Find the distance between (8,0) and (4,-4)
1 answer:
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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_a {\frac{1}{a- (-a)} } \, dx
=1/2a[x]^a_-a= 2a/2a= 1 (applying the limits to the function)
P(−a<X<a) ==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.