Question

algebra i b-cr
pre-test complete
1 2 3 4 5 6 7 8 9
the interquartile range of the data set is 4.
2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12
which explains whether or not 12 is an outlier?
twelve is an outlier because it is greater than the sum of 7 and 4.
twelve is an outlier because it is less than the sum of 7 and 6.
twelve is not an outlier because it is greater than the sum of 7 and 4.
twelve is not an outlier because it is less than the sum of 7 and 6.

Explanation:

Step1: Identify quartiles and IQR

Given data set: $2, 2, 3, 3, 4, 5, 5, 6, 7, 9, 12$ (11 values).

• Median (Q2) = 5 (6th value)
• Q1 = median of lower half: $2,2,3,3,4$ → $3$
• Q3 = median of upper half: $5,6,7,9,12$ → $7$
• Given IQR = 4 (matches $7-3=4$)

Step2: Calculate upper outlier threshold

Upper bound for non-outliers: $Q3 + 1.5 \times IQR$
$= 7 + 1.5 \times 4 = 7 + 6 = 13$

Step3: Compare 12 to threshold

$12 < 13$, so 12 is not an outlier. The sum $7+6$ is the upper bound.

Answer:

Twelve is not an outlier because it is less than the sum of 7 and 6.