The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages ( x ) in the cit
y 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 _____.
2 answers:
Answer:
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
You might be interested in
Answer: The probability that the project will be completed in 33 days or less is equal to<u> </u><u>0.73</u> .
Step-by-step explanation:
We assume that this is a normally distribution.
Given : The critical path length is 30 days .
i.e. Population mean =
Variance :
Then, standard deviation :
Z-score :
The value of z corresponding to 33 =
Now, the p-value =
Hence, the probability that the project will be completed in 33 days or less is equal to 0.73.