"Find the​ z-scores that separate the middle 48​% of the distribution from the area in the tails of the standard normal distribution. The​ z-scores are?"

latinoisraelm1

latinoisraelm1

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2022-08-18

Find the​ z-scores that separate the middle 48​% of the distribution from the area in the tails of the standard normal distribution.
The​ z-scores are?

Answer & Explanation

Isaias Archer

Isaias Archer

Beginner2022-08-19Added 11 answers

Computation of probability of the area above and below the 48% of the distribution from the area:
From the given information, middle area is 48%.
Since it is a standard normal distribution, the area above and below the 48% area are equal.
Thus, the probability of the area above and below the 48% area is given as follows:
P ( z 0 < Z < z 0 ) = 0.48 = ( 1 0.48 ) 2 = 0.26
Thus,
P ( Z < z 0 ) = 0.26 P ( Z > z 0 ) = 1 0.26 = 0.74
Computation of z-scores that separate the middle 48% of the area under standard normal curve:
Using EXCEL formula, “=NORM.S.INV(0.26)” P ( Z < z 0 ) = 0.6434.
Using EXCEL formula, “=NORM.S.INV(0.74)” P ( Z > z 0 ) = 0.6434.
The z-scores that separate the middle 48% of the distribution from the area under standard normal curve are −0.6434 and 0.6434 respectively.

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