Answer to Question #262125 in Statistics and Probability for Neutral

Question

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

Answer 1

$$x,20y, 4, 25, z$$ $$GM=\sqrt[6]{x(16)(20y)(4)(25)(z)}=8.433$$ $$AM=\dfrac{x+16+20y+4+25+z}{6}=11.25$$ $$HM=\dfrac{6}{\dfrac{1}{x}+\dfrac{1}{16}+\dfrac{1}{20y}+\dfrac{1}{4}+\dfrac{1}{25}+\dfrac{1}{z}}=6.026$$ $$x(20y)z=224.787739$$ $$x+20y+z=22.5$$ $$\dfrac{1}{x}+\dfrac{1}{20y}+\dfrac{1}{z}=0.643185$$ $$x+20y+z=22.5$$ $$x(20y)+x(z)+20y(z)=144.580101$$

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

Vieta's Theorem

$$t^3-22.5t^2+144.580101t-224.787739=0$$

Solve graphically

$$t_1=2.285, t_2=8.166, t_3=12.049$$

Answer

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

Or

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

Or

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

Or

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

Or

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

Or

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