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Given-a-set-data-2-2-3-4-5-5-5-8-8-9-11-14-find-the-value-of-Q-1-




Question Number 113919 by bemath last updated on 16/Sep/20
Given a set data    : 2,2,3,4,5,5,5,8,8,9,11,14.  find the value of Q_1
$${Given}\:{a}\:{set}\:{data}\:\: \\ $$$$:\:\mathrm{2},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{5},\mathrm{5},\mathrm{8},\mathrm{8},\mathrm{9},\mathrm{11},\mathrm{14}. \\ $$$${find}\:{the}\:{value}\:{of}\:{Q}_{\mathrm{1}} \\ $$
Answered by bemath last updated on 16/Sep/20
i got Q_1 = X_((12+2)/4) =X_(3+0.5) = 3+0.5(4−3)      = 3+0.5 = 3.5
$${i}\:{got}\:{Q}_{\mathrm{1}} =\:{X}_{\frac{\mathrm{12}+\mathrm{2}}{\mathrm{4}}} ={X}_{\mathrm{3}+\mathrm{0}.\mathrm{5}} =\:\mathrm{3}+\mathrm{0}.\mathrm{5}\left(\mathrm{4}−\mathrm{3}\right) \\ $$$$\:\:\:\:=\:\mathrm{3}+\mathrm{0}.\mathrm{5}\:=\:\mathrm{3}.\mathrm{5}\: \\ $$
Commented by bemath last updated on 16/Sep/20
but my friend answer is   Q_1 = X_((12+1)/4)  = X_(3+0.25) =3+0.25(4−3)    = 3.25
$${but}\:{my}\:{friend}\:{answer}\:{is}\: \\ $$$${Q}_{\mathrm{1}} =\:{X}_{\frac{\mathrm{12}+\mathrm{1}}{\mathrm{4}}} \:=\:{X}_{\mathrm{3}+\mathrm{0}.\mathrm{25}} =\mathrm{3}+\mathrm{0}.\mathrm{25}\left(\mathrm{4}−\mathrm{3}\right) \\ $$$$\:\:=\:\mathrm{3}.\mathrm{25} \\ $$$$ \\ $$
Answered by MJS_new last updated on 16/Sep/20
q_1  is the median of the lower half  q_1 =((3+4)/2)=3.5
$${q}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\mathrm{the}\:\mathrm{lower}\:\mathrm{half} \\ $$$${q}_{\mathrm{1}} =\frac{\mathrm{3}+\mathrm{4}}{\mathrm{2}}=\mathrm{3}.\mathrm{5} \\ $$
Commented by bemath last updated on 16/Sep/20
thank you. i′m correct
$${thank}\:{you}.\:{i}'{m}\:{correct} \\ $$
Commented by ShakaLaka last updated on 16/Sep/20
yes, if we follow the formula  for even  (1/2)[((n/4))^(th) value + ((n/4)+1)^(th) value]  then then the answer will be  3.5
$${yes},\:{if}\:{we}\:{follow}\:{the}\:{formula} \\ $$$${for}\:{even} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\left(\frac{{n}}{\mathrm{4}}\right)^{{th}} {value}\:+\:\left(\frac{{n}}{\mathrm{4}}+\mathrm{1}\right)^{{th}} {value}\right] \\ $$$${then}\:{then}\:{the}\:{answer}\:{will}\:{be} \\ $$$$\mathrm{3}.\mathrm{5} \\ $$

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