Answer:
![E(X) = 0*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%200%2A0.92%20%2B%201%2A0.03%20%2B2%2A0.03%20%2B3%2A0.02%20%3D%200.1500)
In order to find the variance we need to find first 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%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%200%5E2%2A0.92%20%2B%201%5E2%2A0.03%20%2B2%5E2%2A0.03%20%2B3%5E2%2A0.02%20%3D%200.3300)
The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:
![Sd(X) = \sqrt{0.3075}= 0.5545](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%20%5Csqrt%7B0.3075%7D%3D%200.5545)
Step-by-step explanation:
Previous concepts
The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.
The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).
Solution to the problem
LEt X the random variable who represent the number of defective transistors. For this case we have the following probability distribution for X
X 0 1 2 3
P(X) 0.92 0.03 0.03 0.02
We can calculate the expected value with the following formula:
![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*0.92 + 1*0.03 +2*0.03 +3*0.02 = 0.1500](https://tex.z-dn.net/?f=%20E%28X%29%20%3D%200%2A0.92%20%2B%201%2A0.03%20%2B2%2A0.03%20%2B3%2A0.02%20%3D%200.1500)
In order to find the variance we need to find first 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%20%5Csum_%7Bi%3D1%7D%5En%20X%5E2_i%20P%28X_i%29)
And replacing we got:
![E(X^2) = 0^2*0.92 + 1^2*0.03 +2^2*0.03 +3^2*0.02 = 0.3300](https://tex.z-dn.net/?f=%20E%28X%5E2%29%20%3D%200%5E2%2A0.92%20%2B%201%5E2%2A0.03%20%2B2%5E2%2A0.03%20%2B3%5E2%2A0.02%20%3D%200.3300)
The variance is calculated with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 0.33 -(0.15)^2 = 0.3075](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%200.33%20-%280.15%29%5E2%20%3D%200.3075)
And the standard deviation is just the square root of the variance and we got:
![Sd(X) = \sqrt{0.3075}= 0.5545](https://tex.z-dn.net/?f=%20Sd%28X%29%20%3D%20%5Csqrt%7B0.3075%7D%3D%200.5545)
Answer:
(15*x)+((25)(x-2))=230 THE ANSWER D. X=7
I GOT IT CORRECT
Step-by-step explanation:
Answer:
https://www.mathpapa.com/algebra-calculator.html
Step-by-step explanation:
:) welcome to cheating in math paradise