The Complete Guide to Normal Distribution

In-depth analysis of the most important probability distributions in statistics, mastering definitions, formulas, properties and DSE maths applications to enhance your maths performance.

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Visualisation of a normal distribution curve showing a typical bell curve

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What is Normal Distribution? --Basic Concepts and Definitions

Normal Distribution, also known as Normal Distribution or Gaussian Distribution, is the most important continuous probability distribution in statistics and probability theory. This distribution has a wide range of applications in the fields of natural science, social science and engineering.

Basic Concepts:

  • A normal distribution is a bell-shaped symmetric distribution with a probability density function that reaches its maximum at the centre (mean).
  • The shape of the distribution is entirely determined by two parameters: the mean (μ) and the variance (σ²)
  • The mean (μ) determines the centre of the curve and the variance (σ²) determines the width of the curve.
  • When the mean μ = 0 and the standard deviation σ = 1, it is called a standard normal distribution.

Why is it important?

  • A large number of random variables in nature and social phenomena approximately obey the normal distribution, e.g. human height, measurement error, etc.
  • The centre pole theorem shows that the sum of a large number of independent identically distributed stochastic variables tends to a normal distribution
  • Normal distribution is the basis of statistical inference, widely used in hypothesis testing, interval estimation and other fields.
  • In the DSE Mathematics Examination, normal distribution is an important test point, and it is important to master its concepts and applications.
Typical bell shaped normal distribution curve with mean and standard deviation marked

Historical Origins of Normal Distribution

The normal distribution was first introduced in 1733 by the French mathematician Abraham de Moivre, who discovered it while studying approximations to the binomial distribution.

Later, German mathematician Carl Friedrich Gauss and French mathematician Pierre-Simon Laplace conducted in-depth studies on the normal distribution, making it a core concept in statistics.

When Gauss studied the measurement errors, he found that these errors tend to show the characteristics of normal distribution, so the normal distribution is also called Gaussian distribution.

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Normal Distribution Equation - Mathematical expression and parameter analysis

The probability density function (PDF) of a normal distribution is the central formula describing its mathematical properties:

Probability density function formulas for normal distributions, with clear mathematical expressions.

Among them:

  • μ(Mean): Determines the centre of the curve.
  • σ(standard deviation): determines the width of the curve, the larger σ is, the wider the curve is.
  • σ²(variance): the square of the standard deviation
  • e: the base of the natural logarithm, which is equal to 2.71828
  • π: Perihelion, which is equal to 3.14159.

This formula may seem complicated, but it accurately describes the probability density of the normal distribution. The appearance of π and e in the formula is not accidental; they reflect the deep mathematical properties of the normal distribution.

Effect of Parameters on Distribution

Comparison of normal distribution curves for different parameters, showing the effect of mean and standard deviation

Effect of Mean (μ)

A change in the mean value μ causes the entire distribution curve to shift along the x-axis, but does not change the shape of the curve. larger values of μ cause the curve to shift to the right, while smaller values of μ cause the curve to shift to the left.

Effect of Standard Deviation (σ)

A change in the standard deviation σ changes the width of the curve; the larger the σ, the flatter the curve (dispersion); the smaller the σ, the sharper the curve (concentration). The standard deviation directly affects the degree of dispersion of the data.

Intuitive understanding of formula derivation

The derivation of the formula for the normal distribution can be realised in a number of ways. An elegant method is based on two assumptions:

  1. The distribution of stochastic variables in different dimensions is independent of each other.
  2. Spatial distribution with rotational symmetry

These two assumptions led to the exponential function, which eventually led to the familiar formula for the normal distribution. This derivation was first proposed by astronomer John Herschel in 1850.

Properties of Normal Distributions - Key Properties and Statistical Significance

Symmetry

Symmetry of a normal distribution curve, showing properties about the symmetry of the mean µ

The normal distribution curve is symmetric with respect to the mean μ, i.e. f(μ+x) = f(μ-x). This means that the probability of being the same distance above and below the mean is equal.

The mean, median and mode are equal.

Visualisation of the property of equality of mean, median and multitude of a normal distribution

In a normal distribution, the mean, median and multitude have the same value and are all equal to the parameter µ. This is one of the unique properties of the normal distribution.

Rule 68-95-99.7

Visualisation of the 68-95-99.7 rule, showing the distribution of data over a range of standard deviations.

About 681 TP6T of the data fall within one standard deviation of the mean μ (μ±σ); about 951 TP6T of the data fall within two standard deviations (μ±2σ); and about 99.71 TP6T of the data fall within three standard deviations (μ±3σ).

Linear Transformations

Visualisation of Linear Transformation Properties of Normal Distributions

If X~N(μ,σ²), then the linear transformation Y=aX+b~N(aμ+b,a²σ²). This property is very important in the standardisation process and is the basis for solving the normal distribution problem.

additivity

Visualisation of Additivity Properties of Normal Distributions

The sum of independent normal random variables still obeys the normal distribution. If X~N(μ₁,σ₁²) and Y~N(μ₂,σ₂²) and X,Y are independent, then X+Y~N(μ₁+μ₂,σ₁²+σ₂²).

Maximum entropy distribution

Visualisation of Normal Distributions as Maximum Entropy Distributions

Among all probability distributions with the same mean and variance, the normal distribution has the greatest entropy, which makes it the best choice when only the mean and variance are known.

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Standard Normal Distribution - Z-score and Probability Table

Standard normal distribution is a special normal distribution with mean μ = 0 and standard deviation σ = 1, which is usually denoted by Z. It is an important tool for solving normal distribution problems. It is an important tool to solve the problem of normal distribution.

Standardisation process

Any normal distribution variable X~N(μ,σ²) can be converted to a standard normal distribution variable Z by the following transformation:

Mathematical formula for the standardisation process Z = (X-μ)/σ

This process is called "normalisation" or "Z-transformation" and the transformed Z~N(0,1). Normalisation is a key step in solving the normal distribution problem.

Meaning of Z-score

The Z-score indicates the number of standard deviations of the original data from the mean. Example:

  • Z=0 means that the original value is equal to the mean value.
  • Z=1 means that the original value is one standard deviation above the mean.
  • Z = -2 indicates that the raw value is two standard deviations below the mean value

Z-scores allow us to compare data from different normal distributions and are an important tool for statistical analysis.

Standard Normal Distribution Table

Part of a standard normal distribution table showing the cumulative probabilities corresponding to z-values

The standard normal distribution table (Z-table) provides the cumulative probability P(Z≤z) corresponding to the value of Z and is an important tool for solving normal distribution problems. Care should be taken when using the Z-table:

  • The values in the table are usually cumulative probabilities, i.e. P(Z ≤ z)
  • To find P(Z>z), we can use symmetry and calculate 1-P(Z≤z).
  • To find the interval probability P(a≤Z≤b), we can calculate P(Z≤b)-P(Z≤a).

Example of Z-score calculation

Example:A class of students scored an average of 70 points in a Maths exam with a standard deviation of 10 points. Ming scored 85. What is his z-score?

Solution:Z = (X-μ)/σ = (85-70)/10 = 1.5

This means that Ming's grade is 1.5 standard deviations above the mean, which is equivalent to outperforming about 93.31 TP6T of his classmates (check Table Z for this).

Application of Normal Distribution - Practical Scenarios and Cases

Normal state distribution has a wide range of applications in natural science, social science and engineering fields, and the following are some typical application scenarios:

Natural phenomena

Normal distribution of anthropometric data, e.g. height distribution graphs
  • Anthropometry:Physiological characteristics such as height, weight, etc. of human beings approximately obey the normal distribution.
  • Measurement error:Measurement errors are normally normally normally distributed when the same physical quantity is measured repeatedly.
  • Natural variation:Variation in many biological features such as petal length, leaf size, etc.

Education Measurement

Normal Distribution of Examination Results
  • Examination results:The distribution of scores on large exams usually approximates a normal distribution.
  • Standardised testing:For example, the IQ test is designed to have a normal distribution with a mean of 100 and a standard deviation of 15.
  • Grading:Grade conversion and grading based on Z-scores

Quality Control

Normal distribution for quality control applications in manufacturing processes
  • Manufacturing process:Variations in quality indicators such as product size and weight are usually normally normally distributed.
  • Six Sigma:A quality management approach based on normalised distribution with a target defect rate of 3.4 parts per million or less.
  • Control Chart:Used to monitor whether the production process is under statistical control.

Finance and Risk Management

Application of Normal Distribution in Financial Markets
  • Asset gains:Modelling of financial asset yields
  • Risk Assessment:Risk measurement tools such as VaR (Value at Risk)
  • Option pricing:Key assumptions in the Black-Scholes model

Practical Case: Height Distribution of Hong Kong Students

Assuming that the average height of male students in a secondary school in Hong Kong is 170 cm and the standard deviation is 5 cm, we can use the normal distribution to answer the following questions:

  • What is the percentage of students whose height is over 180cm?
  • What is the percentage of students whose height is between 165cm and 175cm?

Answer:

  1. Z = (180-170)/5 = 2, and the table shows that P(Z>2) ≈ 0.0228, i.e. about 2.28% students were taller than 180cm.
  2. Z₁ = (165-170)/5 = -1, Z₂ = (175-170)/5 = 1, and the table shows that P(-1<Z<1) ≈ 0.6826, which means that the heights of students in TP6T ranged from 165cm to 175cm.
Normal curves of height distribution of Hong Kong students, with relevant districts labelled

Normal Distributions in DSE Maths - Exam Tips and Examples

Normal distribution is an important part of the statistics module of DSE Mathematics (Compulsory Part) and is a common question type in the examination. It is also a common question type in the examination. Mastering the techniques of solving normal distribution questions is crucial to improving your performance in DSE Mathematics.

Normal Distribution Question Types in DSE Exam

  1. Given the mean and standard deviation, find the probability of a given interval
  2. Given the probability, find the corresponding range of values.
  3. Calculation and application of standardised scores (Z-scores)
  4. Calculations related to measures of dispersion (standard deviation, variance, etc.)

Key Steps

  1. Identify the mean μ and standard deviation σ in the questions.
  2. Convert the original value to a Z-score: Z = (X-μ)/σ
  3. Use a calculator or Z-table to find the corresponding probabilities.
  4. Further calculations according to the requirements of the question
Scenes related to the DSE examination, the students' revision attitude is distributed.

DSE Style Example

Example question:The Mathematics test scores of Form 6 students in a school approximately obey the normal distribution, with a mean score of 65 and a standard deviation of 12 points.

(a) Find the probability that a randomly selected student will score more than 80 marks.

(b) If there are 200 S6 students in the school, what is the estimated number of students whose scores are between 50 and 70?

Answer:

(a) First calculate the Z-score: Z = (80-65)/12 = 1.25

Find P(X>80) = P(Z>1.25) ≈ 0.1056 (using calculator or Z-table)

Therefore, the probability of randomly selecting a student with a score of more than 80 is about 0.1056, i.e. 10.56%.

(b) Calculate the Z-score: Z₁ = (50-65)/12 = -1.25, Z₂ = (70-65)/12 = 0.42.

Find P(50<X<70) = P(-1.25<Z<0.42) ≈ 0.6293 (using calculator or Z-table)

Therefore, the number of students with scores between 50 and 70 is estimated to be 200 × 0.6293 ≈ 126.

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Calculator Tips - Quickly Solve Normal Distribution Problems

Proficient use of calculators for normal distribution problems in DSE exams can save a lot of time. Below is a list of commonly used scientific calculators:

CASIO fx-50FH II / fx-991EX

Schematic diagram of the steps to operate the Normal Distribution function on a CASIO calculator.

Find the probability of a normal distribution P(X<a)

  1. Press [SHIFT] → [STAT] → [DIST].
  2. Select [NORM] → [Norm CD].
  3. Enter the lower value (Lower), the upper value (Upper), the mean (μ), the standard deviation (σ)
  4. Select [Execute] to get the result

Find the inverse function of the normal distribution (with known probability, find the value)

  1. Press [SHIFT] → [STAT] → [DIST].
  2. Select [NORM] → [Inv Norm].
  3. Enter the probability value (Tail:Left), mean (μ), standard deviation (σ)
  4. Select [Execute] to get the result

TI-84 Plus

Schematic diagram of the steps to operate the Normal Distribution function of the TI-84 Plus Calculator

Find the probability of a normal distribution P(X<a)

  1. Press [2nd] → [DISTR]
  2. Select [normalcdf(]
  3. Input format: normalcdf(lower limit, upper limit, μ, σ)
  4. Press [ENTER] to get the result

Find the inverse function of the normal distribution (with known probability, find the value)

  1. Press [2nd] → [DISTR]
  2. Select [invNorm(]
  3. Input format: invNorm(probability, μ, σ)
  4. Press [ENTER] to get the result

Tips for using the calculator

  • To find P(X>a), you can use symmetry and calculate 1-P(X<a).
  • To find the interval probability P(a<X<b), we can calculate P(X<b)-P(X<a).
  • The calculation of the standardised normal distribution can be done directly by setting µ = 0 and σ = 1.
  • The "Tail" option in the calculator usually refers to the direction of the cumulative probability.
  • Proficient use of the calculator can greatly improve the efficiency of solving questions in the DSE exams.

Practice Questions and Explanations - Consolidate what you've learnt

Test your understanding of the normal distribution with the following exercise questions:

Exercise 1

Title:The life of a particular brand of light bulb obeys the normal distribution with an average life of 1000 hours and a standard deviation of 100 hours.

(a) Find the probability that a randomly selected light bulb will last more than 1200 hours.

(b) How many hours of lamp life does the brand claim that the 95% has exceeded?

Answer:

(a) Z = (1200-1000)/100 = 2, P(X>1200) = P(Z>2) ≈ 0.0228, i.e., about 2.28%

(b) Need to find the value of x such that P(X>x) = 0.95, i.e., P(X<x) = 0.05

The corresponding Z value is -1.645 (look up the table or use the calculator).

x = μ + Z-σ = 1000 + (-1.645)-100 = 835.5 hours

Exercise 2

Title:The results of a university entrance examination approximately obey a normal distribution with a mean score of 500 and a standard deviation of 100. The university plans to admit the top 20% candidates.

(a) What should be the cut-off point for admission?

(b) If a candidate scores 650 marks, what is his Z-score? What is the per centile of this score among all candidates?

Answer:

(a) Need to find the value of x such that P(X>x) = 0.2, i.e. P(X<x) = 0.8

The corresponding Z-value is 0.84 (look up the table or use the calculator).

x = μ + Z-σ = 500 + 0.84-100 = 584 points

(b) Z = (650-500)/100 = 1.5

P(X<650) = P(Z<1.5) ≈ 0.9332, i.e. about 93.32%

This means that the candidate outperformed about 93.321 TP6T candidates and was in the 93.32nd percentile.

Mastering Normal Distributions to Improve Maths Achievement!

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