Title: Default for The Normal Distribution
Course: HELM Workbooks 2023-2024

Exercise

The random variable \(X\) follows a normal distribution with mean \([value]\) and variance \([value]\). When \(X\) takes values between \([value]\) and \([value]\), between which values does the standardised normal variable \(Z\) lie?

\( \leq Z \leq \)
Feedback: Solution The transformation to standardise a Normal random variable is \(Z = \dfrac{X - \mu}{\sigma}\) where \(\mu\) is the expectation of \(X\) and \(\sigma\) is the standard deviation of \(X\). When \(X\) lies in the interval \([value]\) to \([value]\), then the corresponding lower bound for \(Z\) is: \[\dfrac{[value] - [value]}{\sqrt{[value]}} = [value],\] and the corresponding upper bound is: \[\dfrac{[value] - [value]}{\sqrt{[value]}} = [value].\] Therefore, when \([value] \leq X \leq [value]\) then \([value] \leq Z \leq [value]\).

Exercise

What are the probabilities that \(Z\) takes values between:

(You must give your answers to 4 decimal places, i.e. keep full accuracy from the Table)

1. \(0\) and \([value]\)?
2. \(0\) and \([value]\)?
3. \(0\) and \([value]\)?
4. \(0\) and \([value]\)?
Feedback: Solution 1. The required value in the table comes from the row \([value]\) and the column \([value]\). Therefore \(\text{P}(0 \leq Z \leq [value]) = [value]\). 2. The required value in the table comes from the row \([value]\) and the column \([value]\). Therefore \(\text{P}(0 \leq Z \leq [value]) = [value]\). 3. The required value in the table comes from the row \([value]\) and the column \([value]\). Therefore \(\text{P}(0 \leq Z \leq [value]) = [value]\). 4. There is no value in the table for \([value]\) and so we must estimate it using linear interpolation based on the the neighbouring values of \([value]\) and \([value]\). From the table, the corresponding probabilities are \([value]\) and \([value]\). This means that our estimate for the required probability is: \[\begin{align*}&\text{P}(0 \leq Z \leq [value])\\ & ~~~\approx [value] + \dfrac{[value] - [value]}{[value] - [value]} ([value] - [value])\\ &~~~= [value]~(\text{to 4 d.p.})\end{align*}\] Using a statistical calculator, we see that the answer is \(\text{P}(0 \leq Z \leq [value]) = [value]\) to 4 decimal places.

Exercise

Find the following probabilities. (A simple sketch of the standard normal curve will help.)

You must give your answers to 4 decimal places, i.e. keep full accuracy from the Table.

1. \(\text{P}(0 < Z < [value]) = \)
2. \(\text{P}(Z >[value]) = \)
3. \(\text{P}([value] < Z < [value]) = \)
4. \(\text{P}(Z <[value]) = \)
5. \(\text{P}(-[value] < Z < 0) = \)
6. \(\text{P}(Z < -[value]) = \)
7. \(\text{P}(-[value] < Z < -[value]) = \)
8. \(\text{P}(-[value] < Z < [value]) = \)
Feedback: Solution 1. Reading directly from the table, \(\text{P}(0<Z<[value]) = [value]\) to 4 decimal places. 2. By the complement rule and the fact that \(\text{P}(Z<0) = 0.5\): \[\begin{align} \text{P}(Z >[value]) & = 1 - \text{P}(Z < [value]) = 1 - \left(0.5 + \text{P}(0<Z<[value])\right) \\ &= 0.5 − [value] = [value] ~~(\text{to 4 d.p.}). \end{align}\] 3. The probability of this interval can be expressed as a difference of two probabilities: \[\begin{align}\text{P}([value] < Z < [value]) & = \text{P}(0 < Z < [value]) − \text{P}(0 < Z < [value]) \\ &= [value] − [value] = [value] ~~(\text{to 4 d.p.}). \end{align}\] 4. By partitioning the interval at \(0\) means that: \[\begin{align} \text{P}(Z<[value]) & = \text{P}(Z<0)+\text{P}(0<Z<[value]) \\ &= 0.5 + [value] = [value] ~~(\text{to 4 d.p.}). \end{align}\] 5. Using the symmetry of the normal distribution around \(0\): \[\text{P}([value] < Z < 0) = \text{P}(0 < Z < [value]) = [value]~~(\text{to 4 d.p.}).\] 6. From the symmetry of the normal distribution: \[\text{P}(Z < -[value]) = \text{P}(Z > [value]) = 0.5 − \text{P}(0<Z<[value]) = [value]~~(\text{to 4 d.p.}).\] 7. Using the symmetry of the normal distribution around \(0\): \[\text{P}(−[value] < Z < −[value]) = \text{P}([value] < Z < [value]) = [value]~~(\text{to 4 d.p.}).\] 8. Partitioning the interval at \(0\) and utilising the symmetry of the normal distribution gives: \[\begin{align*}\text{P}([value] < Z < [value]) &= \text{P}([value]<Z<0) + \text{P}(0<Z<[value])\\ &= \text{P}(0 < Z < [value]) + \text{P}(0 < Z < [value])\\ &= [value] ~~(\text{to 4 d.p.}).\end{align*}\]

Exercise

If a random variable \(X\) has a standard normal distribution. Find the probability, to 4 decimal places, that it takes a value:

1. less than \([value]\)
2. greater than \([value]\)
3. between \(0\) and \([value]\)
4. between \([value]\) and \([value]\)
Feedback: Solution 1. By partitioning the interval at \(0\): \[\text{P}(Z < [value]) = \text{P}(Z < 0) + \text{P}(0 < Z < [value]) = 0.5 + [value] = [value]~~(\text{to 4 d.p.}).\] 2. By the complement rule: \[\begin{align*}\text{P}(Z > [value]) &= 1 - \text{P}(Z <[value]) \\&= 1 - \left(0.5 + \text{P}(0 < Z < [value])\right) \\&= 0.5 - [value] \\&= [value]~~(\text{to 4 d.p.}).\end{align*}\] 3. Reading directly from the table, \(\text{P}(0<Z<[value]) = [value]~~(\text{to 4 d.p.})\). 4. Utilising the symmetry of the normal distribution gives: \[\begin{align*}\text{P}([value] < Z < [value]) &= \text{P}([value] < Z < [value]) \\&= \text{P}(0 < Z < [value]) - \text{P}(0 < Z < [value]) \\&= [value] - [value] \\&= [value]~~(\text{to 4 d.p.}). \end{align*}\]

Exercise

If \(Z\) has a standard normal distribution, find \(k\)  to 2 decimal places in each of the following cases:

1. If \(\text{P}(Z < k) = [value]\), then \(k =\)
2. If \(\text{P}(Z < k) = [value]\), then \(k =\)
3. If \(\text{P}(0 < Z < k) = [value]\), then \(k=\)
Feedback: Solution 1. As \(\text{P}(Z < k) < 0.5\) then the value for \(k\) will be negative. The standard Normal table only provides values, so we first need to transform the probability using the symmetry of the Normal distribution: \(\text{P}(Z < k) = \text{P}(Z > -k) = 1 - \text{P}(Z < -k)\) where the last equality comes from applying the complement rule. Next we need to partition the probability with respect to the mean of \(0\): \(\text{P}(Z < k) = 1 - \left[\text{P}(Z<0) + \text{P}(0<Z<-k)\right] = 0.5 - \text{P}(0<Z<-k)\) since \(\text{P}(Z<0) = 0.5\). Therefore: \[\begin{align*}\text{P}(Z < k) &= [value]\\0.5 - \text{P}(0<Z<-k) & = [value]\\ \text{P}(0<Z<-k) & = [value]\end{align*}\] Reading from the standard Normal table we see that \(-k = [value]\) and so \(k = [value]\) to 2 decimal places. 2. As \(\text{P}(Z < k) > 0.5\) then the value for \(k\) will be positive. Partitioning the probability with respect to the mean of \(0\) gives: \[\begin{align*} \text{P}(Z < k) &= [value]\\ \text{P}(Z < 0) + \text{P}(0 <Z < k) &= [value]\\ \text{P}(0 <Z < k) &= [value] - \text{P}(Z < 0)\\ \text{P}(0 <Z < k) &= [value] \end{align*}\] Reading from the standard Normal table we see that \(k = [value]\) to 2 decimal places. 3. The structure of the probability is in the correct for to apply the standard Normal table, which we can read directly to find that \(k = [value]\) to 2 decimal places.

Exercise

A worn, poorly set-up machine is observed to produce components whose length \(X\) follows a normal distribution with mean \([value]\text{cm}\) and variance \([value]\text{cm}^2\). Calculate, to 4 decimal places:

1. the probability that a component is at least \([value]\text{cm}\) long:
2. the probability that the length of a component lies between \([value]\text{cm}\) and \([value]\text{cm}\):
Feedback: Solution Let \(X\) be the random variable for the length of the component produced by the machine, then \(X \sim N([value], [value])\). The standardised Normal distribution \(Z \sim N(0,1)\) is obtained through the transformation \(Z = \dfrac{X - [value]}{\sqrt{[value]}} = \dfrac{X - [value]}{[value]}\). 1. The first stage is to transform the probability with respect to the standard Normal random variable: \[\text{P}(X \geq [value]) = \text{P}\left(\dfrac{X - [value]}{[value]} \geq \dfrac{[value] - [value]}{[value]}\right) = \text{P}(Z \geq [value]).\] Next, transform the probability so that we can use the standard Normal table: \[\text{P}(Z \geq [value]) = 1 - \text{P}(Z < [value]) = 0.5 - \text{P}(0 < Z < [value]).\] Reading this value from the table obtains the required probability to be \(\text{P}(X \geq [value]) = 0.5 - [value] = [value]\) to 4 decimal places. 2. Again, the first stage is to standardise the probability: \[ \begin{align} \text{P}([value] < X < [value]) & = \text{P}\left(\dfrac{[value] - [value]}{[value]} < \dfrac{X - [value]}{[value]} < \dfrac{[value] - [value]}{[value]}\right) \\ & = \text{P}([value] < Z < [value]). \end{align}\] Note that this probability is symmetric around the mean of \(0\), so: \[\text{P}([value] < Z < [value]) = 2 \times \text{P}(0 < Z < [value]).\] The value for \([value]\) is not in the standard Normal table, but it is the mid-point between \([value]\) and \([value]\). So an approximate of this probability is: \[\text{P}(0 < Z < [value]) \approx \dfrac{[value]+[value]}{2}.\] Using this value, the required probability is approximated to be: \(\text{P}([value] < X < [value]) \approx 2 \times \dfrac{[value]+[value]}{2} = [value]\)to 4 decimal places. When using a statistical calculator, a more accurate calculation of the probability returns \(\text{P}([value] < X < [value]) = [value] \) to 4 decimal places.

Exercise

The resistance of a strain gauge is normally distributed with a mean of \([value]\) ohms and a standard deviation of \([value]\) ohms. To meet the specification, the resistance must be within the range \([value] \pm [value]\) ohms.

1. What percentage of gauges are unacceptable?
(You must keep full accuracy throughout the question).

First, state the upper and lower limits of acceptable resistance, \(x_1\) and \(x_2\) respectively, and find the corresponding \(Z\)-values, \(z_1\) and \(z_2\):
\(x_1 = \) and \(x_2 = \)
\(z_1 = \) and \(z_2 = \)
Now, using a suitable sketch, calculate the probability that \(z_1 < Z < z_2\):
\(\text{P}(z_1 < Z < z_2) = \)
Therefore, the proportion of unacceptable gauges is \(%\)
2. To what value must the standard deviation be reduced if the proportion of unacceptable gauges is to be no more than \([value]\%\)?
(Keep full accuracy throughout the question, and give your final answer to 2 decimal places.)

First sketch the standard normal curve, marking on it the lower and upper values, \(z_1\) and \(z_2\), and appropriate areas.
Now use the Table to find \(z_2\), and hence write down the value of \(z_1\):
\(z_2\) = , so that \(z_1\) =
Finally, rewrite \(Z = \dfrac{X-\mu}{\sigma}\) to make \(\sigma\) the subject. Put in values for \(z_2\), \(x_2\) and \(\mu\), and hence evaluate \(\sigma\) to 2 decimal places:
\(\sigma = \)
Feedback: Solution 1. The upper and lower limits of acceptable resistance are \(x_1 = [value]-[value] = [value]\), and \(x_2 = [value]+[value] = [value]\). Now, \(Z = \dfrac{X - [value]}{[value]}\), so that \(z_1 = \dfrac{[value]-[value]}{[value]} = [value]\) and \(z_2 = \dfrac{[value]-[value]}{[value]} = [value]\). Using symmetry, \[\begin{align*} \text{P}([value] < Z < [value]) &= 2 \times \text{P}(0 < Z < [value])\\ & = 2 \times [value] \\& = [value] ~~(\text{to 4 d.p.})\end{align*}\] Hence, the proportion of acceptable gauges is \([value]\%\) and the proportion of unacceptable gauges is \([value]\%\). When working to a higher degree of accuracy using a statistical calculator, the required probability is \(\text{P}([value] < X < [value]) = [value]\) to 4 decimal places, meaning that the proportion of unacceptable gauges is \([value]\%\). 2. We require that \(\text{P}(z_1 < Z < z_2) = 1-\dfrac{[value]}{100} = [value].\) Hence, we want \(\text{P}(0 < Z < z_2) = \dfrac{[value]}{2} = [value].\) Using the Table, an entry of \([value]\) corresponds to a \(Z\)-value of \([value]\). So we get: \[z_2 = [value] \quad \text{and} \quad z_1 = -z_2 = [value].\] Finally, we have \(\sigma = \dfrac{x_2- \mu}{z_2} = \dfrac{[value]-[value]}{[value]} = [value] ~~(\text{to 2 d.p.})\).

Exercise

Find the \([value]\%\) probability interval for the standard Normal random variable \(Z\) centred at its mean.

(You must give all answers to 2 decimal places).

\(\text{P}(\) \(< Z < \) \() = [value]\%\)
Feedback: Solution As the interval is centred on about the mean of the standard Normal random variable \(Z \sim N(0,1)\), then \(z_1 = -z_2\) and the probability of this interval can be expressed as follows using the symmetry of the distribution: \[ \text{P}(z_1 < Z < z_2) = 2 \times \text{P}(0 < Z < z_2)\] For this probability to be \([value]\%\), we need to find \(z_2\) such that \[\text{P}(0 < Z < z_2) = \dfrac{[value]}{2} = [value].\] Reading from the standard Normal table we see that \(z_2 = [value]\), hence: \[ \text{P}([value] < Z < [value]) = [value]\%.\]

The Normal Distribution

The normal distribution is the most widely used model for the distribution of a random variable. There is a very good reason for this. Practical experiments involve measurements and measurements involve errors. However you go about measuring a quantity, inaccuracies of all sorts can make themselves felt. For example, if you are measuring a length using a device as crude as a ruler, you may find errors arising due to:

- the calibration of the ruler itself;
- parallax errors due to the relative positions of the object being measured, the ruler and your eye;
- rounding errors;
- ‘guesstimation’ errors if a measurement is between two marked lengths on the ruler;
- mistakes.

If you use a meter with a digital readout, you will avoid some of the above errors but others, often present in the design of the electronics controlling the meter, will be present. Errors are unavoidable and are usually the sum of several factors. The behaviour of variables which are the sum of several other variables is described by a very important and powerful result called the Central Limit Theorem which we will study later in this Workbook. For now we will quote the result so that the importance of the normal distribution will be appreciated.

The Central Limit Theorem

Let \(X\) be the sum of \(n\) independent random variables \(X_i, i=1,2,\dots n\) each having a distribution with mean \(\mu_i\) and variance \(\sigma^2_i\quad (\sigma_i^2<\infty)\), respectively, then the distribution of \(X\) has expectation and variance given by the expressions

\(\displaystyle{\qquad\text{E}(X)=\sum_{i=1}^n\mu_i\qquad\text{and}\qquad \text{V}(X)=\sum_{i=1}^n\sigma^2_i}\)

and becomes normal as \(n\rightarrow\infty\).

Essentially we are saying that a quantity which represents the combined effect of a number of variables will be approximately normal no matter what the original distributions are provided that \(\sigma^2<\infty\). This statement is true for the vast majority of distributions you are likely to meet in practice. This is why the normal distribution is crucially important to engineers. A quotation attributed to rof. G. Lippmann, (1845-1921, winner of the Nobel prize for hysics in 1908) ‘Everybody believes on the law of errors, experimenters because they think it is a mathematical theorem and mathematicians because they think it is an experimental fact.’

You may think that anything you measure follows an approximate normal distribution. Unfortunately this is not the case. While the heights of human beings follow a normal distribution, weights do not. Heights are the result of the interaction of many factors (outside one’s control) while weights principally depend on lifestyle (including how much and what you eat and drink!) In practice, it is found that weight is skewed to the right but that the square root of human weights is approximately normal.

Defining a Normal Distribution

The probability density function of a normal distribution with mean \(\mu\) and variance \(\sigma^2\) is given by the formula \[f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}\] This curve is always bell-shaped with the centre of the bell located at the value of \(\mu\). See Figure below. The height of the bell is controlled by the value of \(\sigma\). As with all normal distribution curves it is symmetrical about the centre and decays as \(x\to \pm \infty\). As with any probability density function the area under the curve is equal to 1.

[Image: The Normal Probability Distribution Function]

Figure: Normal Probability Density Function (PDF)

A normal distribution is completely defined by specifying its mean (the value of \(\mu\)) and its variance (the value of \(\sigma^2\).) The normal distribution with mean \(\mu\) and variance \(\sigma^2\) is written \(N(\mu,\sigma^2)\). Hence the distribution \(N(20,25)\) has a mean of \(20\) and a standard deviation of \(5\); remember that the second parameter is the variance which is the square of the standard deviation.

Key Point

A normal distribution has mean \(\mu\) and variance \(\sigma^2\). A random variable \(X\) following this distribution is usually denoted by \(N(\mu, \sigma^2)\) and we often write \[X ∼ N(\mu, \sigma^2)\]

Clearly, since \(\mu\) and \(\sigma^2\) can both vary, there are infinitely many normal distributions and it is impossible to give tabulated information concerning them all.

For example, if we produce piston rings which have a target mean internal diameter of \(45\) mm then we may realistically expect the actual diameter to deviate from this value. Such deviations are well-modelled by the normal distribution. Suppose we decide that diameters in the range \(44.95\) mm to \(45.05\) mm are acceptable, we may then ask the question ‘What proportion of our manufactured output is satisfactory?’

Without tabulated data concerning the appropriate normal distribution we cannot easily answer this question (because the integral used to calculate areas under the normal curve is intractable.)

Since tabulated data allow us to apply the distribution to a wide variety of statistical situations, and we cannot tabulate all normal distributions, we tabulate only one - the standard normal distribution - and convert all problems involving the normal distribution into problems involving the standard normal distribution.

The Standard Normal Distribution

At this stage we shall, for simplicity, consider what is known as a standard normal distribution which is obtained by choosing particularly simple values for \(\mu\) and \(\sigma\).

Key Point

The standard normal distribution has a mean of zero and a variance of one

In the Figure below, we show the graph of the standard normal distribution which has probability density function \(y=\dfrac{1}{\sqrt{2\pi}}\ e^{-x^2/2}\).

[Image: The Standard Normal Distribution Curve]

Figure: The standard normal distribution curve

The result which makes the standard normal distribution so important is as follows:

Key Point

If the behaviour of a continuous random variable X is described by the distribution \(N(\mu, \sigma^2)\), then the behaviour of the random variable \(Z = \dfrac{X-\mu}{\sigma}\) is described by the standard normal distribution \(N(0, 1)\).

We call \(Z\) the standardised normal variable and we write \(Z ∼ N(0, 1)\)

Example

If the random variable \(X\) is described by the distribution \(N(45,0.000625)\), then what is the transformation required to obtain the standardised normal variable?

Solution

Here, \(\mu = 45\) and \(\sigma^2 = 0.000625\), so that \(\sigma = 0.025\). Hence \(Z = \dfrac{X-45}{0.025}\) is the required transformation.

Example

When the random variable \(X\sim N(45,0.000625)\) takes values between \(44.95\) and \(45.05\), between which values does the random variable \(Z\) lie?

Solution

When \(X = 45.05, Z = \dfrac{45.05 − 45}{0.025} = 2\)

When \(X = 44.95, Z = \dfrac{44.95 − 45}{0.025} = -2\)

Hence \(Z\) lies between \(−2\) and \(2\).

Probabilities and the Standard Normal Distribution

Since the standard normal distribution is used so frequently, a table of values has been produced to help us calculate probabilities. This is located at the end of this quiz and at the end of this page. You should keep a copy open in another tab for later questions but for now you can refer to the copy below. It is based upon the following diagram:

[Image: The normal distribution curve is shaded between 0 and the positive-valued z1.]

Since the total area under the curve is equal to 1, it follows from the symmetry in the curve that the area under the curve in the region \(Z>0\) is equal to \(0.5\). In the figure above, the shaded area is the probability that \(Z\) takes values between \(0\) and \(z_1\). When we ‘look-up’ a value in the table we obtain the value of the shaded area.

Note from the Table that as \(Z\) increases from \(0\), the entries increase, rapidly at first and then more slowly, toward \(5000\) i.e. a probability of \(0.5\). This is consistent with the shape of the curve.

After \(Z=3\) the increase is quite slow so that we tabulate entries for values of \(Z\) rising by increments of \(0.1\) instead of \(0.01\), as in the rest of the Table.

Note: There are different versions of the table of standard normal probabilities depending on the interval of interest. The table used in this section give the values of \(\text{P}(0<Z<z)\) for \(z \geq 0\), this is the area shaded in the figure above. Another common table of standard probabilities are of \(\text{P}(Z<z)\) for \(z \geq 0\), also known as the table of standard normal cumulative probabilities. If graphed, you would see everything left of 0 also shaded. The relationship between the two tables is that the cumulative probabilities are \(0.5\) greater than the table included here since it includes all of the left half of the distribution. Pay attention to which distribution a table is based upon.

Example

What is the probability that \(Z\) takes values between \(0\) and \(1.9\)? (Refer to the table of standard normal probabilities at the end of the quiz.)

Solution

The row beginning ‘\(1.9\)’ and the column headed ‘\(0\)’ is the appropriate choice and its entry is \(4713\). This is to be read as \(0.4713\) (we omitted the ‘\(0.\)’ in each entry for clarity.) The interpretation is that the probability that \(Z\) takes values between \(0\) and \(1.9\) is \(0.4713\).

Example

What is the probability that \(Z\) takes values between \(0\) and \(1.96\)?

Solution

This time we want the row beginning \(1.9\) and the column headed \('6'\).

The entry is \(4750\) so that the required probability is \(0.4750\).

Example

What is the probability that \(Z\) takes values between \(0\) and \(1.965\)?

Solution

There is no entry corresponding to \(1.965\), so we take the average of the values for \(1.96\) and \(1.97\). (This linear interpolation is not strictly correct but is acceptable.)

The two values are \(4750\) and \(4756\) with an average of \(4753\). Hence the required probability is \(0.4753\).

Calculating Other Probabilities

In this Section, we see how to calculate probabilities represented by areas other than those of the type shown in the previous Section.

Case 1

The below Figure illustrates what we do if both \(Z\) values are positive. By using the properties of the standard normal distribution we can organise matters so that any required area is always of ‘standard form’.

[Image: The shaded area between two positive z-values is shown as equal to the area between 0 and the larger z-value, minus the]

Example

Find the probability that \(Z\) takes values between \(1\) and \(2\).

Solution

Using the Standard Normal Table:

\(\text{P}(0 < Z < z_2)\) i.e. \(\text{P}(0 < Z < 2) = 0.4772\)

\(\text{P}(0 < Z < z_1)\) i.e. \(\text{P}(0 < Z < 1) = 0.3413\)

Hence \(\text{P}(1 < Z < 2) = 0.4772 − 0.3413 = 0.1359\)

Remember that with a continuous distribution, \(\text{P}(Z = 1)\) is meaningless (will have zero probability) so that \(\text{P}(1 ≤ Z ≤ 2)\) is interpreted as \(\text{P}(1 < Z < 2)\).

Case 2

The following diagram illustrates the procedure to be followed when finding probabilities of the form \(\text{P}(Z>z_1)\).

[Image: This time the shaded area is the difference between the total area of the right-hand half, 0.5, and an area between 0 an]

Example

What is the probability that \(Z>2\)?

Solution

\(\text{P}(0 < Z < 2) = 0.4772\) (from the Table). Hence the probability is \(0.5 − 0.4772\) \(= 0.0228\).

Case 3

Here we consider the procedure to be followed when calculating probabilities of the form \(\text{P}(Z<z_1)\). Here the shaded area is the sum of the left-hand half of the total area and a ‘standard’ area.

[Image: The shaded area is 0.5 plus the area between 0 and the positive z1.]

Example

What is the probability that \(Z<2\)?

Solution

\(\text{P}(Z < 2) = 0.5 + 0.4772 = 0.9772.\)

Case 4

Here we consider what needs to be done when calculating probabilities of the form
\(\text{P}(-z_1<Z<0)\) where \(z_1\) is positive. This time we make use of the symmetry in the standard normal distribution curve.

[Image: By symmetry, the shaded area between 0 and the positive z1 is equal to the shaded area between 0 and -z1.]

Example

What is the probability that \(-2<Z<0\)?

Solution

The area is equal to that corresponding to \(\text{P}(0 < Z < 2) = 0.4772.\)

Case 5

Finally we consider probabilities of the form \(\text{P}(-z_1<Z<z_2)\). Here we use the sum property and the symmetry property.

[Image: The area between a negative-valued z2 and a positive-valued z1 is equal to the area between 0 and z1 plus the area betwe]

Example

What is the probability that \(-1<Z<2\)?

Solution

\(\text{P}(−1 < Z < 0) = \text{P}(0 < Z < 1) = 0.3413\)

\(\text{P}(0 < Z < 2) = 0.4772\)

Hence the required probability \(\text{P}(−1 < Z < 2)\) is \(0.8185.\)

Other cases can be handed by a combination of the ideas already used.

The Cumulative Distribution Function

We know that the normal probability density function \(f(x)\) is given by the formula \[\qquad f(x)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^2/2\sigma^2}\] and so the cumulative distribution function \(F(x)\) is given by the formula \[\qquad F(x)= \dfrac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^xe^{-(u-\mu)^2/2\sigma^2}d u\] In the case of the cumulative distribution for the standard normal curve, we use the special notation \(\Phi (z)\) and, substituting \(0\) and \(1\) for \(\mu\) and \(\sigma^2\), we obtain \[\qquad \Phi(z)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-u^2/2}d u\] The shape of the curve is essentially ‘\(S\)’ -shaped as shown in the Figure below. Note that the curve runs from \(-\infty\) to \(+\infty\) . As you can see, the curve approaches the value \(1\) asymptotically.

[Image: The Normal Cumulative Distribution Function]

Figure: the Normal Cumulative Distribution Function (CDF)

Comparing the integrals \[\qquad F(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^xe^{-(u-\mu)^2/2\sigma^2}d u\qquad\text{and}\qquad \Phi(z)=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^ze^{-v^2/2}d v\] shows that \[\qquad v=\dfrac{u-\mu}{\sigma}\qquad \text{and so}\qquad d v =\dfrac{d u}{\sigma}\]

and \(F(x)\) may be written as \[\qquad F(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{(x-\mu)/\sigma}e^{-v^2/2} \sigma d v\] \[\qquad \phantom{FFFFFFF(x)}=\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{(x-\mu)/\sigma}e^{-v^2/2}d v=\Phi \left(\dfrac{x-\mu}{\sigma}\right)\] We already know, from the basic definition of a cumulative distribution function, that \[\qquad \text{P}(a<X<b)=F(b)-F(a)\] so that we may write the probability statement above in terms of \(\Phi (z)\) as \[\qquad \text{P}(a<X<b)=F(b)-F(a)=\Phi \left(\dfrac{b-\mu}{\sigma}\right)- \Phi \left(\dfrac{a-\mu}{\sigma}\right).\]

The value of \(\Phi(z)\) is measured from \(z=-\infty\) to any ordinate \(z=z_1\) and represents the probability \(\text{P}(Z<z_1)\).

The values of \(\Phi(z)\) start as shown below: \[\qquad \begin{array}{c|cccccccccc} z & 0.00 & 0.01 & 0.02 & 0.03 & 0.04 & 0.05 & 0.06 & 0.07 & 0.08 & 0.09\\ \hline 0.0 & .5000 &5040 &5080 &5120 &5160 &5199 &5239 &5279 &5319 &5359\\ 0.1 & .5398 &5438 &5478 &5517 &5577 &5596 &5636 &5675 &5714 &5753\\ 0.2 & .5793 &5832 &5871 &5909 &5948 &5987 &6026 &6064 &6103 &6141 \end{array}\] You should compare the values given here with the values given for the standard normal probability integral (the Table at the end of the quiz). Simply adding \(0.5\) to the values in the latter table gives the values of \(\Phi(z)\). You should also note that the diagrams shown at the top of each set of tabulated values tells you whether you are looking at the values of \(\Phi(z)\) or the values of the normal probability integral.

Applications of the Normal Distribution

We have, in the previous subsection, noted that the probability density function of a normal distribution \(X\) is \[y=\dfrac{1}{\sigma\sqrt{2\pi}}e^{\frac{-(x-\mu )^2}{2\sigma^2}}\] This curve is always ‘ bell-shaped’ with the centre of the bell located at the value of \(\mu\). The height of the bell is controlled by the value of \(\sigma\). See the Figure below.

[Image: The Probability Density Function of the Normal Distribution]

Figure: the Probability Density Function of the Normal Distribution

We now show, by example, how probabilities relating to a general normal distribution \(X\) are determined. We will see that being able to calculate the probabilities of a standard normal distribution \(Z\) is crucial in this respect.

Example

Given that the variate \(X\) follows the normal distribution \(X\sim N(151,15^2)\), calculate:

1. \(\text{P}(120\le X\le 155)\)
2. \(\text{P}(X\ge 185)\)

Solution

The transformation used in this problem is \(Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-151}{15}\)

1. \(\text{P}(120\le X\le 155)=\text{P}\left(\dfrac{120-151}{15}\le Z\le \dfrac{155-151}{15}\right)\)
\(\phantom{\text{P555}}=\text{P}(-2.07\le Z\le 0.27)\)
\(\phantom{\text{P}(120\le)}=0.4808+0.1064=0.5872\)
2. \(\text{P}(X\ge 185)=\text{P}\left(Z\ge \dfrac{185-151}{15}\right)\)
\(\phantom{\text{P55}}=\text{P}( Z\ge 2.27)\)
\(\phantom{\text{P}(X\ge 1855)}=0.5-0.4884=0.0116\)

We note that, as for any continuous random variable, we can only calculate the probability that

\(\quad\bullet\quad\) \(X\) lies between two given values;

\(\quad\bullet\quad\) \(X\) is greater than a given value;

\(\quad\bullet\quad\) \(X\) is less that a given value,

rather than for individual values.

Applications

Example

Piston rings are mass-produced. The target internal diameter is \(45\) mm but records show that the diameters are normally distributed with mean \(45\) mm and standard deviation \(0.05\) mm. An acceptable diameter is one within the range \(44.95\) mm to \(45.05\) mm. What proportion of the output is unacceptable?

Solution

There are many words in the statement of the problem; we must read them carefully to extract the necessary information. If \(X\) is the diameter of a piston ring, then \(X\ \sim\ N (45,(0.05)^2)\).

The transformation is \(Z=\dfrac{X-\mu}{\sigma}=\dfrac{X-45}{0.05}.\)

The upper limit of acceptability is \(x_2=45.05\) so that \(z_2=(45.05 -45)/0.05=1.\)

The lower limit of acceptability is \(x_1=44.95\) so that \(z_1=(44.95-45)/0.05=-1\)

The range of `acceptable' \(Z\) values is therefore \(-1\) to \(1\), as shown in the Figure below.

[Image: The Normal distribution curve with vertical lines drawn at z1=-1 and z2=1.]

Using the symmetry of the curves, \(\text{P}(-1 < Z < 1) = 2\times \text{P}(0 < Z < 2) = 2 \times 0.3413 = 0.6826\).

Thus the proportion of unacceptable items is \(1-0.6826=0.3174\), or \(31.74\%\).

Example

If the standard deviation is halved by improved production practices, what is now the proportion of unacceptable items?

Solution

Now \(\sigma=0.025\) so that:

\(\qquad z_2=\dfrac{45.05-45}{0.025}=2\qquad\text{and}\qquad z_1=-2.\)

Then, \(\text{P}(-2 < Z < 2) = 2 \times \text{P}(0 < Z < 2)=2\times 0.4772=0.9544\). Hence the proportion of unacceptable items is reduced to \(1-0.9544=0.0456\) or \(4.56\%\).

We observe that less of the area under the curve now lies outside the interval \((44.95, 45.05)\).

[Image: The normal distribution curve is shaded z=-2 and z=2.]

Probability Intervals - Standard Normal Distribution

We use probability models to make predictions in situations where there is not sufficient data available to make a definite statement. Any statement based on these models carries with it a risk of being proved incorrect by events. Notice that the normal probability curve extends to infinity in both directions. Theoretically any value of the normal random variable is possible, although, of course, values far from the mean position (zero) are very unlikely.

Consider the diagram in the Figure below:

[Image: The Normal distribution curve with 95% of the area shaded in, between z-values z1 = -1.96 and z2 = 1.96.]

The shaded area is \(95\%\) of the total area. If we look at the entry in the Standard Normal Table (at the end of the quiz) corresponding to \(Z=1.96\) we see the value \(4750\). This means that the probability of \(Z\) taking a value between \(0\) and \(1.96\) is \(0.475\). By symmetry, the probability that \(Z\) takes a value between \(-1.96\) and \(0\) is also \(0.475\). Combining these results we see that

\(\qquad \text{P}(-1.96<Z<1.96)=0.95\ \ \text{or}\ \ 95\%\)

We say that the \(95\%\) probability interval for \(Z\) (about its mean of \(0\)) is \((-1.96, 1.96)\). It follows that there is a \(5\%\) chance that \(Z\) lies outside this interval.

Exercise

For the standard Normal random variable \(Z\), find the value \(k\) to 2 decimal places where:

1. \(Z\) exceeds \(k\) on \([value]\%\) of occasions.
\(k = \)
2. \(Z\) exceeds \(k\) on \([value]\%\) of occasions.
\(k = \)
Feedback: Solution 1. The probability for \(\text{P}(Z>k)\) is less than \(0.5\), so the value for \(k\) will be positive. Applying the complement rule and partitioning the interval at \(0\) gives: \[\begin{align*}\text{P}(Z>k) &= 1 - \text{P}(Z < k)\\& = 1 - \left[\text{P}(Z<0) + \text{P}(0<Z<k)\right] \\&= 0.5 - \text{P}(0<Z<k). \end{align*}\] If this probability is to equal \([value]\%\) then \(\text{P}(0<Z<k) = [value].\) Reading from the table of standard Normal probabilities, we see that \(k = [value]\) to 2 decimal places. 2. On the contrary, in this case \(\text{P}(Z>k)\) is greater than \(0.5\) so the value for \(k\) will be negative. Using the symmetry of the standard Normal distribution we get: \[\begin{align*}\text{P}(Z>k) &= \text{P}(Z< -k) \\&= \text{P}(Z< 0) + \text{P}(0 < Z< -k) \\&= 0.5 + \text{P}(0 < Z< -k).\end{align*}\] For this probability to be \([value]\%\) then we require that \(\text{P}(0 < Z< -k) = [value]\). Reading from the table of standard Normal probabilities, we see that \(-k = [value]\) and so \(k = [value]\) to 2 decimal places.

Probability Intervals - General Normal Distribution

We saw in a previous subsection that \(95\%\) of the area under the standard normal curve lay between \(z_1=-1.96\) and \(z_2=1.96\). Using the formula \(Z=\dfrac{X-\mu}{\sigma}\) in the re-arrangement \(X=\mu+Z\sigma\), we can see that \(95\%\) of the area under the general normal curve lies between \(x_1=\mu-1.96\sigma\) and \(x_2=\mu+1.96\sigma\). This is shown in the Figure below.

[Image: The Normal distribution curve with 95% of the area shaded in between x-values, x1 = mu - 1.96 times sigma, and x2 = mu +]

Example

Suppose that the internal diameters of mass-produced pipes are normally distributed with mean \(50\) mm and standard deviation \(2\) mm. What are the \(95\%\) probability limits on the internal diameter of a single pipe?

Solution

Here \(\mu=50\) and \(\sigma=2\) so that the \(95\%\) probability limits are

\(\qquad 50\pm 1.96\times 2 = 50\pm 3.92\) mm,

i.e. \(46.08\) mm and \(53.92\) mm.

Hence the probability interval is \((46.08, 53.92)\).

Exercise

What is the centred \([value]\%\) probability interval for the lifetime of a bulb, \(T\), when the lifetimes of such bulbs are normally distributed with a mean of \([value]\) hours and standard deviation of \([value]\) hours?

(You must keep full accuracy throughout the question).

First sketch the standard normal curve marking the values \(z_1, z_2\) between which \([value]\%\) of the area under the curve is located.

Now, find the corresponding interval \(z\)-values, \((z_1, z_2)\), to 2 decimal places using the Table.

\(\text{P}(\) \(< Z <\) \() = [value]\%\)

Now, calculate the corresponding interval values, \((t_1, t_2)\), to 1 decimal place that correspond to normal distribution for the bulb lifetime.

\(\text{P}(\) \(< T <\) \() = [value]\%\)
Feedback: Solution For the interval \((z_1, z_2)\) to be centred, we have that \(z_1 = -z_2\). This means that \(\text{P}(z_1 < Z < z_2) = 2 \times \text{P}(0 < Z < z_2)\). For this probability to be equal to \([value]\%\) then we need to find the value of \(z_2\) such that \(\text{P}(0 < Z < z_2) = [value]\). Reading from the table, we see that \(z_2 = [value]\) to 2 decimal places. Therefore: \[\text{P}( [value] < Z < [value]) = [value]\%\] The distribution for the bulb life time is \(T \sim N([value], [value]^2)\). The transformation from the standard normal to this distribution is \(T = [value] + [value]Z\). This means that the corresponding lower and upper bounds of the centred interval, rounded to the nearest whole hour, are: \[\begin{align*} t_1 & = [value] + [value] z_1 = [value],\\ t_2 & = [value] + [value] z_2 = [value]. \end{align*}\] Hence, the \([value]\%\) centred probability interval for the lifetimes is \(([value],[value])\) hours.

The Standard Normal Probability Table

The Table below shows the Standard Normal Probability Table of values.

Use this Table to find \(Z\)-values when needed. It is recommended that you open this Table in another tab to refer to throughout the quiz.

[Image: The Standard Normal Probability Table]

The Standard Normal Probability Table

The Table below shows the Stable Normal Probability Table.

Use this Table to find \(Z\)-values when needed. It is recommended that you open this Table in another tab to refer to throughout the quiz.

[Image: The Standard Normal Probability Table]