The Normal distribution

CIMA-BA2 · Chapter 16

The Normal distribution

In a histogram the area of each bar represents the frequency, so the proportion of the total area in any range is the probability of being in that range. The normal distribution is a smooth, symmetrical, bell-shaped version of a histogram for which these areas are already worked out in normal distribution tables.

To use the tables, convert a value into a z-score – the number of standard deviations it is from the mean: z = (x − mean) ÷ standard deviation. The table gives the area between the mean and z; add or subtract 0.5 to find any other area. The tables can also be used backwards, to find the value that gives a stated probability.

1 Introduction

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We discussed probabilities in a previous chapter. In this chapter we will see how tables may be used to calculate probabilities for certain frequency distributions.

2 The histogram revisited

You should remember from the last chapter what the histogram is, and how it is drawn. Importantly, it is the area of the bars that is proportional to the frequency.

The following table shows the annual salaries earned by 150 workers.

Salary	Frequency
$0 – $1,000	25
$1,000 – $2,000	35
$2,000 – $3,000	40
$3,000 – $5,000	50
Total	150

Show this frequency table in the form of a histogram
Calculate the probability of a worker earning between $1,000 and $2,000
Calculate the probability of a worker earning more than $2,000

(a) The first three classes are $1,000 wide, but the last is $2,000 wide. Because the area of a bar represents the frequency, the last bar is drawn at half the height its frequency suggests: 50 ÷ 2 = 25 (frequency per $1,000).

Salary	Class width	Frequency	Height of bar (frequency per $1,000)
$0 – $1,000	$1,000	25	25
$1,000 – $2,000	$1,000	35	35
$2,000 – $3,000	$1,000	40	40
$3,000 – $5,000	$2,000	50	25

If the last bar were drawn 50 high, it would look as if far more workers earned over $3,000 than really do.

(b) 35 of the 150 workers earn between $1,000 and $2,000: 35/150 = 0.23 (23%)

(c) More than $2,000 = 40 + 50 = 90 workers: 90/150 = 0.60 (60%)

We can easily calculate the probabilities from the frequency table. However, if we were presented with an accurately drawn histogram, then even without the original table we could still calculate the probability of a worker earning within a specific range.

The area of all the bars is proportional to the total number of workers, and the area of the bars representing any specific range is proportional to the number of workers earning within that range. We could calculate the probability by dividing the area of the bar(s) representing the specific range by the total area of all the bars.

In Example 1, the two bars above $2,000 have areas of 40 × 1 + 25 × 2 = 90 units out of a total area of 25 + 35 + 40 + 50 = 150 units – the same 0.60 as from the table.

3 The normal distribution

The normal distribution is effectively a ‘smoothed-out’ histogram with a very specific shape.

If the classes of a histogram were made narrower and narrower, the tops of the bars would gradually become a smooth curve – a frequency curve. The area under any part of the curve, as a proportion of the total area, is still the proportion of observations (the probability) in that range. For one particular shape of curve – the normal distribution – these areas have been worked out and published in tables.

The main features of a normal distribution are:

it is symmetrical about the mean
it is continuous
the mean coincides with the mode (and with the median) – there is only one peak
it is ‘bell-shaped’
it never quite touches the horizontal axis – very large and very small values are possible, just very unlikely
the total area under the curve is 1 (100% of the observations), so half of it (0.5) lies on each side of the mean

The curve above is drawn in standard deviations from the mean. The area between the mean and 1 standard deviation above it is 0.3413 of the total; between the mean and 2 standard deviations above it, 0.4772. Because the curve is symmetrical, the same areas lie below the mean.

For a distribution that is shaped normally, we can calculate the areas under the curve as a proportion of the total area (and therefore the probabilities) by using normal distribution tables (which you will be provided with in the exam).

The normal distribution tables give us the area under the curve between the mean and some other point above or below the mean.

The size of the areas we are looking for will depend on how great or small the spread of the distribution is, and therefore when we use the tables we look at the number of standard deviations we are from the arithmetic mean.

We call this distance the z-score.

z = (x − mean) ÷ standard deviation

where x is the value we are interested in. A value below the mean gives a negative z; because the curve is symmetrical, look it up as if it were positive.

Normal distribution table (extract)

The table gives the area under the curve between the mean and z standard deviations from the mean. Find the first decimal of z down the left-hand side and the second decimal along the top: for z = 1.64, use the 1.6 row and the 0.04 column (0.4495). The full table is given in the exam; this extract holds the rows used in this chapter.

z	0.00	0.01	0.02	0.03	0.04
0.5	0.1915	0.1950	0.1985	0.2019	0.2054
1.0	0.3413	0.3438	0.3461	0.3485	0.3508
1.5	0.4332	0.4345	0.4357	0.4370	0.4382
1.6	0.4452	0.4463	0.4474	0.4484	0.4495
1.9	0.4713	0.4719	0.4726	0.4732	0.4738
2.0	0.4772	0.4778	0.4783	0.4788	0.4793
2.5	0.4938	0.4940	0.4941	0.4943	0.4945
3.0	0.4987	0.4987	0.4987	0.4988	0.4988

z	0.05	0.06	0.07	0.08	0.09
0.5	0.2088	0.2123	0.2157	0.2190	0.2224
1.0	0.3531	0.3554	0.3577	0.3599	0.3621
1.5	0.4394	0.4406	0.4418	0.4429	0.4441
1.6	0.4505	0.4515	0.4525	0.4535	0.4545
1.9	0.4744	0.4750	0.4756	0.4761	0.4767
2.0	0.4798	0.4803	0.4808	0.4812	0.4817
2.5	0.4946	0.4948	0.4949	0.4951	0.4952
3.0	0.4989	0.4989	0.4989	0.4990	0.4990

Finding other areas

Proportion wanted	Method	Using Example 2 (mean 10 cm, standard deviation 0.2 cm)
Between the mean and a value	Read the table for z	10 to 10.4 cm: z = 2, area 0.4772
Beyond a value (further from the mean)	0.5 − table area	Less than 9.8 cm: 0.5 − 0.3413 = 0.1587
Up to a value on the far side of the mean	0.5 + table area	Less than 10.4 cm: 0.5 + 0.4772 = 0.9772
Between two values on opposite sides of the mean	Add the two table areas	9.8 to 10.4 cm: 0.3413 + 0.4772 = 0.8185
Between two values on the same side of the mean	Subtract the smaller table area from the larger	10.2 to 10.4 cm: 0.4772 − 0.3413 = 0.1359

A quick sketch of the curve, with the area you want shaded, makes it clear whether to add or subtract. The table also shows that about 68% of observations lie within 1 standard deviation of the mean (2 × 0.3413 = 0.6826), about 95% within 2 (2 × 0.4772 = 0.9544) and almost all (99.7%) within 3 (2 × 0.4987 = 0.9974).

A company produces units with an average length of 10 cm, and a standard deviation of 0.2 cm.

What proportion of the units will have a length of:

more than 10 cms
between 10 and 10.4 cms
less than 9.8 cms

(a) 10 cm is the mean, and the curve is symmetrical, so half the units are longer: 0.5 (50%)

(b) z = (10.4 − 10) ÷ 0.2 = 0.4 ÷ 0.2 = 2.00. From the table (2.0 row, 0.00 column), the area between the mean and z = 2.00 is 0.4772. Proportion = 0.4772 (47.72%)

(c) z = (9.8 − 10) ÷ 0.2 = −1.00 (1 standard deviation below the mean). The table gives 0.3413 between 9.8 cm and 10 cm. Half the units are below 10 cm, so the proportion below 9.8 cm = 0.5 − 0.3413 = 0.1587 (15.87%)

It is also possible to use the normal distribution tables ‘backwards’.

For the same information as in Example 2, there is a 0.95 (or 95%) probability that the length will be more than X cms.

Calculate a value for X (remember that there is a 0.5 (or 50%) probability of the length being more than the average of 10 cms, because the normal distribution is symmetrical).

0.5 of the lengths are more than 10 cm, so for 0.95 to be more than X, X must be below 10 cm, with 0.95 − 0.5 = 0.45 of the area between X and 10 cm.

Look in the body of the table for 0.45: z = 1.64 gives 0.4495 and z = 1.65 gives 0.4505, so z = 1.64 (it is slightly nearer; 1.65 is also acceptable).

X is 1.64 standard deviations below the mean: X = 10 − (1.64 × 0.2) = 10 − 0.328 = 9.672 cm. There is a 95% probability that a unit is longer than 9.672 cm.

4 Test your knowledge

Two quick checks before you move on: work through the flashcards to fix this chapter’s terms and definitions, then sit the objective questions for exam-style practice. Both mark themselves and explain the answers as you go.

Answer the questions one at a time. Your progress is saved so you can leave and come back.

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