ ) and an interquartile range of $I$ ( $I > 0$ ). A new dataset, Dataset $Y$ , is created by transforming each observation $x$ in Dataset $X$ using the linear formula $y = -4x + 25$ . Which of the following correctly gives the standard deviation and the interquartile range of Dataset $Y$ in terms of $s$ and $I$ ?

Standard deviation: $4s$ ; Interquartile range: $4I$ Answer
B
Standard deviation: $-4s + 25$ ; Interquartile range: $-4I + 25$
C
Standard deviation: $4s + 25$ ; Interquartile range: $4I + 25$
D
Standard deviation: $-4s$ ; Interquartile range: $-4I$
E
Standard deviation: $16s$ ; Interquartile range: $16I$

Answer

The standard deviation of Dataset

Y

4s

and the interquartile range is

4I

For any linear transformation of the form

y = ax + b

, the measures of dispersion (such as standard deviation, interquartile range, and range) scale by

|a|

and are unaffected by the additive constant

b

. Here

a = -4

and

b = 25

, so both standard deviation and IQR scale by

|-4| = 4

, resulting in

4s

and

4I

Step-by-Step Solution

Analyze the impact of adding a constant to data values.

Adding

+25

to each data value shifts the position of the data points along the number line, but the relative distances between data points remain constant. Thus, constant addition has zero effect on standard deviation or interquartile range.

Measures of dispersion measure the spread of data around a central value, which is invariant under pure horizontal translations.

Analyze the impact of multiplying data values by a scalar factor.

Multiplying each value by

k = -4

expands the distances between points by a factor of

|k| = |-4| = 4

Standard deviation and IQR are defined as non-negative distance quantities; scaling data by

k

scales dispersion by

|k|

Combine the scale and shift transformations.

New Standard Deviation =

|-4| \times s = 4s

, and New IQR =

|-4| \times I = 4I

Applying the transformation

y = ax + b

transforms standard deviation

\sigma_y = |a|\sigma_x

and

\text{IQR}_y = |a|\text{IQR}_x

Key Concept

Effect of Linear Transformations on Measures of Dispersion

Question

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