The Bernoulli distribution is a distribution whose random variable can only take 0 or 1
- The value of E(x2) is p
- The value of V(x) is p(1 - p)
- The value of E(x79) is p
<h3>How to compute E(x2)</h3>
The distribution is given as:
p(0) = 1 - p
p(1) = p
The expected value of x2, E(x2) is calculated as:
![E(x^2) = \sum x^2 * P(x)](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20%5Csum%20x%5E2%20%2A%20P%28x%29)
So, we have:
![E(x^2) = 0^2 * (1- p) + 1^2 * p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%5E2%20%2A%20%281-%20p%29%20%2B%201%5E2%20%2A%20p)
Evaluate the exponents
![E(x^2) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
Multiply
![E(x^2) = 0 +p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%200%20%2Bp)
Add
![E(x^2) = p](https://tex.z-dn.net/?f=E%28x%5E2%29%20%3D%20p)
Hence, the value of E(x2) is p
<h3>How to compute V(x)</h3>
This is calculated as:
![V(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
Start by calculating E(x) using:
![E(x) = \sum x * P(x)](https://tex.z-dn.net/?f=E%28x%29%20%3D%20%5Csum%20x%20%2A%20P%28x%29)
So, we have:
![E(x) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
![E(x) = p](https://tex.z-dn.net/?f=E%28x%29%20%3D%20p)
Recall that:
![V(x) = E(x^2) - (E(x))^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20E%28x%5E2%29%20-%20%28E%28x%29%29%5E2)
So, we have:
![V(x) = p - p^2](https://tex.z-dn.net/?f=V%28x%29%20%3D%20p%20-%20p%5E2)
Factor out p
![V(x) = p(1 - p)](https://tex.z-dn.net/?f=V%28x%29%20%3D%20p%281%20-%20p%29)
Hence, the value of V(x) is p(1 - p)
<h3>How to compute E(x79)</h3>
The expected value of x79, E(x79) is calculated as:
![E(x^{79}) = \sum x^{79} * P(x)](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%20%5Csum%20x%5E%7B79%7D%20%2A%20P%28x%29)
So, we have:
![E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%5E%7B79%7D%20%2A%20%281-%20p%29%20%2B%201%5E%7B79%7D%20%2A%20p)
Evaluate the exponents
![E(x^{79}) = 0 * (1- p) + 1 * p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%20%2A%20%281-%20p%29%20%2B%201%20%2A%20p)
Multiply
![E(x^{79}) = 0 + p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%200%20%2B%20p)
Add
![E(x^{79}) = p](https://tex.z-dn.net/?f=E%28x%5E%7B79%7D%29%20%3D%20p)
Hence, the value of E(x79) is p
Read more about probability distribution at:
brainly.com/question/15246027
Answer:
i dont know sorry
Step-by-step explanation:
If the circumference is half of the perimeter than the area for both circles would be half of them find one circles area and half it
What do you mean? can you please elaborate.
Answer: just divide and its prett easy from there
Step-by-step explanation: