11.4 The Normal Distribution: Z-Scores, Normal Distribution Tables and Business Probabilities

Why this topic is examined

Syllabus area D1(d) asks you to demonstrate the use of the normal distribution, using "graphs/diagrams and use of normal distribution tables". Normal distribution tables are provided in the exam. Questions typically ask for the probability that a variable exceeds, falls below or lies between given values, or for the value that will be exceeded with a stated probability. A quick sketch of the curve with the area shaded prevents most mistakes.

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Properties of the normal distribution

• It is symmetrical about the mean and bell-shaped.
• The mean, median and mode are equal and sit at the centre.
• The total area under the curve is 1 (100%), so the area to each side of the mean is 0.5.
• It is fully described by two figures: the mean ($\mu$) and the standard deviation ($\sigma$).
• The curve extends indefinitely in both directions, but almost all the area lies within 3 standard deviations of the mean.

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Z-scores and the table

To use the table, convert the value of interest ($x$) into a z-score:

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

The z-score is the number of standard deviations between $x$ and the mean. Normal distribution tables of the kind provided in CIMA exams give the area between the mean and $z$. Extracts:

Because the curve is symmetrical, a negative z-score has the same area as the corresponding positive one; the sign simply tells you which side of the mean the value lies.

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Four standard question types

Weekly demand for a product is normally distributed with a mean of 1,000 units and a standard deviation of 100 units.

1. Probability of exceeding a value

What is the probability that demand exceeds 1,150 units?

$$z = \frac{1{,}150 - 1{,}000}{100} = 1.5 \quad \Rightarrow \quad \text{area} = 0.4332$$

The area above 1,150 is the upper half (0.5) minus the area between the mean and 1,150:

$$P(X > 1{,}150) = 0.5 - 0.4332 = 0.0668 \text{ (6.68\%)}$$

2. Probability of falling below a value

What is the probability that demand is less than 850 units?

$$z = \frac{850 - 1{,}000}{100} = -1.5 \quad \Rightarrow \quad P(X < 850) = 0.5 - 0.4332 = 0.0668$$

By symmetry this equals the previous answer.

What is the probability that demand is less than 1,150 units? Here the area includes the whole lower half plus the area from the mean to 1,150:

$$P(X < 1{,}150) = 0.5 + 0.4332 = 0.9332$$

3. Probability of lying between two values

Either side of the mean. What is the probability that demand is between 900 and 1,150 units?

• 900 gives $z = -1.0$, area 0.3413.
• 1,150 gives $z = 1.5$, area 0.4332.

$$P(900 < X < 1{,}150) = 0.3413 + 0.4332 = 0.7745$$

Same side of the mean. What is the probability that demand is between 1,050 and 1,150 units?

• 1,050 gives $z = 0.5$, area 0.1915.
• 1,150 gives $z = 1.5$, area 0.4332.

$$P(1{,}050 < X < 1{,}150) = 0.4332 - 0.1915 = 0.2417$$

4. Finding the value for a given probability

What level of demand will be exceeded only 5% of the time? The area between the mean and the required value is 0.5 − 0.05 = 0.45, which corresponds to $z = 1.645$:

$$x = \mu + z\sigma = 1{,}000 + (1.645 \times 100) = 1{,}164.5 \text{ units}$$

If the business holds about 1,165 units of stock each week, it should run out in only about 5% of weeks.

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Business applications

Worked example: probability of a loss

A division's annual profit is normally distributed with a mean of $40,000 and a standard deviation of $25,000. What is the probability of a loss (profit below zero)?

$$z = \frac{0 - 40{,}000}{25{,}000} = -1.6 \quad \Rightarrow \quad \text{area} = 0.4452$$ $$P(\text{loss}) = 0.5 - 0.4452 = 0.0548 \text{ (about 5.5\%)}$$

This links directly to Section 11.3: a larger standard deviation would widen the curve and raise the probability of a loss, even with the same mean.