10E

The general normal distribution

Normal distributions with any mean and standard deviation as shifted and stretched standard normals, z-scores, finding probabilities by standardising, comparing scores from different distributions, and working backwards from a probability.

Shifting and stretching

Every normal curve is the standard normal curve shifted so its centre is at $\mu$ and stretched sideways by a factor of $\sigma$. It is still bell-shaped and symmetric, with total area $1$.

Normal curves

The dashed curve is the standard normal. A larger standard deviation gives a wider, lower bell, because the area stays 1.

Try this

What happens to the height of the peak when $\sigma$ is halved?

z-scores

z-score

$$z = \frac{x - \mu}{\sigma} \qquad x = \mu + z\sigma$$

The $z$-score says how many standard deviations a value is above (positive) or below (negative) the mean. If $X$ is normal, then $Z = \dfrac{X - \mu}{\sigma}$ is standard normal, so $P(X \le x) = P(Z \le z)$.

Example 1●●

Heights are normal with mean $170$ cm and standard deviation $8$ cm. Find the probability that a person is shorter than $182$ cm, and taller than $162$ cm.

$z = \dfrac{182 - 170}{8} = 1.5$, so $P(X \lt 182) = P(Z \lt 1.5) = 0.9332$.

$z = \dfrac{162 - 170}{8} = -1$, so $P(X \gt 162) = P(Z \gt -1) = P(Z \lt 1) = 0.8413$.

Example 2: comparing scores●●

Mia scored $72$ in Maths (mean $60$, sd $8$) and $80$ in English (mean $70$, sd $10$). Which result is better compared with the class?

Maths: $z = \dfrac{72 - 60}{8} = 1.5$. English: $z = \dfrac{80 - 70}{10} = 1$.

Maths is better: it is further above the mean, measured in standard deviations.

Example 3: working backwards●●●

For the heights in Example 1, find the height exceeded by only $2.28\%$ of people.

$P(Z \gt z) = 0.0228$ means $P(Z \le z) = 0.9772$, so $z = 2$.

$x = 170 + 2 \times 8 = 186$ cm.

Flashcards

z-score

$z = \dfrac{x - \mu}{\sigma}$

Score from a z-score

$x = \mu + z\sigma$

Negative z-score

The value is below the mean.

Practice

$\mu = 170$ and $\sigma = 8$. Find the $z$-score of $182$.
$\mu = 170$ and $\sigma = 8$. Find the $z$-score of $162$.
Heights are normal with $\mu = 170$ and $\sigma = 8$. Find $P(X \lt 182)$, given $P(Z \lt 1.5) = 0.9332$.
For the same heights, find $P(X \gt 162)$, given $P(Z \lt 1) = 0.8413$.
For the same heights, find the height with $z = 2$.
$\mu = 65$ and $\sigma = 4$. Which score has $z = -1.5$?
Maths: $72$ (mean $60$, sd $8$). English: $80$ (mean $70$, sd $10$). Which is the better result?
Increasing $\sigma$ makes the normal curve:
A $z$-score of $0$ means the value: