Revise The normal distribution for Cambridge 9709 (5.5)

A continuous random variable $X$ that is normally distributed is defined by two parameters: its mean, $\mu$, and its variance, $\sigma^2$. We write this as $X \sim N(\mu, \sigma^2)$. The graph of its probability density function is a symmetric, bell-shaped curve. The mean $\mu$ determines the location of the centre of the curve, while the standard deviation $\sigma$ (the square root of the variance) determines its spread. A larger $\sigma$ results in a wider, flatter curve, while a smaller $\sigma$ gives a taller, narrower curve.

Since there are infinitely many possible normal distributions (one for each combination of $\mu$ and $\sigma^2$), it would be impossible to have probability tables for all of them. Instead, we standardise any normal variable to convert it to the standard normal distribution, denoted by $Z$, which has a mean of 0 and a variance of 1. We write this as $Z \sim N(0, 1^2)$. The formula for this transformation is:

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

This $Z$-score tells us how many standard deviations an original value $x$ is away from its mean $\mu$. Once we have a $Z$-score, we can use the standard normal tables (provided in the exam formula book MF19/MF20) to find the required probabilities. The tables give $\Phi(z) = P(Z < z)$, the area to the left of $z$. For other areas, we use symmetry properties, such as $P(Z > z) = 1 - \Phi(z)$ and $P(Z < -z) = 1 - \Phi(z)$.

For a binomial distribution $X \sim B(n, p)$, calculating probabilities can be very tedious if $n$ is large. Fortunately, if $n$ is large enough and $p$ is not too close to 0 or 1, the shape of the binomial distribution is very similar to a normal distribution. We can use a normal distribution as an approximation, provided the conditions $np > 5$ and $n(1-p) > 5$ are met.