Question

Difficulty: EasyMeasures of Dispersion and Position (Range, IQR, Standard Deviation, Percentiles)

A dataset of employee salaries has a standard deviation of 66. If every salary in the dataset is multiplied by 33 and then increased by 1010, what is the standard deviation of the resulting dataset?

  1. A
    66
  2. 1818Answer
  3. C
    2828
  4. D
    1616
  5. E
    5454

Answer

The standard deviation of the resulting dataset is 1818.
Multiplying each data point by 33 increases the spread around the mean by a factor of 33, changing the standard deviation from 66 to 1818. Adding 1010 to each value shifts all values by the same amount, which leaves the distance between data points and the mean unchanged. Thus, the standard deviation remains 1818.

Step-by-Step Solution

1
Evaluate the effect of multiplying each value in the dataset by a positive constant c=3c = 3.
The standard deviation is scaled by 33, giving a new standard deviation of 6×3=186 \times 3 = 18.
Scaling every data value by a constant factor cc multiplies the measures of dispersion (such as standard deviation, range, and IQR) by c|c|.
2
Evaluate the effect of adding a constant k=10k = 10 to each value in the dataset.
The standard deviation remains 1818.
Adding a constant to every data value shifts the entire distribution along the number line without changing the relative distances between values or the overall spread.

Key Concept

Linear Transformations on Standard Deviation
Estimated Time:45s
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