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Question-201972




Question Number 201972 by Lekhraj last updated on 17/Dec/23
Answered by mr W last updated on 18/Dec/23
𝚽_(≀x) =(1/2)[1+erf(((xβˆ’π›)/( (√2) 𝛔)))]  with erf(x)=(2/( (βˆšπ›‘)))∫_0 ^x e^(βˆ’t^2 ) dt    (i)(a)  (1/2)[1+erf(((45βˆ’60)/(10(√2))))]     =(1/2)(1βˆ’0.886)β‰ˆ6%  (i)(b)  (1/2)[erf(((75βˆ’60)/(10(√2))))βˆ’erf(((50βˆ’60)/(10(√2))))]     =(1/2)(0.866βˆ’(βˆ’0.683))β‰ˆ77%  (ii)  (1/2)[1+erf(((xβˆ’60)/(10(√2))))]=0.9  erf(((xβˆ’60)/(10(√2))))=0.8  ((xβˆ’60)/(10(√2)))=erf^(βˆ’1) (0.8)=0.9062  β‡’x=60+0.9062Γ—10(√2)β‰ˆ73(%)
$$\boldsymbol{\Phi}_{\leqslant\boldsymbol{{x}}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\boldsymbol{{erf}}\left(\frac{\boldsymbol{{x}}βˆ’\boldsymbol{\mu}}{\:\sqrt{\mathrm{2}}\:\boldsymbol{\sigma}}\right)\right] \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{erf}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}}{\:\sqrt{\boldsymbol{\pi}}}\int_{\mathrm{0}} ^{\boldsymbol{{x}}} \boldsymbol{{e}}^{βˆ’\boldsymbol{{t}}^{\mathrm{2}} } \boldsymbol{{dt}} \\ $$$$ \\ $$$$\left({i}\right)\left({a}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{erf}\left(\frac{\mathrm{45}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}βˆ’\mathrm{0}.\mathrm{886}\right)\approx\mathrm{6\%} \\ $$$$\left({i}\right)\left({b}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[{erf}\left(\frac{\mathrm{75}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)βˆ’{erf}\left(\frac{\mathrm{50}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right] \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}.\mathrm{866}βˆ’\left(βˆ’\mathrm{0}.\mathrm{683}\right)\right)\approx\mathrm{77\%} \\ $$$$\left({ii}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+{erf}\left(\frac{{x}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)\right]=\mathrm{0}.\mathrm{9} \\ $$$${erf}\left(\frac{{x}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}\right)=\mathrm{0}.\mathrm{8} \\ $$$$\frac{{x}βˆ’\mathrm{60}}{\mathrm{10}\sqrt{\mathrm{2}}}={erf}^{βˆ’\mathrm{1}} \left(\mathrm{0}.\mathrm{8}\right)=\mathrm{0}.\mathrm{9062} \\ $$$$\Rightarrow{x}=\mathrm{60}+\mathrm{0}.\mathrm{9062}Γ—\mathrm{10}\sqrt{\mathrm{2}}\approx\mathrm{73}\left(\%\right) \\ $$

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