Question

A probability distribution has a mean of 35 and a standard deviation of 3. Use Chebychev's inequality to find a bound on the probability that an outcome of the experiment lies between the following (a) 30 and 40 at least 0.64 % (b) 25 and 45 at least 0.91 X% Need Help? Read Watch

Answer 1

(a)

Now X is less than 30 or greater than 40 if and only if,

$| X - \mu| \geq k$

where, $\mu$ = 35 and $k = 5$. So the probability that X <= 30 or X >= 40 is,

$P(|X-\mu| \geq k) \leq \frac{\sigma^2}{k^2} = \frac{3^2}{5^2} = \frac{9}{25}$

So X will be between 30 and 40 with probability = 1 - (9/25) = 16/25 = 64% (not 0.64% as written in the answer box)

(b)

Now X is less than 25 or greater than 45 if and only if,

$| X - \mu| \geq k$

where, $\mu$ = 35 and $k = 10$. So the probability that X <= 25 or X >= 45 is,

$P(|X-\mu| \geq k) \leq \frac{\sigma^2}{k^2} = \frac{3^2}{10^2} = \frac{9}{100}$

So X will be between 30 and 40 with probability = 1 - (9/100) = 91/100 = 91% (not 0.91% as written in the answer box)

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