Determine the two z-scores that separate the middle 87.4% of the distribution from the area in the tails of the standard normal distribution.

Assume that z-score is between -c and c ⇒ P(-c < z < c) = the given value
Find the value of P(z < -c or z > c) = 1 - the given value
P(z < -c) + P(z > c) = the answer of the previous step
2*P(z < -c) = the answer of the previous step
Find the value of P(z < -c)
In the z-table having area to the left of z, look for the value closest to the value of P(z < -c) inside the table to find the closest value of z

∵ P(-c < z < c) = 87.4%

∵ 87.4% = 87.4 ÷ 100 = 0.874

∴ P(-c < z < c) = 0.874

∵ P(z > c) = 1 - 0.874 = 0.126

∴ P(z < -c) + P(z > c) = 0.126

∵ P(z > c) = P(z < -c)

∴ P(z < -c) + P(z < -c) = 0.126

∴ 2*P(z < -c) = 0.126

- Divide both sides by 2

∴ P(z < -c) = 0.063

Let us use the z-table to find the corresponding values of

z to 0.063

∵ The corresponding value of z to 0.063 = -1.53

∴ The two z-scores = -1.53 and 1.53

The attached figure for more understand

The two z-scores are -1.53 and 1.53

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Determine the two z-scores that separate the middle 87.4% of the distribution from the area in the tails of the standard normal distribution.​