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A manager 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.

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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

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