The scores of a high school entrance exam are approximately normally distributed with a given mean mc023-1.jpg = 82.4 and standard deviation mc023-2.jpg = 3.3. What percentage of the scores are between 75.8 and 89?

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2016-04-08 23:34:56

The score corresponding to the random variable x is z = (x - μ)/σ Given: μ = 82.4, population mean σ = 3.3, populaton standard deviation. We want to determine the percentage of scores between 75.8 and 89. For x₁ = 75..8, the z-score is z₁ = (75.8 - 82.4)/3.3 = -2 (2 std. deviations to the left of the mean) For x₂ = 89, the z-score is z₂ = (89 - 82.4)/3.3 = 2 (2 std. deviations to the right of the mean) From standard tables, P(z <= z1) = 0.02275 P(z <= 2) = 0.9772 Therefore, P(75. < x < 89) = 0.9772 - 0.02275 = 0.9544 = 95.4% This result is correct, because the bell curve contains about 95% of the total area within 2 standard deviatins to the lest and to the right of the mean. Answer: 95.4%

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