Question Number 201972 by Lekhraj last updated on 17/Dec/23
Answered by mr W last updated on 18/Dec/23
$$\boldsymbol{\Phi}_{\leqslant\boldsymbol{{x}}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\boldsymbol{{erf}}\left(\frac{\boldsymbol{{x}}β\boldsymbol{\mu}}{\:\sqrt{\mathrm{2}}\:\boldsymbol{\sigma}}\right)\right] \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{erf}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}}{\:\sqrt{\boldsymbol{\pi}}}\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{e}}^{β\boldsymbol{{t}}^{\mathrm{2}} } \boldsymbol{{dt}} \\ $$$$ \\ $$$$\left({i}\right)\left({a}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{erf}\left(\frac{\mathrm{45}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}β\mathrm{0}.\mathrm{886}\right)\approx\mathrm{6\%} \\ $$$$\left({i}\right)\left({b}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{erf}\left(\frac{\mathrm{75}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)β{erf}\left(\frac{\mathrm{50}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{866}β\left(β\mathrm{0}.\mathrm{683}\right)\right)\approx\mathrm{77\%} \\ $$$$\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{erf}\left(\frac{{x}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right]=\mathrm{0}.\mathrm{9} \\ $$$${erf}\left(\frac{{x}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)=\mathrm{0}.\mathrm{8} \\ $$$$\frac{{x}β\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}={erf}^{β\mathrm{1}} \left(\mathrm{0}.\mathrm{8}\right)=\mathrm{0}.\mathrm{9062} \\ $$$$\Rightarrow{x}=\mathrm{60}+\mathrm{0}.\mathrm{9062}Γ\mathrm{10}\sqrt{\mathrm{2}}\approx\mathrm{73}\left(\%\right) \\ $$