Question

a psychologist has designed a questionnaire to measure individuals' aggressiveness. Suppose that the scores on the questionnaire are normally distributed with a standard deviation of 100. Suppose also that exactly 5% of the scores exceed 700. Find the mean of the distribution of scores.

Answer 1

`\begin{gathered} x\text{ = 700} \\ \\ \mu\text{ mean = ?} \\ \\ \sigma\text{ = 100 standard deviation} \end{gathered}`
`z\text{ value = }\frac{x\text{ - }\mu}{\sigma}`

If 5% exceed 700

`\begin{gathered} \text{P}_r\text{ (X > 700) = 0.05} \\ P_r\text{ ( X }\leq\text{ 700 ) = 0.95} \end{gathered}`
`\begin{gathered} P_r(Z\leq\frac{700\text{ - }\mu}{100}\text{ ) = 0.95} \\ \\ P_r(0\leq Z\leq\frac{700\text{ - }\mu}{100}\text{ ) = 0.95 - 0.5} \\ \frac{700\text{ - }\mu}{100}\text{ = 0.45} \\ From\text{ normal distribution table 0.45 = 1.645} \end{gathered}`
`\begin{gathered} \frac{700\text{ - }\mu}{100}\text{ = 1.645} \\ 700\text{ - }\mu\text{ = 100 }*\text{ 1.645} \\ 700\text{ - }\mu\text{ = 164.5} \\ \mu\text{ = 700 - 164.5} \\ \mu\text{ = 535.5} \end{gathered}`

Final answer

mean = 535.5