Question

Suppose a distribution has a mean of 111 and standard deviation of 7.6. If Chebyshev’s Theorem tells us that 81.1% of the values are between a and b (symmetrical about the mean), then what are these values?

Answer 1

Hi,

Explanation:

Within k standard deviations of the mean,

`\mu\ \pm\ k*\sigma`, lie at least

`1-(1)/(k^2)`

proportion of the values ( k>1)

`1-(1)/(k^2) =(81.1)/(100) \\\\(1)/(k^2) =(100-81.1)/(100) \\\\(1)/(k^2) =(18.9)/(100) \\\\k^2=5.29100...\\\\k=2.30021853...\approx{2.3}\\\\a=111-2.3*7.6\approx{93.5}\\\\b=111+2.3*7.6\approx{128.5}\\`