Question
4.67 A random variable $X$ has a mean $p=10$ and variance $\sigma^{2}=4$. Using Chebyshev's theorem, find(a) $P(X-10 \mid \geq 3)$(b) $\mathrm{P}(|\mathrm{X}-10|<3)$(c) $P(b<X<15)$(d) the value of the constant $\mathrm{c}$ such that$$P(X-10 \mid \geq \mathrm{c}) \leq 0.04$$
Step 1
According to Chebyshev's theorem, the probability of a random variable $X$ being within $k$ standard deviations of the mean is at least $1 - \frac{1}{k^{2}}$. That is, $P(\mu - k\sigma \leq X \leq \mu + k\sigma) \geq 1 - \frac{1}{k^{2}}$. Show more…
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