I have been battling with the question for a while. I need help.

Given a normal distribution graph see graph here:

enter image description here

A) Find how many standard deviation from the mean, A needs to be so the shaded area will amount to 0.5%.

What i did: I added all the values on the y-axis and divide the sum by 9 to get the mean value 0.2. And then calculated the variance to be 0.11, and the standard deviation to be 0.01222 and used the normal distribution formula to get x, which gives me 0.2.

Is this correct?

B) Given 1000 observation of the variable, use Poisson distribution to find the possibility that 5 of this observations are in the shaded area.

What i did: $\Pr(X = 5) = \dfrac{e^{-500} \cdot 500^5}{5!}$

This i know is wrong, but i don't know what next to do.

  • $\begingroup$ Keep in mind that the y-axis has values between the labeled tick marks. Can't forget about those. (For example, 0.08 is on the y-axis too.) $\endgroup$ Oct 25, 2016 at 20:34
  • $\begingroup$ true that. Thanks. $\endgroup$
    – Juwon
    Oct 25, 2016 at 21:10

1 Answer 1


In part B, the probability of success on each of $1000$ independent trials is $0.5\% = 0.005$, so the number of successes has a binomial distribution with $n=1000$ and $p=0.005$. The expected value is $np = 1000\times 0.005 = 5.$ The Poisson distribution is used as an approximation to the binomial distribution, and you get $$ \Pr(X=5) = \frac{e^{-\lambda} \lambda^5}{5!} = \frac{e^{-5}\cdot 5^5}{5!} \approx 0.175467. $$ If you used the exact distribution, you'd have \begin{align} \Pr(X=5) & = \binom{1000} 5 p^5 (1-p)^{1000-5} = \frac{1000\cdot999\cdot998\cdot997\cdot996}{5\cdot4\cdot3\cdot2\cdot1} \cdot 0.005^5\cdot 0.995^{995} \\[10pt] & \approx 0.1759076. \end{align}

  • $\begingroup$ Thanks for the input. I've now realized the silly mistake i made with your help. $\endgroup$
    – Juwon
    Oct 25, 2016 at 20:52

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