SOLUTION
Given the question in the image, the following are the solution steps to answer the question.
STEP 1: Write the given data
[tex]\lbrace14.9,21.1,21.2,8.4,14.5,5.9,7.6,10.0,4.8,3.2,28.7,29.5\rbrace[/tex]STEP 2: Find the mean ofthe data
[tex]\begin{gathered} The\:arithemtic\:mean\:\left(average\right)\:is\:the\:sum\:of\:the\:values\:in\:the\:set\:divided\:by\:the\:number\:of\:elements\:in\:that\:set. \\ \mathrm{If\:our\:data\:set\:contains\:the\:values\:}a_1,\:\ldots \:,\:a_n\mathrm{\:\left(n\:elements\right)\:then\:the\:average}=\frac{1}{n}\sum _{i=1}^na_i\: \\ Sum=169.8 \\ n=12 \\ mean=\frac{169.8}{12} \\ mean=14.15 \end{gathered}[/tex]STEP 3: Find the median
[tex]\begin{gathered} \mathrm{The\:median\:is\:the\:value\:separating\:the\:higher\:half\:of\:the\:data\:set,\:from\:the\:lower\:half.} \\ \:the\:number\:of\:terms\:is\:odd,\:then\:the\:median\:is\:the\:middle\:element\:of\:the\:sorted\:set \\ If\:the\:number\:of\:terms\:\:is\:even,\:then\:the\:median\:is\:the\:arithmetic\:mean\:of\:the\:two\:middle\:elements\:of\:the\:sorted\:set \\ \\ \mathrm{Arrange\:the\:terms\:in\:ascending\:order} \\ 3.2,\:4.8,\:5.9,\:7.6,\:8.4,\:10,\:14.5,\:14.9,\:21.1,\:21.2,\:28.7,\:29.5 \\ median=12.25 \end{gathered}[/tex]Hence, it can be seen here that the mean is larger than median.
STEP 4: Find the Interquartile range
[tex]\begin{gathered} The\:interquartile\:range\:is\:the\:difference\:of\:the\:first\:and\:third\:quartiles \\ First\text{ Quartile}=6.75 \\ Third\text{ quartile}=21.15 \\ IQR=14.4 \end{gathered}[/tex]STEP 5: Find the standard deviation
[tex]\begin{gathered} \mathrm{The\:standard\:deviation,\:}\sigma \left(X\right)\mathrm{,\:is\:the\:square\:root\:of\:the\:variance:\quad }\sigma \left(X\right)=\sqrt{\frac{\sum _{i=1}^n\left(x_i-\bar{x}\right)^2}{n-1}} \\ Standard\text{ deviation}=9.11836 \end{gathered}[/tex]Hence, it can be seen from above that the interquartile range is larger than the standard deviation.
STEP 6: Find the range
[tex]\begin{gathered} \mathrm{The\:range\:of\:the\:data\:is\:the\:difference\:between\:the\:maximum\:and\:the\:minimum\:of\:the\:data\:set} \\ Minimum=3.2 \\ Maximum=29.5 \\ Range=26.3 \end{gathered}[/tex]STEP 7: Fnd the variance
[tex]\begin{gathered} \mathrm{The\:sample\:variance\:measures\:how\:much\:the\:data\:is\:spread\:out\:in\:the\:sample.} \\ \mathrm{For\:a\:data\:set\:}x_1,\:\ldots \:,\:x_n\mathrm{\:\left(n\:elements\right)\:with\:an\:average}\:\bar{x}\mathrm{,\:}Var\left(X\right)=\sum _{i=1}^n\frac{\left(x_i-\bar{x}\right)^2}{n-1} \\ Variance=83.14454 \end{gathered}[/tex]Hence, it can be seen that the range is not larger than the variance.
Therefore, the answer is I and II only.
