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.​

The two z-scores are -1.53 and 1.53Step-by-step explanation:To find the two z-scores that separate the middle 87.4% of the distribution from the area in the tails of the standard normal distributionAssume that z-score is between -c and c ⇒ P(-c < z < c) = the given valueFind the value of P(z < -c or z > c) = 1 - the given valueP(z < -c) + P(z > c) = the answer of the previous step2*P(z < -c) = the answer of the previous stepFind 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) = 0.063Let us use the z-table to find the corresponding values ofz to 0.063∵ The corresponding value of z to 0.063 = -1.53∴ The two z-scores = -1.53 and 1.53The attached figure for more understandThe two z-scores are -1.53 and 1.53Learn more:You can learn more about z-score in brainly.com/question/6270221#LearnwithBrainly