The running times, in minutes, for a fleet of delivery vehicles have a 25th percentile of … — GRE General Test

The running times, in minutes, for a fleet of delivery vehicles have a 25th percentile of $134$ minutes and a 75th percentile of $242$ minutes. To analyze the data in hours, a manager converts each vehicle's running time to hours by dividing by $60$ and then adds a fixed delay of $1.25$ hours to every converted value. What is the interquartile range, in hours, of the adjusted running times?

Answer: 1.8 hours

Answer

The interquartile range of the adjusted running times is 1.8 hours.

The interquartile range (IQR) measures the spread of the middle 50% of a dataset (

Q_3 - Q_1

). For the original dataset in minutes,

IQR = 242 - 134 = 108

minutes. When data is transformed linearly via

Y = aX + b

, scaling all values by a positive constant

a

scales the IQR by

a

, whereas adding a constant

b

shifts all values equally and does not change the distance between percentiles. Here

a = \frac{1}{60}

and

b = 1.25

. Therefore, the new IQR in hours is

108 \times \frac{1}{60} = 1.8

hours.

Step-by-Step Solution

Find the original interquartile range (IQR) in minutes

IQR = 242 - 134 = 108 minutes

The interquartile range is defined as the difference between the 75th percentile (Q3) and the 25th percentile (Q1).

Apply the scaling factor for unit conversion

108 / 60 = 1.8 hours

Dividing every value by 60 scales the data by a factor of 1/60, which scales measures of dispersion like IQR by |1/60|.

Evaluate the effect of adding the constant shift

Adjusted IQR remains 1.8 hours

Adding a constant shift to every data point shifts the entire distribution without changing the spread or distance between percentiles.

Key Concept

Linear Transformations on Measures of Dispersion

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