Answer:
Mean = 0
Variance = 4/3
Standard Deviation √4/3
a= 0.9
Step-by-step explanation:
If X has a uniform distribution over [a,b] then its Mean is a+b/2 and variance is (b-a)²/12
Here a= -2 and b= 2
Now finding the mean = a+b/2=-2+2/2= 0
Variance = (b-a)²/12=( 2-(-2))²/12= 4²/12= 16/12= 4/3
Standard Deviation = √Variance= √4/3
b) [tex]\int\limits^a_b f({x}) \, dx[/tex]= \int\limits^a_a {\frac{1}{a- (-a)} } \, dx
=1/2a[x]^a_-a= 2a/2a= 1 (applying the limits to the function)
P(−a<X<a) =[tex]\int\limits^a_b {x} \, dx[/tex]=1/2 * 2a= a (applying the limits to the function)
P(−a<X<a)= 0.9
a= 0.9
In the given question the limits are -a to a . When we apply these in the above instead of [a,b] we get the above answer.
Using the uniform distribution, we have that:
An uniform distribution has two bounds, a and b.
In this problem, interval [−2,2], hence [tex]a = -2, b = 2[/tex].
The mean is:
[tex]E(X) = \frac{a + b}{2}[/tex]
Hence:
[tex]E(X) = \frac{-2 + 2}{2} = 0[/tex]
The variance is:
[tex]V(X) = \frac{(b - a)^2}{12}[/tex]
Hence
[tex]V(X) = \frac{(2 - (-2))^2}{12} = \frac{16}{12} = 1.25[/tex]
The standard deviation is the square root of the variance, hence [tex]\sqrt{1.25} = 1.12[/tex]
A similar problem is given at https://brainly.com/question/14767944
