Question

Difficulty: MediumPercentiles and Quartiles

A dataset consists of 20 distinct numerical values arranged in increasing order. The lower quartile (Q1Q_1) of the dataset is equal to the average of the 5th and 6th values, and the upper quartile (Q3Q_3) is equal to the average of the 15th and 16th values. Given that the lower quartile Q1=42Q_1 = 42, the interquartile range (IQR=Q3Q1\text{IQR} = Q_3 - Q_1) is 38, and the 15th value in the dataset is 74, what is the value of the 16th term?

Answer: 86

Answer

86
To find the 16th value, first determine the upper quartile Q3Q_3 using the interquartile range: Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80. Since Q3Q_3 is the average of the 15th and 16th terms, set up the equation 74+x162=80\frac{74 + x_{16}}{2} = 80. Multiplying by 2 gives 74+x16=16074 + x_{16} = 160, so x16=86x_{16} = 86.

Step-by-Step Solution

1
Calculate the upper quartile (Q3Q_3) using the given lower quartile (Q1Q_1) and interquartile range (IQR).
Q3=Q1+IQR=42+38=80Q_3 = Q_1 + \text{IQR} = 42 + 38 = 80
By definition, the interquartile range is the difference between the upper and lower quartiles (IQR=Q3Q1\text{IQR} = Q_3 - Q_1).
2
Relate the upper quartile (Q3Q_3) to the 15th and 16th values of the ordered dataset.
Q3=x15+x162=80Q_3 = \frac{x_{15} + x_{16}}{2} = 80
For an ordered dataset of 20 elements, Q3Q_3 is the median of the upper half of the data (the 11th through 20th terms), which equals the arithmetic mean of the 15th and 16th terms.
3
Substitute the known value of the 15th term (x15=74x_{15} = 74) into the equation and solve for the 16th term (x16x_{16}).
\frac{74 + x_{16}}{2} = 80 \implies 74 + x_{16} = 160 \implies x_{16} = 86
Multiplying both sides by 2 gives 160, and subtracting 74 yields 86.

Key Concept

Interquartile Range and Quartile Calculations
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