Question

29. There are 2m+n workers in a bank, Mr. Smith is one of them. Each day, the headquarters randomly choose m individuals to work in the customer service area. Using the de Moivre-Laplace theorem, approximate the probability that during 100 consecutive days, Mr. Smith will spend at most 2k days in the customer service area. The answer should be provided as b(t), where Ф is the CDF of the standard normal distribution. Next, the value of the CDF should be obtained from distribution tables. Both t and the value of the CDF should be provided as numbers in decimal form, rounded to 4 digits. ANSWER:

Answer 1

Statement of de Moivre-Laplace central limit theorem :-

Let {X_i} be a sequence of random variables where P(X_i=1)=p=1-P(X_i=0).

$Define:\; \; S=\sum_{i=1}^{l}X_{i}\;\;\;\;\;\;\;\;\;\;Then\;\;\;\frac{S-lp}{\sqrt{lp(1-p)}}\sim N(0,1)\;\;for\;large\;value\;of\;l$

According to this problem;

$X_{i}=\left\{\begin{matrix} 1 & if\;Mr.\;Smith\;is \;selected\;on\;i^{th}\;day \\ 0 & if\;Mr.\;Smith\;is\;not\; \;selected\;on\;i^{th}\;day \end{matrix}\right.\;\;\;\;\;\;\;\;\;i=1,2,.....,100$

p = P(X_i=1) = Probability that Mr. Smith is selected on i^th day ; i=1,2,....,100

   $=\frac{No.\;of\;ways\;in\;which\;m-1\;workers\;other\;than\;Mr.\;Smith\;are\;selected}{No.\;of\;ways\;in\;which\;m\;workers\;are\;selected}$ $\!\!\!\!\!\!\!\!\!\!\!\!\!\!=\frac{(2m+n)_{(m-1)}}{(2m+n)_{m}}=\frac{(2m+n)!/(2m+n-m+1)!}{(2m+n)!/(2m+n-m)!}=\frac{1}{2m+n-m+1}=\frac{1}{m+n+1}$

$Total\;no.\;of\;days\;Mr.\;Smith\;is\;selected=S=\sum_{i=1}^{100}X_{i}$

Thus according to the de Moivre-Laplace central limit theorem :-

$\frac{S-100p}{\sqrt{100p(1-p)}}\sim N(0,1)$

Probability that Mr. Smith will spend at most 2k days in the customer service area

$=P(S\leqslant 2k)=P(\frac{S-100p}{\sqrt{100p(1-p)}}\leqslant \frac{2k-100p}{\sqrt{100p(1-p)}})=\Phi (\frac{2k-100p}{\sqrt{100p(1-p)}})$

$Now\;;\;\;\frac{2k-100p}{\sqrt{100p(1-p)}}=\frac{2k-\frac{100}{m+n+1}}{\sqrt{\frac{100}{m+n+1}(1-\frac{1}{m+n+1})}}=\frac{2k(m+n+1)-100}{10\sqrt{m+n}}$

$\therefore \;P(S\leqslant 2k)=\Phi(t)\;\;\;where\;\;t=\frac{2k(m+n+1)-100}{10\sqrt{m+n}}$

Now the value of t and $\Phi$(t) depends on the values of m and n and k.