The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages ( x ) in the city has the following probability distribution. x f (x) 0 0.80 1 0.15 2 0.04 3 0.01 The mean and the standard deviation for the number of electrical outages (respectively) are .

The mean is 0.26

The standard deviation is 0.5765

Step-by-step explanation:

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0.26 and 0.577

mean =0.26

Standard deviation = 0.577

Step-by-step explanation:

x       P(x)        x*P(x)              (x-μ)²                             (x -μ)²*P(x)

0       0.80       0            (0 - 0.26)² = 0.0676            0.0676*0.8 = 0.0541

1        0.15         0.15       (1 - 0.26)² = 0.5476             0.5476*0.15 = 0.0821

2       0.04        0.08       (2 - 0.26)² = 3.0276           3.0276*0.04 = 0.1211

3       0.01         0.03       (3 - 0.26)² = 7.5076           7.5076*0.01 = 0.0751

The mean μ = 0 + 0.15 + 0.08 + 0.03 =0.26

The mean μ =0.26

Variance σ² = 0.0541 + 0.0821 + 0.1211 + 0.0751 = 0.3324

Standard Deviation σ² = √σ = = 0.5765 ~ 0.577

Standard Deviation σ =  0.577

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)