Question

Smoking males in a given area have a mean life expectancy of 68.5 years, with a standard deviation of 5.3 years. The distribution of life expectancy is not assumed to be symmetric.

Between what two life expectancies does Chebyshev's Theorem guarantee that we will find at least 89% of smoking males?

Round your answers to the nearest tenth. Enter the bounds in ascending order.

Answer 1

Answer:

52.6, 84.4

Step-by-step explanation:

Answer 2

Chebyshev's Theorem guarantee that we will find at least 89% of smoking males in the life expectancy range [52.6, 84.4].

What is Chebyshev's inequality?

In probability theory, Chebyshev’s inequality guarantees that, for a wide class of probability distributions, no more than a certain fraction of values can be more than a certain distance from the mean.

Let X be a random variable with mean μ with a finite variance σ², then for any real number k > 0, P(|X-μ| < kσ) ≥ 1-1/k².

Given,

Mean life expectancy = μ = 68.5 years

Standard deviation = σ = 5.3 years

P(|X-μ| < kσ) ≥ 1-1/k²

P(|X-68.5| < 5.3k) ≥ 1-1/k²

P(|X-68.5| < 5.3k) ≥ 0.89

$1 - \frac{1}{k^{2} } = 0.89$

$\frac{1}{k^{2} } = 0.11\\ \\\frac{1}{k^{2} } = \frac{1}{9} \\\\k = 3$

Between the following two life expectancies does Chebyshev's Theorem guarantee that we will find at least 89% of smoking males:

= [μ - 3σ, μ + 3σ]

= [68.5 - 3*5.3, 68.5 + 3*5.3]

= [52.6, 84.4]

Learn more about Chebyshev's inequality here

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