Answer:
a) This is a continuous random variable because the set of possible values is an interval.
b) Height = 0.5
Graph attached.
c) P(Y≤1) = 0.5
Step-by-step explanation:
a) If the density curve of the outcomes has a constant height, we have a uniform distribution for values between 0 and 2 (outside this range, the density function has a value of 0).
This function is continous and has a value for each real number. The set of possible values has a is an interval between 0 and 2, all with equal probability (the ones that are otuside this interval have 0 probability, so they are not possible values).
b) The height is calculated so that the total area under the density curve has to be equal to 1.
Then, we have to calculate the integral between 0 and 2 of the density function:
[tex]\int\limits^2_0 {U(x)} \, dx =\int\limits^2_0 {H} \, dx =H\int\limits^2_0 dx=H(x_2-x_1)=H(2-0)=1\\\\2H=1\\\\H=0.5[/tex]
The height is H=0.5
(Graph attached)
c) The probability can be calculated by integrating the density function (which is equal to the area under the curve) between 0 and 1, or using the graph.
With the graph, we see that the area is equal to the height (0.5) by the width of the interval (1), so the total area is 0.5 x 1 = 0.5.
The probability P(Y≤1) is equal to 0.5.
Using the uniform distribution, it is found that:
a)
This is a continuous random variable because the set of possible values is an interval.
b)
The height is [tex]\frac{1}{2}[/tex], and the sketch is given at the end of this answer.
c)
P(Y ≤ 1) = 0.5
An uniform distribution has two bounds, a and b.
The probability of finding a value of at lower than x is:
[tex]P(X < x) = \frac{x - a}{b - a}[/tex]
In this problem, uniformly distributed between 0 and 2, thus [tex]a = 0, b = 2[/tex].
Item a:
Values in an interval, thus continuous. Discrete are for sample spaces, which is not the case here.
Item b:
The height is:
[tex]h = \frac{1}{b - a} = \frac{1}{2}[/tex]
The graph is sketched at the end of this answer.
Item c:
[tex]P(Y \leq 1) = \frac{1 - 0}{2 - 0} = \frac{1}{2} = 0.5[/tex]
Thus, P(Y ≤ 1) = 0.5.
A similar problem is given at https://brainly.com/question/13547683
