Answer:
The expected value of X is 
and the variance of X is 
The expected value of Y is 
and the variance of Y is 
Step-by-step explanation:
(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or 
, is
The probability mass function 
of X is given by
Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function 
of Y is given by
The expected value of X is
![E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)](https://tex.z-dn.net/?f=E%28X%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5Ccdot%20p_%7BX%7D%28x%29)
The expected value of Y is
![E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)](https://tex.z-dn.net/?f=E%28Y%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5Ccdot%20p_%7BY%7D%28x%29)
(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is
![V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2](https://tex.z-dn.net/?f=V%28X%29%3D%5Csum_%7Bx%5Cin%20D%7D%20%28x-%5Cmu%29%5E2%5Ccdot%20p%28x%29%3DE%28X%5E2%29-%5BE%28X%29%5D%5E2)
The variance of X is
![E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)](https://tex.z-dn.net/?f=E%28X%5E2%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5E2%5Ccdot%20p_%7BX%7D%28x%29)
The variance of Y is
![E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)](https://tex.z-dn.net/?f=E%28Y%5E2%29%3D%5Csum_%7Bx%5Cin%20%5B28%2C32%2C42%2C44%5D%7D%20x%5E2%5Ccdot%20p_%7BY%7D%28x%29)
Answer:
4,-3
Step-by-step explanation:
Answer:
A Type I error would occur if there was no evidence of an improvement on the national exam but there really was improvement.
Step-by-step explanation:
A type 1 error simply occurs when we incorrectly reject a true null hypothesis. In the scenario above, An experiment was conducted in other it know if there is sufficient evidence to support a claim that a new teaching method developed improves students score. If this claim is actually true in the real sense. However, after conducting a statistical test, we conuded that there was no sufficient evidence to support her claim of improvement using the new method, hence, the claim was rejected. By rejecting the claim, a true null has been rejected. Hence, a type 1 error has been committed.
The path that Gloria follows when she jumped is a path of parabola.
The equation of the parabola that describes the path of her jump is 
The given parameters are:
<em>Assume she starts from the origin (0,0)</em>
The midpoint would be:
So, the vertex of the parabola is:
Express properly as:
A point on the graph would be:
The equation of a parabola is calculated using:
Substitute in
Substitute in
Collect like terms
Solve for a
Simplify
Substitute in
Hence, the equation of the parabola that describes the path of her jump is
See attachment for the graph
Read more about equations of parabola at:
brainly.com/question/4074088
