Find the zscores that separate the middle 30 of the distribu

Find the z-scores that separate the middle 30% of the distribution from the area in the tails of the standard normal distribution.

Please show work/explain

Thank you

Solution

The middle area of the standard normal curve is 30%.

Therefore, the remaining areas (to the left of -z and to the right of z) is 70%.

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 35% or 0.35.
This means the area below z (to the left of z) is 0.65.

From the normal probability table, find z such that P( z < ? ) = 0.65
Look for 0.65 under the area and read z that corresponds to it.
The closes z is 0.39
Therefore, z-scores that separate the middle 30% of the distribution from the tails of standard normal distribution are -0.39 and 0.39