Question

A normally distributed variable, x, has a mean of 52. p(x<51.15)= 0.446. Find the standard deviation of x.

Answer 1

`z=(x-\mu)/(\alpha)`
`\begin{gathered} \Rightarrow z=(51.15-52)/(\alpha) \\ \end{gathered}`

z-score for a p-value of 0.446, left tailed p(x<51.15) is -0.1358

`\begin{gathered} -0.1358=(51.15-52)/(\alpha) \\ \alpha=(51.15-52)/(-0.1358)=6.2592 \end{gathered}`

Hence the standard deviation of x is 6.2592.

z-score can be obtained from the z-probability table