Question

$\displaystyle X$ has a normal distribution with mean 32 and variance $\displaystyle \sigma^2$. Given that the probability $\displaystyle X$ is less than 33.14 is 0.6406, find $\displaystyle \sigma^2$. Give your answer correct to 2 decimal places.

Attempt:

$\displaystyle X \sim N (32,\sigma^2)$

$\displaystyle P\left( \frac{X-\mu}{\sigma} \leq \frac{33.14-32}{\sigma} \right) = 0.6046$

$\displaystyle P\left( Z \leq \frac{33.14 - 32}{ \sigma } \right) = 0.6046$

$\displaystyle \frac{33.14 - 32}{ \sigma } = 0.27$

$\displaystyle \frac{33.14 - 32}{0.27} = \sigma$

$\displaystyle \sigma = 4.22$

$\displaystyle \sigma^2 = 17.83 (2 DP)$

Where did I go wrong?