Question

Difficulty: MediumRange and Standard Deviation

Data Set PP consists of five distinct integers with a range of 2020 and a standard deviation of d1d_1, where d1>0d_1 > 0. A new Data Set QQ is formed by multiplying each integer in Data Set PP by 2-2 and then adding 77 to the product. What are the range and standard deviation of Data Set QQ in terms of d1d_1?

  1. Range is 4040; standard deviation is 2d12d_1Answer
  2. B
    Range is 4040; standard deviation is 2d1-2d_1
  3. C
    Range is 40-40; standard deviation is 2d1-2d_1
  4. D
    Range is 4747; standard deviation is 2d1+72d_1 + 7
  5. E
    Range is 2020; standard deviation is 2d12d_1

Answer

Range is 4040; standard deviation is 2d12d_1
The correct answer accounts for both linear transformation rules for measures of dispersion: multiplying every value in a dataset by a scalar cc multiplies both the range and standard deviation by c|c|, and adding a constant kk to every value does not alter either measure. Here, multiplying by 2-2 scales the range from 2020 to 4040 and the standard deviation from d1d_1 to 2d12d_1. Adding 77 has no effect on either measure.

Step-by-Step Solution

1
Analyze the effect of multiplying each element by a constant c=2c = -2.
The range scales by c=2=2|c| = |-2| = 2, giving a new range of 20×2=4020 \times 2 = 40. The standard deviation scales by c=2=2|c| = |-2| = 2, giving a new standard deviation of 2d12d_1.
Measures of dispersion (range and standard deviation) are non-negative distance metrics, so they scale by the absolute value of any multiplicative constant.
2
Analyze the effect of adding a constant k=7k = 7 to each element.
The range remains 4040, and the standard deviation remains 2d12d_1.
Adding a constant shifts every data point by the exact same distance, leaving all pairwise distances between data points unchanged.

Key Concept

Linear Transformations on Range and Standard Deviation
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