Answer:
Point 1: (-5.03, 0.001)
Point 2: (0, 0.333)
Point 3: (3.068, 9.698)
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:
![Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%5Csqrt%7B2.3779%7D%3D%201.542%20%5Capprox%201.54)
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:
![E(X) = \sum_{i=1}^n X_i P(X_I)](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20X_i%20P%28X_I%29)
And replacing we got:
![E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61](https://tex.z-dn.net/?f=%20E%28X%29%3D%200%2A0.3%20%2B1%2A0.25%20%2B2%2A0.2%20%2B3%2A0.12%20%2B4%2A0.07%2B%205%2A0.04%20%2B6%2A0.02%3D1.61)
Now we can find the second moment given by:
![E(X^2) =\sum_{i=1}^n X^2_i P(X_i)](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29%20)
And replacing we got:
![E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97](https://tex.z-dn.net/?f=%20E%28X%5E2%29%3D%200%5E2%2A0.3%20%2B1%5E2%2A0.25%20%2B2%5E2%2A0.2%20%2B3%5E2%2A0.12%20%2B4%5E2%2A0.07%2B%205%5E2%2A0.04%20%2B6%5E2%2A0.02%3D4.97)
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:
![Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%5Csqrt%7B2.3779%7D%3D%201.542%20%5Capprox%201.54)
What is part A? I feel their is something you are missing in your answer
The answer is No, they are not proportional. This is because 8/42 is not equal to 20/105.