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Edexcel IAL - Statistics 1- 6.1 Normal Distribution, Mean and Variance- Study notes  - New syllabus

Edexcel IAL – Statistics 1- 6.1 Normal Distribution, Mean and Variance -Study notes- New syllabus

Edexcel IAL – Statistics 1- 6.1 Normal Distribution, Mean and Variance -Study notes -Edexcel A level Maths- per latest Syllabus.

Key Concepts:

  • 6.1 Normal Distribution, Mean and Variance

Edexcel IAL Maths-Study Notes- All Topics

The Normal Distribution

The normal distribution is a continuous probability distribution that is widely used to model real-life data such as heights, weights, measurement errors, and examination marks.

A random variable that follows a normal distribution is written as  

\( X \sim N(\mu, \sigma^2) \)

where:

\( \mu \) is the mean

\( \sigma^2 \) is the variance

\( \sigma \) is the standard deviation

Shape and Symmetry

The normal distribution has the following key features:

  • It is symmetric about the mean \( \mu \)
  • The mean, median, and mode are equal
  • The curve is bell-shaped

Because of the symmetry, probabilities on either side of the mean are equal.

Mean and Variance

For a normally distributed random variable \( X \sim N(\mu, \sigma^2) \):

\( E(X) = \mu \)

\( \mathrm{Var}(X) = \sigma^2 \)

These results may be used directly without proof.

Standardisation

Probabilities for the normal distribution are found by converting values of \( X \) to a standard normal variable \( Z \), where

\( Z = \dfrac{X – \mu}{\sigma} \)

The standard normal distribution is written as

\( Z \sim N(0,1) \)

Use of Normal Distribution Tables

Tables give values of the cumulative distribution function

\( P(Z \leq z) \)

To find probabilities involving \( X \):

  • Convert values of \( X \) to \( Z \) using standardisation
  • Use the table to find probabilities for \( Z \)
  • Use symmetry when appropriate

Interpolation is not required in this syllabus.

Z- table value:

z0.000.010.020.030.040.050.060.070.080.09
0.00.50000.50400.50800.51200.51600.51990.52390.52790.53190.5359
0.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.5754
0.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.6141
0.30.61790.62170.62550.62930.63310.63680.64060.64430.64800.6517
0.40.65540.65910.66280.66640.67000.67360.67720.68080.68440.6879
0.50.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224
0.60.72580.72910.73240.73570.73890.74220.74540.74860.75180.7549
0.70.75800.76110.76420.76730.77040.77340.77640.77940.78230.7852
0.80.78810.79100.79390.79670.79950.80230.80510.80780.81060.8133
0.90.81590.81860.82120.82380.82640.82890.83150.83400.83650.8389
1.00.84130.84380.84610.84850.85080.85310.85540.85770.85990.8621
1.10.86430.86650.86860.87080.87290.87490.87700.87900.88100.8830
1.20.88490.88690.88880.89070.89250.89440.89620.89800.89970.9015
1.30.90320.90490.90660.90820.90990.91150.91310.91470.91620.9177
1.40.91920.92070.92220.92360.92510.92650.92790.92920.93060.9319
1.50.93320.93450.93570.93700.93820.93940.94060.94180.94300.9441
1.60.94520.94630.94740.94850.94950.95050.95150.95250.95350.9545
1.70.95540.95640.95730.95820.95910.95990.96080.96160.96250.9633
1.80.96410.96490.96560.96640.96710.96780.96860.96930.97000.9706
1.90.97130.97190.97260.97320.97380.97440.97500.97560.97610.9767
2.00.97730.97780.97830.97880.97930.97980.98030.98080.98120.9817

Using Symmetry

For the standard normal distribution:

\( P(Z \leq 0) = 0.5 \)

\( P(Z \leq -z) = 1 – P(Z \leq z) \)

These properties are frequently used to simplify calculations.

Solving Problems Involving Equations

Some questions may require:

  • Finding unknown values of \( \mu \) or \( \sigma \)
  • Solving simultaneous equations involving probabilities

These problems rely on correct use of standardisation and normal tables, not on derivation of formulas.

Example :

The mass of packets of rice is normally distributed with mean \( \mu = 5 \) kg and standard deviation \( \sigma = 0.2 \) kg.

Find the probability that a randomly selected packet has a mass less than 4.8 kg.

▶️ Answer/Explanation

Let \( X \sim N(5, 0.2^2) \).

Standardise:

\( Z = \dfrac{4.8 – 5}{0.2} = -1 \)

Using normal tables:

\( P(Z \leq -1) = 1 – P(Z \leq 1) = 1 – 0.8413 = 0.1587 \)

Conclusion: The required probability is 0.1587.

Example :

The heights of students in a college are normally distributed with mean \( \mu = 170 \) cm and standard deviation \( \sigma = 6 \) cm.

Find the probability that a randomly selected student has a height between 164 cm and 176 cm.

▶️ Answer/Explanation

Let \( X \sim N(170, 6^2) \).

Standardise both values:

\( Z_1 = \dfrac{164 – 170}{6} = -1,\quad Z_2 = \dfrac{176 – 170}{6} = 1 \)

Using normal tables and symmetry:

\( P(-1 \leq Z \leq 1) = 0.8413 – 0.1587 = 0.6826 \)

Conclusion: The required probability is 0.6826.

Example :

A random variable \( X \) is normally distributed. It is given that

\( P(X \leq 50) = 0.5 \) and \( P(X \leq 58) = 0.9772 \).

Find the mean \( \mu \) and the standard deviation \( \sigma \).

▶️ Answer/Explanation

Since \( P(X \leq 50) = 0.5 \), the value 50 corresponds to the mean.

\( \mu = 50 \)

From normal tables,

\( P(Z \leq 2) = 0.9772 \)

So,

\( \dfrac{58 – 50}{\sigma} = 2 \)

\( \sigma = 4 \)

Conclusion: The distribution is \( X \sim N(50, 4^2) \).

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