How can i find the expected value?

Question

The textbook was published in a circulation of 100000 copies. The probability that the textbook is not properly stitched is 0.0001. Compile a series of the distribution of the number of textbooks that are not properly stitched. Find E(ξ) of this random variable.

My solution. But can i find E(ξ) faster?

Answer 1

• The exact distribution of the number of improperly stitched textbooks is binomial: \begin{align} P_{100000}(k)&={100000 \choose k}0.0001^k\,0.9999^{100000-k}\\ &={N\choose k}p^k(1-p)^{N-k}\, \end{align} where $\ N=100000\ $ and $\ p=0.0001\ $. For this distribution, $\ E(\xi)=Np=10\ $, as lulu's comment notes.
• The distribution you've used is the Poisson distribution, which can be a good approximation to the binomial distribution when $\ N\ $ is large and $\ p\ $ or $\ 1-p\ $ is small. Since that is the case here, using the Poisson approximation is reasonable: $$ P_{100000}(k)\approx\frac{\lambda^k}{k!}e^{-\lambda}\ , $$ but $\ \lambda $ here must be the expected number, $\ Np\ $ ($\ =10\ $ for your example), so it would appear that you already knew this???
• You have some typos in your expression for $\ P_{100000}(k)\ $. It should be $$ P_{100000}(k)\approx\frac{\lambda^{\color{red} k}}{k!}e^{-\lambda}=\frac{10^{\color{red} k}}{10!}e^{-10}\ , $$ and I have no idea what $\ k=\overline{0,100000}\ $ means.
• In your formula for $\ E(\xi)\ $ you need to have $\ x_k=k\ $. That is, $\ E(\xi)=\sum_\limits{k=1}^{100000}kP_{100000}(k)\ $. For the binomial distribution this becomes \begin{align} E(\xi)&= \sum_{k=1}^N k {N\choose k}p^k(1-p)^{N-k}\\ &=Np \sum_{k=1}^N {N-1\choose k-1}p^{k-1}(1-p)^{(N-1)-(k-1)}\\ &=Np \sum_{j=0}^{N-1} {N-1\choose j}p^j(1-p)^{(N-1)-j}\\ &=Np\ , \end{align} and for the Poisson distribution, it is \begin{align} E(\xi)&= \sum_{k=1}^Nk \frac{\lambda^k}{k!}e^{-\lambda}\\ &=\lambda e^{-\lambda} \sum_{k=1}^N \frac{\lambda^{k-1}}{(k-1)!}\\ &= \lambda e^{-\lambda} \sum_{j=0}^{N-1} \frac{\lambda^j}{j!}\\\\ &\approx\lambda \end{align} when $\ N\ $ is very large.