Normal Distribution

    As we continue onwards in probability application and theory, we encounter the Normal Distribution, or oftentimes called the Gaussian Distribution, which is a continuous probability distribution characterized by being centered around its mean, implying that data near the mean occur more frequently than data further away. Its continuous probability density function is given by:

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

where,

$$\begin{align*} x &= \text{the random varaiable} \ \mu &= \text{the population mean} \ \sigma &= \text{the population standard deviation} \ \end{align*}$$

Properties of A Normal Distribution

1. The normal distribution is symmetric around its mean, $\mu$.
2. The distribution tends from $-\infty$ to $+\infty$.
3. The curve of the distribution is bell-shaped.
4. The mean, median and mode of the distribution are all equal and located at the center of the distribution.
5. The distribution's spread is determined by its standard deviation, $\sigma$, where $68%$ of the distribution lies in $\mu \pm \sigma$, $95%$ of the distribution lies in $\mu \pm 2\sigma$ and approximately $99.7%$ of the distribution in $\mu \pm 3\sigma$. Everything past this point is typically considered an outlier.

Defining a Normal Distribution

    For a continous random variable, $X$, which follows a normal distribution with mean $\mu$ and variance, $\sigma^2$, it can then be stated that:

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

where,

$$P(X < a) = P(Z < z)$$ $$P(X > a) = P(Z > z)$$ $$P(a < X < b) = P(z_1 < Z < z_2)$$ $$P(X=a) \approx P(a-0.5 < X < a+0.5) \approx P(z_1 < Z < z_2)$$

where $Z$ represents the standardized normal distribution, and $z$, $z_1$, and $z_2$ represented standardized normal scores. Note: In a continous probability distribution, $\le$ and $<$ are equivalent likewise with $\ge$ and $>$.

Standardized Normal Distribution

    A standardized normal distribution allows for comparability of different normal distributions with varying population means and standard deviations. It is given by:

Let $Z$ be the standardized normal random variable.

$$Z \sim \mathcal{N}(0, 1)$$

And, for a normal distribution with continuous random variable $X$, $Z$ is found using:

$$Z = \frac{X-\mu}{\sigma}$$

    In most textbooks, at the back, you may find a cumulative normal distribution table which shows all probabilities for a standardized normal distribution for $P(Z \le z)$. This cumulative probability mass is denoted by $\Phi(z)$ read as 'phi' of $z$. It is also noteworthy that most tables go up to 3 decimal places, therefore when calculating the standardized z-score of $a$, round it to 3 decimal places.

Example:

Let $X$ be a continuous random variable with mean $\mu$ and variance $\sigma^2$, where $\mu = 5$ and variance = $4$. Then, it follows that:

$$X \sim \mathcal{N}(\mu, \sigma^2)$$ $$X \sim \mathcal{N}(3, 4)$$

Calculate: $P(X < 4)$

$$\begin{align*} P(X < 4) \ = P(Z < \frac{4-3}{\sqrt{4}}) \ = P(Z < \frac{4-3}{2}) \ = P(Z < \frac{1}{2}) \ = \Phi(\frac{1}{2})\ = 0.6915 \ = 0.692 \text{ (to 3 s.f.)} \ \end{align*}$$

Please note that this review does not cover how to read your z-tables. A variety of resources are available online, such as: https://www.youtube.com/watch?v=3M6pQq3Qko4

Normal Approximation to Binomial

    The normal distribution can be utilized to approximate the discrete binomial distribution. It follows for a discrete random variable, $X$, modelled binomially with parameters, $n$ and $p$, such that $np \ge 5$ and $n(1-p) \ge 5$ then:

$$X \sim \mathcal{N}(np, npq), \quad \text{approximately.}$$

    It is crucial to note that we are utilizing a continuous distribution to approximate a discrete distribution; therefore, the random variable $X$ must undergo continuity corrections to account for this discrepancy, ensuring the probability mass is accurately captured:

1. $P(X > a) \rightarrow P(X > a+0.5)$
2. $P(X \ge a) \rightarrow P(X > a-0.5)$
3. $P(X < a) \rightarrow P(x < a-0.5)$
4. $P(X \le a) \rightarrow P(X < a+0.5)$

Questions

1. A scientist is investigating the lengths of the leaves of a certain type of plant. The scientist assumes that the lengths of the leaves of this type of plant are normally distributed. He measures the lengths, $x cm$, of the leaves of a random sample of $8$ plants of this type. His results are as follows. $$3.5, 4.2, 3.8, 5.2, 2.9, 3.7, 4.1, 3.2$$ i. Calculate the mean, $\mu$, and variance $\sigma^2$ (Ans: $\mu = 3.825$, $\sigma^2 \approx 0.434$) ii. State the distribution and its parameters. (Ans: $X \sim \mathcal{N}(3.825, 0.434)$) iii. Calculate $P(X < 4)$. (Ans: $0.606$)

(Adapted from CIE Further Probability & Statistics, October/November 2024, Cambridge Assessment Internal Education)

2. Paul's training session times, in minutes, are modelled as a normal variable, distributed with a standard deviation of $18$. In $3$ sessions out of $4$, Paul takes more than $80$ minutes to complete his training. a. Determine to the nearest minute, Paul's mean training time. (Ans: $\mu \approx 92$) b. Find the probability that one of Paul's sessions will last more than $2$ hours. (Ans: $0.05991$)

Paul trained 4 times last week, and all his sessions lasted more than $100$ minutes.

c. Find the probability that at least one of his sessions last week lasted more than $2$ hours. (Ans: $0.1824$)

(Mr Astbury Maths, Adapted from A-Level Mathematics Paper)