Question

Suppose X has a continuous uniform distribution over the interval [−2,2]

Answer 1

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) $\int\limits^a_b f({x}) \, dx$=  \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) =$\int\limits^a_b {x} \, dx$=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.