Question Number 113919 by bemath last updated on 16/Sep/20
$${Given}\:{a}\:{set}\:{data}\:\: \\ $$$$:\:\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{8},\mathrm{8},\mathrm{9},\mathrm{11},\mathrm{14}. \\ $$$${find}\:{the}\:{value}\:{of}\:{Q}_{\mathrm{1}} \\ $$
Answered by bemath last updated on 16/Sep/20
$${i}\:{got}\:{Q}_{\mathrm{1}} =\:{X}_{\frac{\mathrm{12}+\mathrm{2}}{\mathrm{4}}} ={X}_{\mathrm{3}+\mathrm{0}.\mathrm{5}} =\:\mathrm{3}+\mathrm{0}.\mathrm{5}\left(\mathrm{4}−\mathrm{3}\right) \\ $$$$\:\:\:\:=\:\mathrm{3}+\mathrm{0}.\mathrm{5}\:=\:\mathrm{3}.\mathrm{5}\: \\ $$
Commented by bemath last updated on 16/Sep/20
$${but}\:{my}\:{friend}\:{answer}\:{is}\: \\ $$$${Q}_{\mathrm{1}} =\:{X}_{\frac{\mathrm{12}+\mathrm{1}}{\mathrm{4}}} \:=\:{X}_{\mathrm{3}+\mathrm{0}.\mathrm{25}} =\mathrm{3}+\mathrm{0}.\mathrm{25}\left(\mathrm{4}−\mathrm{3}\right) \\ $$$$\:\:=\:\mathrm{3}.\mathrm{25} \\ $$$$ \\ $$
Answered by MJS_new last updated on 16/Sep/20
$${q}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{half} \\ $$$${q}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{4}}{\mathrm{2}}=\mathrm{3}.\mathrm{5} \\ $$
Commented by bemath last updated on 16/Sep/20
$${thank}\:{you}.\:{i}'{m}\:{correct} \\ $$
Commented by ShakaLaka last updated on 16/Sep/20
$${yes},\:{if}\:{we}\:{follow}\:{the}\:{formula} \\ $$$${for}\:{even} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{{n}}{\mathrm{4}}\right)^{{th}} {value}\:+\:\left(\frac{{n}}{\mathrm{4}}+\mathrm{1}\right)^{{th}} {value}\right] \\ $$$${then}\:{then}\:{the}\:{answer}\:{will}\:{be} \\ $$$$\mathrm{3}.\mathrm{5} \\ $$