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

Question

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

Answer 1

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 $(μ)$ and population variance $(σ^{2})$ . It is symmetric with respect to its mean.
Given information:
$X \sim N (6, 4)$
Therefore,
$μ = 6$
$σ^{2} = 4$

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

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

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