Respuesta :
8) The interquartile range for A is less than that of B. This means data in A vary less than in B.
Answer is A
9) The interquartile range is given as
[tex]\begin{gathered} Q_3-Q_{1\text{ }}\text{where} \\ Q_3term=\text{ third quartile} \\ Q_1term=\text{ First quartile} \\ Q_{3\text{ }}term=\text{ }\frac{3}{4}(n+1)\text{ where} \\ n\text{ is frequency =5} \\ Q_3term=\text{ }\frac{3}{4}(5+1)=4\frac{1}{2} \\ Q_1term=\text{ }\frac{1}{4}(5+1)=1\frac{1}{2} \end{gathered}[/tex][tex]\begin{gathered} To\text{ get the interquartile range, we arrange the data in ascending order to be} \\ A\colon\text{ 6,11,15,20,23} \\ TheQ_1in\text{ the dataset lies halfway between the 1st and 2nd t}erm\text{ (1}\frac{1}{2}\text{) } \\ \text{Thus it is given as the midpoint = }\frac{6+11}{2}=\frac{17}{2}=8.5 \\ \text{The Q}_3\text{ lie between the 4th and 5th term, thus, }\frac{20+23}{2}=\frac{43}{2}=21.5 \end{gathered}[/tex]Thus, the interquartile range = 21.5 - 8.5 = 13
[tex]\begin{gathered} To\text{ get the interquartile range, we arrange the data in ascending order to be} \\ B\colon\text{ 9,10,14,15,18} \\ TheQ_1in\text{ the dataset lies halfway between the 1st and 2nd t}erm\text{ (1}\frac{1}{2}\text{) } \\ \text{Thus it is given as the midpoint = }\frac{9+10}{2}=\frac{19}{2}=9.5 \\ \text{The Q}_3\text{ lie between the 4th and 5th term, thus, }\frac{15+18}{2}=\frac{33}{2}=16.5 \end{gathered}[/tex]Thus, the interquartile range = 16.5 - 9.5 = 7
