Amanager of a large computer network has developed the following probability distribution of the number of interruptions per day: interruptions (x), p(x). 0, 0.32. 1, 0.35. 2, 0.18. 3, 0.08. 4, 0.04. 5, 0.02. 6, 0.01. compute the expected number of interruptions per day. compute the standard deviation.

Step-by-step explanation:

Given that a manager of a large computer network has developed the following probability distribution of the number of interruptions per day

X denotes interruptions and P(X) probability

 x*p(X)x^2p(x)

00.3200

10.350.350.35

20.180.360.72

30.080.240.72

40.040.160.64

50.020.10.5

60.010.060.36

Total11.273.29

Variance1.6771  

std dev1.295028957  

Expected value = sum of all products of x with corresponding p

= 1.27

Answer is 1.27

Std dev = sqrt of variance = 1.2950

d? maybe

step-by-step explanation:

no this is a nonlinear negative