Multiply the GCF of the numerical part 3 and the GCF of the variable part x^2y to get
3x^2y.
Answer:
Yeah same here
Step-by-step explanation:
Answer:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

Step-by-step explanation:
For this case we have the following distribution given:
X 0 1 2 3 4 5 6
P(X) 0.3 0.25 0.2 0.12 0.07 0.04 0.02
For this case we need to find first the expected value given by:

And replacing we got:

Now we can find the second moment given by:

And replacing we got:

And the variance would be given by:
![Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.97%20-%281.61%29%5E2%20%3D2.3779)
And the deviation would be:

3.92
= 1.00 + 1.00 + 1.00 + 0.92
= 100/100 + 100/100 + 100/100 + 92/100
= 300/100 + 92/100
= 392/100
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392/100 = 196/50 = 98/25