A dataset of 25 student test scores has a mean of 70, a standard deviation of 8, and an in… — GRE General Test

A dataset of 25 student test scores has a mean of $70$ , a standard deviation of $8$ , and an interquartile range ( $\text{IQR}$ ) of $12$ . A instructor creates a modified dataset by multiplying each original test score by $1.5$ and then adding $10$ to the result. What is the standard deviation of the modified dataset?

A
$8$
$12$ Answer
C
$18$
D
$22$
E
$28$

Answer

For any dataset, if each value

x

is transformed into

a \cdot x + b

, where

a

and

b

are constants, the new standard deviation is given by

|a| \cdot \sigma_{\text{old}}

. In this problem,

a = 1.5

and

b = 10

. Therefore, the new standard deviation is

1.5 \times 8 = 12

. The addition of

10

affects measures of center (such as the mean and median) but has no effect on measures of dispersion (such as standard deviation, IQR, and range).

Step-by-Step Solution

Analyze the effect of multiplying each data point by a constant on standard deviation.

Multiplying every value in a dataset by a constant

c = 1.5

multiplies the standard deviation

\sigma

|c|

. The new intermediate standard deviation becomes

1.5 \times 8 = 12

Standard deviation measures dispersion around the mean, so scaling all data points by a factor scales the distance between each point and the mean by that same factor.

Analyze the effect of adding a constant to each data point on standard deviation.

Adding a constant

k = 10

to every value shifts the entire distribution without changing the distances between data points or the relative spread around the mean. Thus, standard deviation remains

12

Adding a constant shifts both the individual values and the mean by the same amount, leaving

(x_i - \bar{x})

unchanged.

Key Concept

Linear Transformations on Measures of Dispersion

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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