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

a) P(1) = 0.1637

b)

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:

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:

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

Where .

Then, and are calculated as:

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

4a. is 3 miles every hour

4b. is 1/3 of an hour per mile

step-by-step explanation:

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this