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I have the following problem :

Determine what age interval will contain at least 95% of the data (Chebyshev's) ?

Now, I have standard deviation of 1.516, mean of 19.211.

The formula is $1-(1/k^2) = .95$ So I solve for k to get $\sqrt 2$.

Now, I calculate the interval by mean + $\sqrt 2$*standard deviation = RIGHT AGE INTERVAL

Why is this wrong? I get approximately $21.3578$.

The left interval, let's just skip it for now.

Is my logic off?

FabolousPotato
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1 Answers1

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Your calculations do have an error: your $\sqrt{2}$ should be $\sqrt{\frac{1}{1-0.95}}=\sqrt{20}$.

You seem to have started with Chebyshev's inequality of the form $$\mathbb P(|X-\mu|\le k \sigma) \ge 1-\frac1{k^2}$$

The right hand side is $0.95$ when $k=\sqrt{20} \approx 4.472136$ so your interval should be $[\mu-k\sigma, \mu+k \sigma]$ and in this case $[12.431,25.991]$

Henry
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