From a sample with nequals12, the mean number of televisions per household is 2 with a standard deviation of 1 television. Using Chebychev's Theorem, determine at least how many of the households have between 0 and 4 televisions.

[tex]P(0\leq X\leq 4)=P(0-2\leq (X-\mu) \leq 4-2 )\\=P(-2\leq (X-\mu) \leq 2)\\=P(|X-\mu|\leq (2\times 1))\\\geq 1-\frac{1}{2^{2}} =1-\frac{1}{4} =1-0.25=0.75[/tex]

This implies that at least 75% of the 12 households have between 0 and 4 televisions.

Compute 75% of n = 12 as follows:

[tex]75\%\ of\ 12=\frac{75}{100}\times12= 9[/tex]

Thus, at least 9 households have between 0 and 4 televisions.

From a sample with nequals12​, the mean number of televisions per household is 2 with a standard deviation of 1 television. Using​ Chebychev's Theorem, determine at least how many of the households have between 0 and 4 televisions.

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From a sample with nequals12, the mean number of televisions per household is 2 with a standard deviation of 1 television. Using Chebychev's Theorem, determine at least how many of the households have between 0 and 4 televisions.