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?

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%

Multiply 9 and 67. then multiply 3 with 67. lastly the answer that you got for 9x67 is to be subtracted by the answer you got for 3x67