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From a sample n=36, the mean duration of a geysers eruption is 3.76 minutes and the standard deviation is 1.21 minutes using chebychevs theorem. Determine at least how many of the eruptions lasted between 0.13 and 7.39 minutes

User Nicka
by
6.4k points

1 Answer

3 votes

\mathbb P(0.13<X<7.39)=\mathbb P\left(-3<(X-3.76)/(1.21)<3\right)=\mathbb P\left(|X-\mu|<3\sigma)

where
\mu is the mean duration and
\sigma is the standard deviation. By Chebychev's theorem, this probability is bounded below by


\mathbb P(|X-\mu|<3\sigma)\ge1-\frac1{3^2}=\frac89
User Neokoenig
by
6.8k points
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