Question

g Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number X has a Poisson distribution with lamda = .2. a) What is the probability that a disk has exactly one missing pulse? b) What is the probability that a disk has at least two missing pulses? c) What is EX

Answer 1

Answer:

a) P(1) = 0.1637

b) $P(x\geq 2) = 0.0176$

c) E(x) = 0.2

Step-by-step explanation:

If X follows a poisson distribution, the probability that a disk has exactly x missing pulses is:

$P(x)=\frac{e^{-m}*m^x}{x!}$

Where m is the mean and it is equal to the value of lambda. So, replacing the value of m by 0.2, we get that the probability that a disk has exactly one missing pulse is equal to:

$P(1)=\frac{e^{-0.2}*0.2^1}{1!}=0.1637$

Additionally, the probability that a disk has at least two missing pulses can be calculated as:

$P(x\geq 2)=1-P(x$

Where $P(x$.

Then, $P(0)$ and $P(x\geq 2)$ are calculated as:

$P(0)=\frac{e^{-0.2}*0.2^0}{0!}=0.8187\\P(x\geq 2) = 1 - (0.8187 + 0.1637)\\P(x\geq 2) = 0.0176$

Finally, In the poisson distribution, E(x) is equal to lambda. So E(x) = 0.2