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From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that (i) all 4 will fire, (ii) at most 2 will not fire? Finally, use Chebyshev theorem to interpret the interval μ±2σ and μ±3σ. (Ans : 1/6, 29/30, (-1.04, 3.44)).

Can somebody explain to me "use Chebyshev theorem to interpret the interval μ±2σ and μ±3σ." ? I know the Chebyshev inequality but I am reading this sentence for the first time.

Thanks in advance.

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    Chebyshev's inequality gives bounds for the probability that a random variable is within $t$ standard deviations of its mean. In this case, $\mu\pm 2\sigma$ is for the probability that $X$ is within $2\sigma$ of its mean $\mu$. – 2017-02-23

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Chebyshev's inequality says that $$P(|X-\mu|>k\sigma) \le \frac{1}{k^2}$$ where $\mu$ is the mean of $X$ and $\sigma$ is its standard deviation. This is the probability of the random variable being more than $k$ standard deviations from the mean, and note that its maximum value goes down as $k$ gets large, as you would expect.

Another way to put this is that the interval $(\mu-k\sigma,\mu +k\sigma)$ will contain more than a $1-1/k^2$ fraction of the data. So for $k=2,$ Chebyshev says that at least $1-1/2^2=75\%$ of the data lie within two standard deviations. For $k=3$ it means that more than $1-1/3^2 \approx 89\%$ of the data lie within three standard deviations.

The answer you give appears to compute an interval, not interpret it. I'm guessing it's an interval for the number of missiles that successfully fire. In order to get that you need to compute the mean $\mu$ and standard deviation $\sigma$ of that random variable and plug it into $(\mu-k\sigma,\mu+k\sigma)$ for $k=2,3$ and then interpret the interval as above.

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    Thank you for the answer. I computed μ±2σ and μ±3σ and the answer seems to be (μ-3σ, μ+3σ). Do you know why? Like, why are we taking the bigger interval? – 2017-02-23
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    @AbhishekAgrawal No, the question doesn't even say to compute. Just know that the bigger interval will contain the number $8/9$ of the time if we repeated the experiment over and over. It's a little weird in this case since the variable is also discrete. So a better way to interpret it would be that there is a greater than $8/9$ probability that there will be $0,1,2,$ or $3$ failed launches. Of course we know there is actually an $100\%$ probability of this, consistent with Chebyshev. – 2017-02-23