Question

A data set is normally distributed with a mean of 18 and astandard deviation of 2.7. Approximately what percent of the data are greaterthan 12.6?

Answer 1

Given;

`x=12.8,\mu=18,\sigma=2.7`

The z-score is;

`\begin{gathered} z=(x-\mu)/(\sigma) \\ \\ z=(12.8-18)/(2.7) \\ \\ z=-1.9259 \end{gathered}`

Thus, the approximate percent of the data greater than 12.6 is;

`P(x>12.8)=97.5\text{ \%}`

ANSWER: 97.5%