Q:

From a sample with n=20​, 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. At least nothing of the households have between 0 and 4 televisions.

Accepted Solution

A:
Answer:At least 15 households have between 0 and 4 televisions.Step-by-step explanation:Chebychev's Theorem:P(|X-μ|>kσ)≤[tex]\frac{1}{k^2}[/tex]μ= meanσ= standard deviationk= number of typical deviationsIt is also expressed as:P(μ-kσ<X<μ+kσ)≥[tex]1-\frac{1}{k^2}[/tex]Using​ Chebychev's Theorem, we determine: P(0<X<4)μ-kσ=0 ⇒ 2-k·1=0 ⇒k=2μ+kσ=4 ⇒ 2+k·1=4 ⇒k=2P(0<X<4)=[tex]1-\frac{1}{2^2}[/tex]=0.75For a sample n=20,0.75×20=15At least 15 households have between 0 and 4 televisions.