Question 973142
mean of X=3
SD =2 (variance is 4);  Would expect most of the values to be within 1 and 5
x> 4.7
x<1.3

Both of those satisfy (|X&#8722;3|>1.7      ;;;;x-3>1.7;  x>4.7 ;;;   x-3 < -1.7;  x<1.3

z+(x-mean)/2 = (1.7)/2   z> +0.85

z=(1.3-3)/2  =- (1.7)/2   z< -0.85

P= 0.40   This seems reasonable.