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=\frac{x-\mu}{\alpha}$$\begin{gathered} \Rightarrow z=\frac{51.15-52}{\alpha} \\ \end{gathered}$

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

$\begin{gathered} -0.1358=\frac{51.15-52}{\alpha} \\ \alpha=\frac{51.15-52}{-0.1358}=6.2592 \end{gathered}$

Hence the standard deviation of x is 6.2592.

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