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

a) P(1) = 0.1637

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

c) E(x) = 0.2

Explanation:

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

`P(x)=(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)=(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<2)`

Where
`P(x<2)=P(0)+P(1)`.

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

`P(0)=(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