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Let x follow a Normal distribution with mean $2$ and variance $4$.

Stats question that I need help on.

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    You have to reduce this to a Normal(0,1) and then tabulate – 2017-02-26
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    Take this one step at a time. What is the probability that $x$ is less than $-1.92$? How many standard deviations below the mean is $-1.92$? – 2017-02-26
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    Related: http://math.stackexchange.com/questions/2162732/what-is-the-probability-that-x-is-less-than-5-92?rq=1 – 2017-02-26
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    @BruceET what are you talking about... just giving the final answer wouldn't be helpful to help the person *learn* how to do this on his own in the first place, but giving an arbitrary number (*which is incorrect*) is not helpful either. Did you forget that he said variance and not standard deviation? – 2017-02-26
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    @JMoravitz What formula should be first used to find out the probability of x? – 2017-02-26
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    @JMoravitz: Sorry, typo. Just returned to fix it. Answer is about .95. Removing previous Comment. – 2017-02-26
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    @Thedonwon look at your previous question and the already existing answer and the comments on that answer. – 2017-02-26
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    Please read [this post](http://meta.math.stackexchange.com/a/9960) and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you [edit] your question to include some motivation, and an explanation of your own attempts. – 2017-02-26

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Normalize the values in order to use the standard Normal distribution (whose values are usually given in a table in a statistics course).

\begin{align} z_1 &= \frac{x_1-\mu}{\sigma} & z_2 &= \frac{x_2-\mu}{\sigma} \\ &= \frac{-1.92-2}{2} & &= \frac{5.92-2}{2} \\ &= -1.96 & &= 1.96 \\ \end{align}

Recall that the Normal distribution is symmetric around the mean.

Let $Z$ be a standard uniformly distributed variable. \begin{align} P(-1.96 \leq Z \leq 1.96) &= P(|Z| \leq 1.96) \\ &= 1 - P(|Z| > 1.96) \\ &= 1 - 2 \cdot P(Z < -1.96) \\ &= 1 - 2 \cdot P(Z < -1.96) \\ &\approx 1 - 2 \dot (0.025) \\ &\approx 1 - 0.05 \\ &\approx 0.95 \end{align}

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    And now we just have to wait for installment number 3 of the same question by this OP... Pity. – 2017-02-26