Based on data from a college, scores on a certain test are normally distributed with a mean of 2524 and a standard deviation of 325 find the percentage of scores greater than 1979

Answer :

Step 1:

Write the given data

[tex]\begin{gathered} \text{mean }\mu\text{ = 2524} \\ \text{Standard deviation }\sigma\text{ = 325} \\ x\text{ = 1979} \end{gathered}[/tex]

Step 2:

Write the z-score formula

[tex]z\text{ = }\frac{\text{x - }\mu}{\sigma}[/tex]

Step 3

Substitute the values in the z score equation

[tex]\begin{gathered} \text{z = }\frac{1979\text{ - 2524}}{325} \\ z\text{ = }\frac{-545}{325} \\ z\text{ = -1.677} \end{gathered}[/tex]

Step 4:

Draw the z-curve

Step 5

Probability that scores greater than 1979 id Pr (z > -1.677)

[tex]=\text{ 0.5 - 0.0475 + 0.5 = }0.953[/tex]

[tex]\begin{gathered} percentageofscoresgreaterthan1979 \\ =\text{ 0.953 }\times\text{ 100 = 95.3\%} \end{gathered}[/tex]

Final answer

= 95.4%

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