Normal Distribution – Numerical Problems with Solutions

Normal Distribution – Numerical Problems with Solutions

Author: Bindeshwar Singh Kushwaha
Platform: PostNetwork Academy


1. Definition of Normal Distribution

A continuous random variable $X$ follows a Normal Distribution with mean $\mu$ and variance $\sigma^2$ if its probability density function (PDF) is:

$$
f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{ -\frac{(x – \mu)^2}{2\sigma^2} }, \quad -\infty < x < \infty
$$

It is denoted as:

$$
X \sim N(\mu, \sigma^2)
$$

  • The curve is symmetric about the mean $\mu$.
  • The spread depends on the standard deviation $\sigma$.

2. Writing the PDF of Given Normal Distributions

Example 1

(a) If $X \sim N(10, 25)$

$$
f(x) = \frac{1}{\sqrt{2 \pi 25}} \exp\left( -\frac{(x – 10)^2}{2 \cdot 25} \right)
$$

(b) If $X \sim N(36, 20)$

$$
f(x) = \frac{1}{\sqrt{2 \pi 20}} \exp\left( -\frac{(x – 36)^2}{2 \cdot 20} \right)
$$

(c) If $X \sim N(0, 2)$

$$
f(x) = \frac{1}{\sqrt{2 \pi 2}} \exp\left( -\frac{x^2}{2 \cdot 2} \right)
$$


3. Effect of Variance on the Curve

  • Larger variance → wider and flatter curve.
  • Smaller variance → narrower and taller curve.

For example:

  • $N(10,25)$ → Wide spread
  • $N(0,2)$ → Narrow distribution


4. Question: Find the PDF

If $X \sim N\left(\frac{1}{2}, \frac{4}{9}\right)$

Solution

Here,

$$
\mu = \frac{1}{2}, \quad \sigma^2 = \frac{4}{9}
$$

Therefore,

$$
\sigma = \frac{2}{3}
$$

Substituting in the standard formula:

$$
f_X(x)=\frac{3}{2\sqrt{2\pi}}e^{-\frac{9(x-\tfrac{1}{2})^2}{8}}
$$


5. Another Example

If $X \sim N(-40,16)$

Here,

$$
\mu=-40, \quad \sigma^2=16
$$

$$
\sigma=4
$$

The PDF becomes:

$$
f_X(x)=\frac{1}{4\sqrt{2\pi}}e^{-\frac{(x+40)^2}{32}}
$$


6. Identifying Parameters from Given PDF

(i)

$$
f(x) = \frac{1}{6\sqrt{2\pi}} \exp\left( -\frac{(x-46)^2}{2 \cdot 36} \right)
$$

Comparing with standard form:

$$
\mu = 46, \quad \sigma^2 = 36
$$


(ii)

$$
f(x) = \frac{1}{\sqrt{2\pi}} \exp\left( -\frac{(x-60)^2}{2 \cdot 16} \right)
$$

$$
\mu = 60, \quad \sigma^2 = 16
$$


7. Finding Mean and Variance

Case (i)

$$
f(x)=\frac{1}{2\sqrt{2\pi}}e^{-\frac{x^2}{8}}
$$

Comparing with standard form:

$$
\mu = 0, \quad \sigma^2 = 4
$$


Case (ii)

$$
f(x)=\frac{1}{2\sqrt{\pi}}e^{-\frac{(x-2)^2}{4}}
$$

$$
\mu = 2, \quad \sigma^2 = 2
$$


Conclusion

The Normal Distribution is one of the most important probability distributions in statistics.
Understanding how to:

  • Write the PDF
  • Identify mean and variance
  • Compare with standard form
  • Understand variance effects

is essential for mastering probability and statistical inference.


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