Question

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 1

Step 1:

Write the given data

`\begin{gathered} \text{mean }\mu\text{ = 2524} \\ \text{Standard deviation }\sigma\text{ = 325} \\ x\text{ = 1979} \end{gathered}`

Step 2:

Write the z-score formula

`z\text{ = }\frac{\text{x - }\mu}{\sigma}`

Step 3

Substitute the values in the z score equation

`\begin{gathered} \text{z = }\frac{1979\text{ - 2524}}{325} \\ z\text{ = }(-545)/(325) \\ z\text{ = -1.677} \end{gathered}`

Step 4:

Draw the z-curve

Step 5

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

`=\text{ 0.5 - 0.0475 + 0.5 = }0.953`

`\begin{gathered} percentageofscoresgreaterthan1979 \\ =\text{ 0.953 }*\text{ 100 = 95.3\%} \end{gathered}`

Final answer

= 95.4%