A random variable X has a mean p = 10 and a variance σ2 = 4. Using Chebyshev's theorem, find. (a) P(X – 10| > 3);. (b) P (| X – 10 | < 3);. (c) P(b < X < 15);. (d) The value of the constant c such that P(X – 10| > c) < 0.04..
Added by Ruth L.
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Mathematically, this is expressed as: \[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2} \] \[ P(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \] Show more…
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