Answer :
Answer:
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)
0 0.32 0 0
1 0.35 0.35 0.35
2 0.18 0.36 0.72
3 0.08 0.24 0.72
4 0.04 0.16 0.64
5 0.02 0.1 0.5
6 0.01 0.06 0.36
Total 1 1.27 3.29
Variance 1.6771
std dev 1.295028957
Expected value = sum of all products of x with corresponding p
= 1.27
Answer is 1.27
Std dev = sqrt of variance = 1.2950