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Suppose that the number of customers that enter
a bank in an hour is a Poisson random variable, and sup-
pose that $P(X=0)=0.05 .$ Determine the mean and
variance of $X .$

$x \rightarrow$ number of customers

$x \rightarrow$ Poisson random variable

$P(x=0)=0.05$

$\mu, \sigma^{2} ? !$

$\mu=\sigma^{2}=\lambda$

$f(x=x)=\frac{\lambda^{x} e^{-\lambda}}{x !}$

$P(x=0)=\frac{\lambda^{\circ} e^{-\lambda}}{0 !}=0.05$

$e^{-\lambda}=0.05$

$-\lambda=\ln 0.05$

$\lambda=2.996$

$\mu=\sigma^{2}=2.996$

The number of flaws in bolts of cloth in textile man-
uracturing is assumed to be Poisson distributed with a mean of
0.1 flaw per square meter.
(a) What is the probability that there are two flaws in 1 square
meter of cloth?
(b) What is the probability that there is one flaw in 10 square
meters of cloth?

(c) What is the probability that there are no flaws in 20 square
meters of cloth?
(d) What is the probability that there are at least two flaws in
10 square meters of cloth?

Flows in bolts $\rightarrow$ Poisson distribution

$\mu=0.1$ Per 1 $\mathrm{m}^{2}$

(a) $X \rightarrow$ n: of flows in 1 $\mathrm{m}^{2}$

$P(x=2)=\frac{\lambda^{x} e^{-\lambda}}{x !}=\frac{0.1^{2} e^{-0.1}}{2 !}=0.0045$

$\lambda= \mu=0.1$

(b) $y_{\rightarrow}$ n: of flows in $10\mathrm m^{2}$

$P\left(y=1\right)=\frac{\lambda^{y} e^{-\lambda}}{y !}, \quad \lambda=0.1 \times 10=1$

$P(y=1)=\frac{1^{1} e^{-1}}{1 !}=0.3679$

$w \rightarrow$ n: of flows in $20 \mathrm m^{2}$

$\lambda=0.1 \times 20=2$

$P(w=0)=\frac{2^{0} e^{-2}}{0!}=0.1353$

(d) $P(y \geq 2)$

$=1-P(y<2)$

$=1-\left[P(y=0)+P\left(y=1\right)\right]$

$=1 - \frac{1^{\circ} e^{-1}}{1}-0.3679$

$=0.2642$