What is the interquartile range (IQR) of this data set? 2, 6, 7, 11, 15, 16, 17

A.9
B.7
C.10
D.15

11 is the median of the entire data set

6 is the median of the lower half and 16 is the median of the upper half data scores...

Subtract 16 from 6 gives 10 so the answer is C) 10

Explanation:

The first step is to sort the data from smallest to largest. Luckily, that's already been done for us.

Next, we find the median. This is the middle most number of the set. In this case, the median is 11 since we have 3 values below it (2,6,7) and 3 values above it (15,16,17).

Let

L = {2,6,7}

U = {15,16,17}

where L and U are the lower and upper sets respectively.

In other words, anything in set L is lower than the median while anything in set U is above the median. The median itself (11) is not part of either set.

Notice how 6 is the midpoint of set L, while 16 is the midpoint of set U.

This means Q1 = 6 and Q3 = 16

Therefore,

IQR = Q3 - Q1

IQR = 16 - 6

IQR = 10

which is why choice C is the final answer.

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.