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Question-209026

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Question Number 209026 by Spillover last updated on 30/Jun/24
Answered by Spillover last updated on 07/Jul/24
f(x)=((4/x))((1/4))^x ((3/4))^(4−x) x=0,1,2,3,4 f(1)=((4/1))((1/4))^1 ((3/4))^(4−1) =((4/1))((1/4))((3/4))^3 =(9/(64)) f(2)=((4/2))((1/4))^2 ((3/4))^(4−2) =((4/2))((1/(16)))((3/4))^2 =((18)/(96)) f(3)=((4/3))((1/4))^3 ((3/4))^(4−3) =((4/3))((1/4))^3 ((3/4))^1 =(1/(64)) f(4)=((4/4))((1/4))^4 ((3/4))=(1/(256)) p(x<4)=Σ_(x=1) ^(4−1) f(x)=f(1)+f(2)+f(3) =(9/(64))+((18)/(96))+(1/(256))=0.34375<0.5 also p(x≤4) f(1)+f(2)+f(3)+f(4) =(9/(64))+((18)/(96))+(1/(64))+(1/(256)) =((2839)/(5888))=0.482 0.482≤0.5 mode=4
$${f}\left({x}\right)=\left(\frac{\mathrm{4}}{{x}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{{x}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−{x}} \:\:\: \\ $$$${x}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$${f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−\mathrm{1}} =\left(\frac{\mathrm{4}}{\mathrm{1}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} =\frac{\mathrm{9}}{\mathrm{64}}\:\:\: \\ $$$${f}\left(\mathrm{2}\right)=\left(\frac{\mathrm{4}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−\mathrm{2}} =\left(\frac{\mathrm{4}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{16}}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{18}}{\mathrm{96}} \\ $$$${f}\left(\mathrm{3}\right)=\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{4}−\mathrm{3}} =\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{64}} \\ $$$${f}\left(\mathrm{4}\right)=\left(\frac{\mathrm{4}}{\mathrm{4}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{256}} \\ $$$${p}\left({x}<\mathrm{4}\right)=\underset{{x}=\mathrm{1}} {\overset{\mathrm{4}−\mathrm{1}} {\sum}}{f}\left({x}\right)={f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right) \\ $$$$=\frac{\mathrm{9}}{\mathrm{64}}+\frac{\mathrm{18}}{\mathrm{96}}+\frac{\mathrm{1}}{\mathrm{256}}=\mathrm{0}.\mathrm{34375}<\mathrm{0}.\mathrm{5} \\ $$$${also}\:\: \\ $$$${p}\left({x}\leqslant\mathrm{4}\right)\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+{f}\left(\mathrm{3}\right)+{f}\left(\mathrm{4}\right) \\ $$$$=\frac{\mathrm{9}}{\mathrm{64}}+\frac{\mathrm{18}}{\mathrm{96}}+\frac{\mathrm{1}}{\mathrm{64}}+\frac{\mathrm{1}}{\mathrm{256}}\:=\frac{\mathrm{2839}}{\mathrm{5888}}=\mathrm{0}.\mathrm{482} \\ $$$$\mathrm{0}.\mathrm{482}\leqslant\mathrm{0}.\mathrm{5} \\ $$$${mode}=\mathrm{4} \\ $$$$ \\ $$

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