Question

Let z be a random variable with a standard normal distribution.

Find “a” such that P(|Z| <A)= 0.95

This is what I have:

P(-A<Z<A) = 0.95
-A = -1.96
How do I use the symmetric property of normal distribution to make A = 1.96?
My answer at the moment is P(|z|< (-1.96) = 0.95

Answer 1

Solution:

We have to find value A such that: P(|Z| <A)= 0.95

That is:

P( - A < Z < A) =0.95

A standard normal distribution is symmetric about the center = 0 or mean = 0.

If area between -A to A is 0.95 then area outside this region is 1 - 0.95 = 0.05

So we divide this 0.05 area in two tails equally as: 0.05 / 2 = 0.025

That Area in left tail = P( Z< -A) = 0.025

and Area in right tail = P( Z > A ) =0.025

We have P( - A < Z < A) =0.95 and P( Z< -A) = 0.025

Thus total Area below Z < A is:

P( Z < A) = P( Z< -A) + P( - A < Z < A)

P( Z < A) = 0.025 + 0.95

P( Z < A) = 0.9750

Now look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : z = 1.96

That is A = 1.96

Thus

P(|Z| <A)= 0.95

P(|Z| < 1.96 )= 0.95

Or

Look in z table for Area = 0.025 or its closest area and find z value:

Area 0.0250 corresponds to -1.9 and 0.06

Thus z = -1.96

Since due to symmetry, P( Z < -z ) = P( Z > + z)

we have P( Z < -1.96 ) = P ( Z > +1.96) = 0.025

Thus A = 1.96