Determine the quartile or interquartile range as specified.
-The normal annual precipitation (in inches)is given below for 21 different U.S. cities. Find the first quartile,
$\mathrm { Q } _ { 1 } \text {. }$
$\begin{array} { l l l l l l l } 39.1 & 16.4 & 25.4 & 18.4 & 27.1 & 27.8 & 30.6 \end{array}$
$\begin{array} { l l l l l l l l } 15.2 & 42.6 & 18.4 & 13.5 & 19.8 & 32.3 & 10.0 \end{array}$
$\begin{array} { l l l l l l l } 14.8 & 33.6 & 12.9 & 35.0 & 22.3 & 11.2 & 51.7 \end{array}$
A)$15.2 \mathrm { in } .$
B) $14.8 \mathrm { in } .$
C) $14.900 \mathrm { in }$.
D) $15.00 \mathrm { in }$.

Question

Determine the quartile or interquartile range as specified.
-The normal annual precipitation (in inches)is given below for 21 different U.S. cities. Find the first quartile, 
$\mathrm { Q } _ { 1 } \text {. }$ 
$\begin{array} { l l l l l l l } 39.1 & 16.4 & 25.4 & 18.4 & 27.1 & 27.8 & 30.6 \end{array}$
$\begin{array} { l l l l l l l l } 15.2 & 42.6 & 18.4 & 13.5 & 19.8 & 32.3 & 10.0 \end{array}$
$\begin{array} { l l l l l l l } 14.8 & 33.6 & 12.9 & 35.0 & 22.3 & 11.2 & 51.7 \end{array}$  
A)$15.2 \mathrm { in } .$
B) $14.8 \mathrm { in } .$
C) $14.900 \mathrm { in }$.
D) $15.00 \mathrm { in }$.

Quizplus · Accepted Answer

The first quartile is the median of the lower half of the data. To find the first quartile, first order the data from least to greatest:$10.0, 11.2, 12.9, 13.5, 14.8, 15.2, 16.4, 18.4, 18.4, 19.8, 22.3, 25.4, 27.1, 27.8, 30.6, 32.3, 33.6, 35.0, 39.1, 42.6, 51.7$The median of the lower half of the data is the average of the 11th and 12th values, which are 15.2 and 16.4, respectively. Therefore, the first quartile is $\mathrm{Q}_1 = 15.2 \mathrm{in}$.

Determine the Quartile or Interquartile Range as Specified Q1. \mathrm { Q } _ { 1 } \text {. }Q1​.

Determine the Quartile or Interquartile Range as Specified $\mathrm { Q } _ { 1 } \text {. }$