Find the zscores that separate the middle 89 of The distribu

Solution

The middle area of the standard normal curve is 89%.

Therefore, the remaining areas (to the left of -z and to the right of z) is 11%.

Because the normal distribution is symmetric, the area to the left of -z is the same as the area to the right of z , each being 5.5% or 0.055.
This means the area below z (to the left of z) is 0.945.

From the normal probability table, find z such that P( z < ? ) = 0.945
Look for 0.945 under the area and read z that corresponds to it.
The closes z is 1.6
Therefore, z-scores that separate the middle 89% of the distribution from the tails of standard normal distribution are -1.6 and 1.6