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

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

Question
Random variables
asked 2021-01-31

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

Answers (1)

2021-02-01

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 \((\mu)\) and population variance \(\displaystyle{\left(\sigma^{{{2}}}\right)}\). It is symmetric with respect to its mean.
Given information:
\(\displaystyle{X}\sim{N}{\left({6},{4}\right)}\)
Therefore,
\(\displaystyle\mu={6}\)
\(\displaystyle\sigma^{{{2}}}={4}\)

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

0

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