MMerorcn ABe Bbl6opKn Ana Hekoroporo npu3Haka. Tlepsas Bbl6opka: 19,45,57,54,16,45,55,70 Bropan Bbl6opka: 42,44,58,60,19,44,60,69,60,58,33,5 Vicnonb3yn U - Kpurepur MaHHa-YuTHu, npu ypoBHe 3HaMMOCTH alpha = 0,05 BbIRCHMTb, 3HauMMO nu pasnnyne MexAy AaHHbIMM. Bbl6paTb unbHbIN OTBeT (OTBer oxpyrnutb Ao Aecarbix): Bbl6epure oAMH orBer: a. U_(tann)=46MU_(kpor)=22,H_(0) orgepraercs, pasnwwe 3HaMMMO b. U_(tand)=46MU_(epstr)=22,H_(0) npummmaerca, pasnmuwe He3HaunMo C. U_(nakn)=40,5MU_(epur)=22,H_(0) pasnuune 3Ha4nMO A d. U_(tand)=40,5MU_(kpotr)=22,H_(0) pa3nmune He3HauHMO Ouncturb Mon Bbl6op

To determine the correct answer, we need to calculate the mean and standard deviation for both sets of numbers and then compare them to the given options.<br /><br />For the first set of numbers ($19, 45, 57, 54, 16, 45, 55, 70$):<br /><br />Mean ($\mu$) = (19 + 45 + 57 + 54 + 16 + 45 + 55 + 70) / 8 = 42.25<br /><br />Standard Deviation ($\sigma$) = $\sqrt{\frac{\sum (x_i - \mu)^2}{n}}$ = $\sqrt{\frac{(19-42.25)^2 + (45-42.25)^2 + (57-42.25)^2 + (54-42.25)^2 + (16-42.25)^2 + (45-42.25)^2 + (55-42.25)^2 + (70-42.25)^2}{8}}$ = 14.85<br /><br />For the second set of numbers ($42, 44, 58, 60, 19, 44, 60, 69, 60, 58, 33, 5$):<br /><br />Mean ($\mu$) = (42 + 44 + 58 + 60 + 19 + 44 + 60 + 69 + 60 + 58 + 33 + 5) / 12 = 44.25<br /><br />Standard Deviation ($\sigma$) = $\sqrt{\frac{\sum (x_i - \mu)^2}{n}}$ = $\sqrt{\frac{(42-44.25)^2 + (44-44.25)^2 + (58-44.25)^2 + (60-44.25)^2 + (19-44.25)^2 + (44-44.25)^2 + (60-44.25)^2 + (69-44.25)^2 + (60-44.25)^2 + (58-44.25)^2 + (33-44.25)^2 + (5-44.25)^2}{12}}$ = 16.25<br /><br />Now, let's compare these values to the given options:<br /><br />a. $U_{tann}=46MU_{kpor}=22,H_{0}$ orgepraercs, pasnwwe 3HaMMMO<br />b. $U_{tand}=46MU_{epstr}=22,H_{0}$ npummmaerca, pasnmuwe He3HaunMo<br />c. $U_{nakn}=40,5MU_{epur}=22,H_{0}$ pasnuune 3Ha4nMO<br />d. $U_{tand}=40,5MU_{kpotr}=22,H_{0}$ pa3nmune He3HauHMO<br /><br />The correct answer is option c. $U_{nakn}=40,5MU_{epur}=22,H_{0}$ pasnuune 3Ha4nMO.