# Let X \sim N(6,4).Find the probabilities P(X<3).

Let $X\sim N\left(6,4\right)$.Find the probabilities $P\left(X<3\right)$.

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Step 1
Introduction:
The normal probability is a type of continuous probability distribution that can take random values. The normal distribution is determined by the two parameters - the population mean $\left(\mu \right)$ and population variance $\left({\sigma }^{2}\right)$. It is symmetric with respect to its mean.
Given information:
$X\sim N\left(6,4\right)$
Therefore,
$\mu =6$
${\sigma }^{2}=4$

Step 2
$P\left(X<3\right)$ is computed as follows:
$P\left(X<3\right)=P\left(\frac{X-\mu }{\sqrt{{\sigma }^{2}}}<\frac{3-\mu }{\sqrt{{\sigma }^{2}}}\right)$
$=P\left(Z<\frac{3-6}{\sqrt{4}}\right)$
$=P\left(Z<-1.5\right)$
$=1-P\left(Z<1.5\right)$
$=1-0.93319=0.06681$
Therefore,
$P\left(X<3\right)=0.0668$