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I have a normal approximation of binomial, $X\sim \mathcal{N}(36,4.6475)$. I have to find the probability that the number of red blocks ($\mu = 36$) differs from its expected value by less than $10\%$. I know how to calculate normal distribution, I just don't understand how exactly I should work with these percentages and expected value?

Heeeelp. I tried calculating it between 32.4 and 39.6 But it should be between 32.5 and 39.5 why is that?

Deo
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1 Answers1

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Hint:

To be found is $P\left[\left|X-36\right|<3.6\right]$ since $X$ differs from its expectation by less than 10% iff $\left|X-36\right|<3.6$.

drhab
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  • Ah thanks, that is all i had trouble visualising. Thank you so much :) – Deo Apr 26 '14 at 14:19
  • Although I do understand that logic, If I try to find the normCdf between 36-3.6 and 36+3.6 the answer i get is 0.561 This is normal approximation of binomial (90,0.4).

    I need my answer to be 0.5486, which is less. Do you have an idea of how to achieve that?

     – Deo Apr 26 '14 at 14:25
  • Did you substitute the right parameter? Sometimes people work with $N\left(\mu,\sigma^{2}\right)$ but other times with $N\left(\mu,\sigma\right)$. If that's not the problem then I cannot help you. – drhab Apr 26 '14 at 14:35
  • Can you just tell me what exactly you get for this question? :) – Deo Apr 26 '14 at 14:39
  • Sorry but I can't. I have not the disposal of tables, calculators or that sort of things. In some way I even 'hate' them. – drhab Apr 26 '14 at 14:41
  • Try this one: $P\left[\left|X-36\right|<0.1\times X\right]$. Here the difference between $X$ and its expectation is less than $10%$ of $X$ (instead of less than $10%$ of its expectation). If it works then let me know. In that case I will adapt my answer. – drhab Apr 26 '14 at 14:57