# The Normal Distribution

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The **Normal Distribution** is:

- A very common probability density function

- A very realistic model of many observed distributions in real life

- A curve with a symmetrical, bell-shape

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Let **X** be a continuous random variable that follows a normal distribution

with

- mean = $\mu$
- standard deviation = $\sigma$

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## Notation:

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

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## WARNING!!

Most of the calculations will involve the **standard deviation**, but the notation shows the **variance**!

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## Properties of the Normal Distribution

The equation for the normal distribution is:

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Notice that the distribution has the following properties:

- Symmetrical

- mean = $\mu$

- median = $\mu$

- mode = $\mu$

- x-axis is an asymptote

- Maximum value is at $\left( \mu, \; \dfrac{1}{\sigma \sqrt{2\pi}} \right)$

- $Pr(a < X < b) = \displaystyle{ \int_a^b \; f(x) \; dx}$

- $\displaystyle{ \int_{-\infty}^{+\infty} \; f(x) \; dx} = 1$

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Notice the curve is very close to the **x-axis** at $\mu \pm 3\sigma$

… but not touching because the x-axis is an asymptote

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**NOTE:**

- $\mu$ represents a translation of the curve to the right $\big( \text{if } \mu > 0 \big)$

- $\sigma$ represents a dilation of the curve from the y-axis (in the x-direction)

- and $\sigma$
**also**represents a dilation of $\dfrac{1}{\sigma}$ from the x-axis (in the y-direction)

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## Standard Normal Distribution

When $\mu = 0 \text{ and } \sigma = 1$ we get the **Standard Normal Distribution**

We use **Z** for the **Standard Normal Distribution**

… … … $Z \sim N \big(0,\; 1\big)$

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Therefore the equation of a Standard Normal Distribution is:

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**Note:** the **Standard Normal Distribution** is

- centred on the y-axis $\big( \mu = 0 \big)$

- the x-axis is an asymptote $\big( y = 0 \big)$

- is close to the x-axis at $x = \pm3$

- Has a mean, median, mode at $x = 0$

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## Confidence Intervals

For the normal distribution, the confidence intervals (introduced in Ch 10 for discrete variables) are accurate enough for most practical purposes (but are still approximate).

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### 68% Confidence Interval

Approximately **68%** of the data lies within **one** standard deviation of the mean.

… … $Pr \big( \mu - \sigma < X < \mu + \sigma \big) \approx 0.68$

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### 95% Confidence Interval

Approximately **95%** of the data lies within **two** standard deviations of the mean.

… … $Pr \big( \mu - 2\sigma < X < \mu + 2\sigma \big) \approx 0.95$

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### 99.7% Confidence Interval

Approximately **99.7%** of the data lies within **three** standard deviations of the mean.

… … $Pr \big( \mu - 3\sigma < X < \mu + 3\sigma \big) \approx 0.997$

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### Example 1

Scores from a maths test are normally distributed with

a mean of 84 and a standard deviation of 4

… … … $X \sim N\big(84, \; 16\big)$

Therefore its equation is this:

… **a)** .. Sketch the graph of **X**

… **b)** .. Find the approximate percentage of scores between **80** and **88**

… **c)** .. Find the approximate percentage of scores below **84**

… **d)** .. Find the approximate percentage of scores above **92**

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**Solution**

… **a)** .. Sketch the graph of **X**

… … … Use $\mu \text{ and } \sigma$ to produce the scale

… … … A vertical line at $\mu$ can help to keep the graph symmetrical

… … … Don't forget to label the turning point

… … … The bell shape should approach the axis close to $\mu \pm 3\sigma$.

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… **b)** .. Find the approximate percentage of scores between **80** and **88**

… … … **80** is 4 below the mean $\big( \mu - \sigma \big)$

… … … **88** is 4 above the mean $\big( \mu + \sigma \big)$

… … … so

… … … $80 < x < 88$ is one standard deviation each side of the mean

… … … so

… … … **68%** (approx) of the scores are between **80** and **88**

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… **c)** .. Find the approximate percentage of scores below **84**

… … … median = 84

… … … so

… … … **50%** of the scores are below **84**

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… **d)** .. Find the approximate percentage of scores above **92**

… … … **92** is 8 above the mean $\big( \mu + 2\sigma \big)$

… … … **95%** are within **two** standard deviations of the mean, so

… … … **5%** are outside **two** standard deviations of the mean

… … … The distribution is symmetrical (half above, half below)

… … … so **2.5%** (approx) are above **92**

… … … {The other **2.5%** are below **76**}

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## Normal Distribution on the Classpad Calculator

… … From the **Main** screen, go to the **Interactive** menu and select **Distribution**

… … Select **Normalcdf** … {Do NOT use normalpdf}

… … Then enter lower limit, upper limit, standard deviation, mean

… … {Using the interactive version will make sure you enter values in the correct order}

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To calculate: $Pr \big( \text{lower } < X < \text{ upper} \big)$

… … **normalcdf (**lower, upper, $\sigma$, $\mu$ **)**

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## Caution

Do not write calculator notation as a part of your answer for SACs or Exams.

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### Example 1 (continued)

For the variable: $X \sim N \big(84,\; 16\big)$

Use the CAS to find these percentages correct to one decimal place

… **a)** .. Find the % of scores between **80** and **88**

… **b)** .. Find the % of scores below **84**

… **c)** .. Find the % of scores above **92**

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**Solution**

… **a)** .. Find the % of scores between **80** and **88**

… … … **enter:** $\text{normalcdf lower } = 80, \text{ upper }= 88,\; \sigma = 4,\; \mu = 84$

… … … $Pr \big(80 < X < 88\big) = 0.6827 = 68.3 \%$

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… **b)** .. Find the % of scores below **84**

… … … **enter:** $\text{normalcdf lr } = -\infty. \text{ ur } = 84,\; \sigma = 4,\; \mu = 84$

… … … $Pr\big( X < 84\big) = 0.5 = 50.0 \%$

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… **c)** .. Find the % of scores above **92**

… … … **enter:** $\text{normalcdf lr } = 92, \text{ ur }= \infty,\; \sigma = 4.\; \mu = 84$

… … … $Pr\big(X > 92\big) = 0.0228 = 2.3\%$

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**Note:** Do not write calculator notation in your answers in tests/exams. I include it here so you can follow what is going on.

Working and answers should be written in mathematical notation:

eg your answer might look like:

… … $\text{For } X \sim N\big(84, \; 16\big)$

… … $Pr(80 < X < 88) = 0.6827$

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### Example 2

Sketch the graph of $X \sim N\big(20, \; 9\big)$

**Solution:**

… … $\mu = 20$

… … $\sigma^2 = 9 \qquad so \quad \sigma = 3$

… … Mark in scale, using steps of **3** either side of **20**

… … Draw the bell-shaped curve, centred at **20**

… … The curve should get close to the axis near **11** and **29** $\big(\mu \pm 3\sigma \big)$

… … Maximum is at $\left( \mu, \; \dfrac{1}{\sigma \sqrt{2\pi}} \right)$

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**Don't Forget!**

… … $Pr(X < a) = Pr(X \leqslant a)$

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## Interactive Approximate Normal Distribution

The following link is to an interactive that demonstrates:

- the difference between an approximate normal distribution and the theoretical normal distribution

- the way that increasing the sample size makes the approximate distribution approach the theoretical normal

- the effect of changing the standard deviation on the shape of the curve

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