This is what do I have to solve:

Byron Book: Exercise 8.3 chapter 8 Verify the use of Chebyshev’s inequality in (8.6) of Example 8.16. Show that if the population mean is indeed 48.2333 and the population standard deviation is indeed 26.5170, then at least 8/9 of all tasks require less than 127.78 seconds of CPU time.

This is how I did it :

So first I calculated: 127.78-48.23 = 79.5

Then I wrote P(X<127) < (26.5^2)/(79.5^2) = 0.11

I don't understand if I did it right because I did not use this part : at least 8/9 of... and I think it is important.

Can anyone help me with this ? Thank you in advance.

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    $\begingroup$ Can you please elaborate, what does the population samples represent? How they are related to tasks and CPU time? $\endgroup$ Dec 18, 2020 at 19:05
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    $\begingroup$ Can you please specify which book or resource are you referring to? $\endgroup$ Dec 18, 2020 at 19:13
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    $\begingroup$ Essentially, you didn't formulate the problem. A lot of things are missing. $\endgroup$
    – user114966
    Dec 18, 2020 at 20:00
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    $\begingroup$ You need to check Chebyshev's inequality. It would give you a bound of $P(X<127.78)$ from below, not from above. You could use this version of the inequality. It is more direct for the type of bound that you need. Just put $\mu=48.2333$, $\sigma=26.5170$, $u=127.78$ and let $l\to0$ from below. Then, hopefully, the value of $\frac{4[(\mu-l)(u-\mu)-\sigma^2]}{(l-u)^2}$ is greater than or equal to $8/9$. $\endgroup$
    – plop
    Dec 19, 2020 at 14:50
  • $\begingroup$ plop yes this is what I was looking for. Thank you. In the end what I did was right I talked with a teacher about that the only thing missing is that I have to do 1- 0.11 = 0.89 which is 8/9. I do that because I need to find the > not < $\endgroup$
    – Alex97
    Dec 19, 2020 at 20:03

1 Answer 1


The exercise above is right but is missing something. You need to do 1-0.11 since you need to find the <, not the >. When I wrote 0.11 I have found the > but I need the < so 1-.. .


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