Suppose that X is a random variable with mean and variance both equal to 20. &nbsp;What can be said about P{0&lt;X&lt;40}?

Assume that X is a random variable with mean μ=20and variance σ2=20. Then, by Chebyshev's inequality,P{|X−20|≥20}≤σ2202=120.Therefore, since P{|X−20|≥20}=1−P{|X−20|&lt;20}, we have:P{|X−20|&lt;20}≥1−120=1920. (*)On the other hand,P{|X−20|&lt;20}=P{−20&lt;X−20&lt;20}=P{0&lt;X&lt;40}⇒(⋅)P{0&lt;X&lt;40}≥1920.Result:P{0&lt;X&lt;40}≥1920

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2022-08-19

Suppose that X is a random variable with mean and variance both equal to 20. What can be said about P{0<X<40}?

Answer & Explanation

Maeve Saunders

Beginner2022-08-20Added 2 answers

Assume that X is a random variable with mean

μ = 20

and variance

σ^{2} = 20

. Then, by Chebyshev's inequality,

P {| X - 20 | \geq 20} \leq \frac{σ^{2}}{20^{2}} = \frac{1}{20}

.
Therefore, since

P {| X - 20 | \geq 20} = 1 - P {| X - 20 | < 20}

, we have:

P {| X - 20 | < 20} \geq 1 - \frac{1}{20} = \frac{19}{20}

. (*)
On the other hand,

P {| X - 20 | < 20} = P {- 20 < X - 20 < 20} = P {0 < X < 40} \Rightarrow^{(\cdot)}

P {0 < X < 40} \geq \frac{19}{20}

.
Result:

P {0 < X < 40} \geq \frac{19}{20}

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