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looi76 · #1 April 16th, 2008, 08:45 AM

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

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

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

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

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

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

$\sigma = 4.22$

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

Where did I go wrong?