Study guide
11.4 The Normal Distribution: Z-Scores, Normal Distribution Tables and Business Probabilities
CIMA BA2: Fundamentals of Management Accounting Exam
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1 Chapter 1: The Context of Management Accounting
1.1 The Need for Management Accounting, the GMAP Definition & Comparison with Financial Accounting 1.2 Qualities of Good Management Information: From Data to Decisions 1.3 Information for Operational, Managerial and Strategic Levels 1.4 The Role of the Management Accountant: CIMA's View, IFAC's Domain & Position in the Organisation 1.5 CIMA as a Professional Body: Members, Candidates, the Profession & Society
2 Chapter 2: Cost Identification, Classification and Behaviour
2.1 Cost Classification by Nature, Function, and Element 2.2 Cost Behaviour Patterns and Relevant Range 2.3 The High-Low Method and Cost Prediction 2.4 Scattergraph and Least-Squares Regression for Cost Estimation
3 Chapter 3: Overheads and Absorption Costing
3.1 Production Overheads: Allocation and Apportionment 3.2 Overhead Absorption Rates (OAR) & Under/Over-Absorption
4 Chapter 4: Marginal Costing, Profit Reconciliation and Cost-Based Pricing
4.1 Marginal Costing vs Absorption Costing Principles 4.2 Reconciling Marginal and Absorption Costing Operating Profits 4.3 Pricing with Cost Information: Full-Cost-Plus, Marginal-Cost-Plus, Mark-Ups, Margins and Return Targets
5 Chapter 5: Budget Preparation: Functional, Master and Cash Budgets
5.1 Purposes of Budgeting & The Budgeting Process 5.2 The Principal Budget Factor & Functional Budgets 5.3 The Master Budget: Budgeted Profit or Loss, Financial Position and Cash Flow 5.4 Cash Budgets & Cash Flow Forecasting
6 Chapter 6: Budgetary Control: Flexible Budgets, Behaviour and Budgeting Approaches
6.1 Fixed vs Flexible Budgeting and Budget Flexing 6.2 Budgetary Control Systems, Participation & Behavioural Aspects 6.3 Alternative Budgeting Approaches: Incremental, ZBB & Rolling Budgets
7 Chapter 7: Standard Costing and Cost Variances
7.1 Standard Costing Principles and Types of Standards 7.2 Direct Material Price and Usage Variances 7.3 Direct Labour Rate, Efficiency, and Idle Time Variances 7.4 Variable and Fixed Overhead Variances
8 Chapter 8: Sales Variances, Operating Statements and Interpreting Variances
8.1 Sales Variances and Operating Statements (Marginal and Absorption Formats) 8.2 Interpreting Variances: Causes, Interrelationships and Deciding What to Investigate
9 Chapter 9: Performance Measurement
9.1 The Need for Performance Measures: Responsibility Accounting, Controllability and Service Industries 9.2 Financial and Non-Financial Performance Measures in Manufacturing and Service Organisations
10 Chapter 10: Integrated Accounts, Job and Batch Costing and Management Reports
10.1 Integrating Cost and Financial Accounts: Control Accounts and the Manufacturing Account 10.2 Integrated Ledgers with Under- and Over-Absorption and Standard Cost Variances 10.3 Job and Batch Costing: Job Cost Cards, Job Accounts and Pricing a Job 10.4 Cost Reports for Manufacturing, Service and Not-for-Profit Organisations
11 Chapter 11: Risk, Probability and Statistics for Decision-Making
11.1 Risk, Uncertainty and Probability: Rules and Joint Probabilities for Decisions 11.2 Expected Values, Expected Value Tables and Their Limitations 11.3 Averages and Dispersion: Mean, Median, Mode, Range, Variance, Standard Deviation and Coefficient of Variation 11.4 The Normal Distribution: Z-Scores, Normal Distribution Tables and Business Probabilities
12 Chapter 12: Relevant Costing and Short-Term Decisions
12.1 Relevant Cost Concepts: Future, Incremental, and Opportunity Costs 12.2 Relevant Costs of Materials, Labour, and Overheads 12.3 Short-Term Operational Decisions: Make-or-Buy, Shutdowns, Special Orders
13 Chapter 13: Breakeven Analysis and Limiting Factor Analysis
13.1 Single-Product Break-Even Analysis & Margin of Safety 13.2 Target Profit Calculations, Operating Gearing & CVP Assumptions 13.3 Multi-Product CVP Analysis & Break-Even Charts 13.4 Limiting Factor Analysis (Key Factor Analysis)
14 Chapter 14: Long-Term Decisions: Time Value of Money, Payback, NPV and IRR
14.1 Time Value of Money: Compounding, Discounting, Annuities & Perpetuities 14.2 Payback Period Method: Constant and Uneven Cash Flows 14.3 Net Present Value (NPV): Methodology and Decision Rules 14.4 Internal Rate of Return (IRR) and Interpolation
11.4 The Normal Distribution: Z-Scores, Normal Distribution Tables and Business Probabilities
Key Takeaways Explain This Topic AI
The normal distribution is a symmetrical bell-shaped curve defined by its mean and standard deviation, with the mean, median and mode all at the centre.
A value is converted to a z-score with z = (x − μ) ÷ σ, the number of standard deviations it lies from the mean.
Normal distribution tables give the area between the mean and a z-score; because each half of the curve has an area of 0.5, tail probabilities are 0.5 minus the table value.
About 68% of values lie within 1 standard deviation of the mean, about 95% within 2 and about 99.7% within 3.
Last updated: September 2026
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.
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 ( μ \mu μ ) and the standard deviation ( σ \sigma σ ).
The curve extends indefinitely in both directions, but almost all the area lies within 3 standard deviations of the mean.
- Distance from the mean Approximate area (probability)
- Within ±1 standard deviation 68.3%
- Within ±2 standard deviations 95.4%
- Within ±3 standard deviations 99.7%
- Z-scores and the table
To use the table, convert the value of interest ( x x x ) into a z-score : z = x − μ σ z = \frac{x - \mu}{\sigma} z = σ x − μ
The z-score is the number of standard deviations between x x x and the mean. Normal distribution tables of the kind provided in CIMA exams give the area between the mean and z z z . Extracts:
- z z z Area between mean and z z z
- 0.50 0.1915
- 1.00 0.3413
- 1.50 0.4332
- 1.60 0.4452
- 1.645 0.4500 (by interpolation)
- 1.96 0.4750
- 2.00 0.4772
- 2.50 0.4938
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.
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 = 1,150 − 1,000 100 = 1.5 ⇒ area = 0.4332 z = \frac{1{,}150 - 1{,}000}{100} = 1.5 \quad \Rightarrow \quad \text{area} = 0.4332 z = 100 1 , 150 − 1 , 000 = 1.5 ⇒ 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 (6.68%) P(X > 1{,}150) = 0.5 - 0.4332 = 0.0668 \text{ (6.68\%)} P ( X > 1 , 150 ) = 0.5 − 0.4332 = 0.0668 (6.68%)
2. Probability of falling below a value
What is the probability that demand is less than 850 units? z = 850 − 1,000 100 = − 1.5 ⇒ P ( X < 850 ) = 0.5 − 0.4332 = 0.0668 z = \frac{850 - 1{,}000}{100} = -1.5 \quad \Rightarrow \quad P(X < 850) = 0.5 - 0.4332 = 0.0668 z = 100 850 − 1 , 000 = − 1.5 ⇒ 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 P(X < 1{,}150) = 0.5 + 0.4332 = 0.9332 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 z = -1.0 z = − 1.0 , area 0.3413.
1,150 gives z = 1.5 z = 1.5 z = 1.5 , area 0.4332. P ( 900 < X < 1,150 ) = 0.3413 + 0.4332 = 0.7745 P(900 < X < 1{,}150) = 0.3413 + 0.4332 = 0.7745 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 z = 0.5 z = 0.5 , area 0.1915.
1,150 gives z = 1.5 z = 1.5 z = 1.5 , area 0.4332. P ( 1,050 < X < 1,150 ) = 0.4332 − 0.1915 = 0.2417 P(1{,}050 < X < 1{,}150) = 0.4332 - 0.1915 = 0.2417 P ( 1 , 050 < X < 1 , 150 ) = 0.4332 − 0.1915 = 0.2417
Important
Add the areas when the two values are on opposite sides of the mean. Subtract them when both values are on the same side.
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 z = 1.645 z = 1.645 : x = μ + z σ = 1,000 + ( 1.645 × 100 ) = 1,164.5 units x = \mu + z\sigma = 1{,}000 + (1.645 \times 100) = 1{,}164.5 \text{ units} x = μ + z σ = 1 , 000 + ( 1.645 × 100 ) = 1 , 164.5 units
If the business holds about 1,165 units of stock each week, it should run out in only about 5% of weeks.
Business applications
Application Normal distribution question
Inventory planning What stock level keeps the probability of a stock-out below a target?
Budget risk If profit is normally distributed, what is the probability of a loss?
Quality control What proportion of output will fall outside tolerance limits?
Project appraisal What is the probability that NPV will be negative?
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 = 0 − 40,000 25,000 = − 1.6 ⇒ area = 0.4452 z = \frac{0 - 40{,}000}{25{,}000} = -1.6 \quad \Rightarrow \quad \text{area} = 0.4452 z = 25 , 000 0 − 40 , 000 = − 1.6 ⇒ area = 0.4452 P ( loss ) = 0.5 − 0.4452 = 0.0548 (about 5.5%) P(\text{loss}) = 0.5 - 0.4452 = 0.0548 \text{ (about 5.5\%)} P ( loss ) = 0.5 − 0.4452 = 0.0548 (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.
Tip
Before calculating, sketch the bell curve, mark the mean and the value(s), and shade the area the question asks for. Decide whether you need 0.5 minus the table value, 0.5 plus it, the sum of two areas or their difference.