 ## S11 Probability and the normal distribution Even when data follows a normal distribution, different data sets will have their own mean and standard deviation and a different bell shaped curve as illustrated below. But every score in a normally distributed data set, regardless of the shape, has an equivalent score in the standard distribution. The mean of a normal distribution corresponds to a standardised score of $$0$$ and we can see that $$\mu\pm\sigma\rightarrow\pm1$$, $$\mu\pm2\sigma\rightarrow\pm2$$ and $$\mu\pm3\sigma\rightarrow\pm3$$.

### $$z-$$Scores

For other values of the mean and standard deviation, we can use the formula \begin{align*} z & =\frac{x-\mu}{\sigma} \end{align*} to find $$z$$ scores (where the $$z$$ score is the number of standard deviations away from the mean).

For example, if we have a Normal Distribution with a mean of $$1.2$$ and a standard deviation of $$0.4$$, then to find the standardised score for $$x=1.7$$: \begin{align*} z & =\frac{(x-\mu)}{\sigma}\\ & =\frac{(1.7-1.2)}{0.4}\\ & =1.25 \end{align*}

A score of $$1.7$$ in the distribution with mean $$1.2$$ and standard deviation $$0.4$$ is equivalent to a standardised score of $$1.25$$. Alternatively, we could say that the score $$1.7$$ is $$1.25$$ standard deviations above the mean for that distribution.

### Calculating Probabilities Using $$z-$$Scores

Once we have converted the scores of our distribution into standard scores or $$z$$-scores we can use normal distribution tables to calculate precise percentages and probabilities.

The normal distribution is a continuous distribution, so we can find the probability that $$x$$ is greater than or less than a particular value, but not that $$x$$ is equal to a particular value.

Because the total area under the standardised curve is $$1$$, $$\Pr\left(z<\beta\right)$$ is equivalent to the area to the left of $$\beta.$$ #### Example 1

If the mean maximum temperature for Melbourne in January is $$25.9^{\circ}$$C with a standard deviation of $$2.1^{\circ}$$ what is the probability that the mean maximum temperature for January $$2015$$ will be above $$28^{\circ}$$C?

First draw a diagram. \begin{align*} x & =28\\ z & =\frac{(x-\mu)}{\sigma}\\ & =\frac{(28-25.9)}{2.1}\\ & =1.\\ \Pr\left(x>28\right) & =\Pr\left(z>1\right)\\ & =1-\Pr\left(z<1\right)\\ & =1-0.8413\ \ \left(0.8413\textrm{ is from the table at end of this module}\right)\\ & =0.1587. \end{align*} That is, the probability that the mean maximum temperature for January $$2015$$ will be above $$28^{\circ}$$C is $$0.1587.$$

#### Example 2

The top $$0.5$$% of students applying for Stato university are given full scholarships. If the mean score on the entrance exam is $$372$$ and the standard deviation is $$40$$, what mark is needed to obtain a scholarship?

First draw a diagram. The area to the right of $$x_{s}=0.005$$.1 We want to find $$x_{s}$$.

We know that $$\Pr\left(x>x_{s}\right)=0.005$$ but we must first find the $$z$$- score corresponding to this. Let’s call it $$z_{s}.$$ Then \begin{align*} \Pr\left(z>z_{s}\right) & =0.005 \end{align*} and \begin{align*} \Pr\left(z<z_{s}\right) & =1-0.005\\ & =0.9950. \end{align*} Using the table at the end of this module, we look up $$0.995$$ in the body of the table and see that the corresponding $$z$$- score is $$z_{s}=2.57$$ or $$z_{s}=2.58$$. We will take $$z_{s}=2.58.$$2 You could take the average of $$2.57$$ and $$2.58$$ in this case, that is $$2.575$$ but two decimal places of accuracy is usually enough. Now using the formula we have \begin{align*} z_{s} & =\frac{x_{s}-\mu}{\sigma}\\ 2.58 & =\frac{x_{s}-372}{40}.\\ \textrm{Rearranging we have, }x_{s} & =2.58\times40+372\\ & =475.2\,. \end{align*} So applicants who score more than $$475.2$$ will get a scholarship.

### Exercises

1. If a population has a mean I.Q. of $$100$$ and a standard deviation of $$15$$, find:

1. the probability that an individual chosen at random will have an I.Q. between $$110$$ and $$130$$.

2. the probability that an individual chosen at random will have an I.Q. greater than $$87$$.

(a) $$0.2297\quad$$(b) $$0.8069$$

2. A coffee machine is regulated to deliver $$200\,mL$$ per cup. In fact, the amount of coffee varies, following a normal distribution with a mean of $$200\,mL$$ and a standard deviation of $$10\,mL$$.

1. What is the probability that a cup contain less than $$195\,mL$$?

2. What is the probability that a cup will contain more than $$220\,mL$$?

3. What is the probability of a cup containing between $$195$$ and $$215\,mL$$?

(a) $$0.3085\quad$$(b) $$0.0228\quad$$(c) $$0.6247$$

3. The heights of a group of men follow a normal distribution with a mean of $$180\,cm$$ and a standard deviation of $$6\,cm$$.

1. What is the probability that a man chosen from this group is less than $$185\,cm$$ tall?

2. If the tallest $$10\%$$ of this group are automatically eligible for a basketball team, what is the qualifying height?

(a) $$0.7977\quad$$(b) $$187.69\,cm$$
The table below gives the probability (the shaded area to the left) for a particular $$z$$-value. 