Answer to Question #262176 in Statistics and Probability for Aashish

Question #262176

How do I find the values of more than one data in a set given the values of harmonic geometric and arithmetic mean i.e


In a study a set of observations were as follows:x,16,20b,4,25,c.Given that the geometric mean is 8.433 the arithmetic mean 11.25 and the harmonic mean 6.026 compute the values x y z



1
Expert's answer
2021-11-15T19:27:09-0500
x,20y,4,25,zx,20y, 4, 25, z

GM=x(16)(20y)(4)(25)(z)6=8.433GM=\sqrt[6]{x(16)(20y)(4)(25)(z)}=8.433

AM=x+16+20y+4+25+z6=11.25AM=\dfrac{x+16+20y+4+25+z}{6}=11.25

HM=61x+116+120y+14+125+1z=6.026HM=\dfrac{6}{\dfrac{1}{x}+\dfrac{1}{16}+\dfrac{1}{20y}+\dfrac{1}{4}+\dfrac{1}{25}+\dfrac{1}{z}}=6.026

x(20y)z=224.787739x(20y)z=224.787739

x+20y+z=22.5x+20y+z=22.5

1x+120y+1z=0.643185\dfrac{1}{x}+\dfrac{1}{20y}+\dfrac{1}{z}=0.643185

x+20y+z=22.5x+20y+z=22.5

x(20y)+x(z)+20y(z)=144.580101x(20y)+x(z)+20y(z)=144.580101



x(20y)z=224.787739x(20y)z=224.787739

Vieta's Theorem


t322.5t2+144.580101t224.787739=0t^3-22.5t^2+144.580101t-224.787739=0

Solve graphically


t1=2.285,t2=8.166,t3=12.049t_1=2.285, t_2=8.166, t_3=12.049

Answer

x=2.285,y=0.4083,z=12.049x=2.285, y=0.4083, z=12.049

Or


x=2.285,y=0.60245,z=8.166x=2.285, y=0.60245, z=8.166

Or


x=8.166,y=0.11425,z=12.049x=8.166, y=0.11425, z=12.049

Or


x=8.166,y=0.60245,z=2.285x=8.166, y=0.60245, z=2.285

Or


x=12.049,y=0.11425,z=8.166x=12.049, y=0.11425, z=8.166

Or


x=12.049,y=0.4083,z=2.285x=12.049, y=0.4083, z=2.285


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