Question

2d. Probability that -1.96 > Z > 1.96 (this is the probability that the value is less than negative 1.96 AND greater than 1.96). This is the value in the two tails

Answer 1

You wrote in the brackets correct way that is -1.96 < Z < 1.96 . But you typed it wrong which is -1.96 > Z > 1.96.

So, I am going to solve using the first form which is -1.96 < Z < 1.96

We know that P(a < X < b) = P(X < b ) - P (X < a) --- (a)

and normal distribution is symmetric. So P(X < -a) = 1 - P(X < a) for any real number a ---- (b)

P ( -1.96 < Z < 1.96 ) = P(Z < 1.96) - P(Z < -1.96) from (a)

= P(Z < 1.96) - (1 - P (Z < 1.96) ) from (b)

= 2*P(Z < 1.96) - 1

= 2(0.975) - 1 from Z score table

= 1.95 - 1 = 0.95

So the value for above expression is 0.95

in two tailed the rejection region is on the both sides . half on the left and half on the right. So, it is such that

P(Z < -1.96) is 0.025 and P(Z > 1.96) is 0.025. So we leave 0.5/2 on the both tails of the probability curve.