IB Mathematics AA Normal distribution Study Notes
IB Mathematics AA Normal distribution Study Notes
IB Mathematics AA Normal distribution Notes Offer a clear explanation of Normal distribution, its mean and variance, including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Normal distribution, its mean and variance.
Normal Distribution
Introduction
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution widely used in statistics to model real-world phenomena. Its characteristic bell-shaped curve represents the distribution of data where most observations cluster around the mean.
Key Concepts
1. Properties of the Normal Distribution
- Symmetrical about the mean (\( \mu \)).
- The mean (\( \mu \)), median, and mode are equal.
- Defined by two parameters:
- Mean (\( \mu \))—the central value.
- Standard deviation (\( \sigma \))—controls the spread of the data.
- The total area under the curve is 1, representing the entire probability space.
- Follows the empirical rule:
- Approximately 68% of the data lies within \( \mu \pm \sigma \).
- Approximately 95% of the data lies within \( \mu \pm 2\sigma \).
- Approximately 99.7% of the data lies within \( \mu \pm 3\sigma \).
2. Diagrammatic Representation
The normal distribution is represented by a bell-shaped curve with the highest point at the mean. The curve tapers symmetrically on both sides, approaching but never touching the x-axis.
3. Natural Occurrence
- The normal distribution is observed in various natural and real-world phenomena, such as:
- Heights of individuals.
- Measurement errors in experiments.
- IQ scores.
- Stock market returns.
4. Normal Probability Calculations
- Probabilities are calculated as the area under the curve for a given interval of \( X \).
- Use technology (graphing calculators or statistical software) to find:
- Probabilities between two values of \( X \).
- The cumulative probability \( P(X \leq x) \).
5. Inverse Normal Calculations
- Used to determine the value of the variable (\( X \)) for a given probability.
- Given the mean (\( \mu \)) and standard deviation (\( \sigma \)), use technology to find:
- \( X \) such that \( P(X \leq x) = p \).
6. Standard Normal Distribution (\( Z \))
- The standard normal distribution has:
- Mean (\( \mu \)) = 0.
- Standard deviation (\( \sigma \)) = 1.
- Transformation of a normal variable \( X \) to the standard normal variable \( Z \):
\( Z = \frac{X – \mu}{\sigma} \)
Solved Examples
Example 1: Calculating Probabilities
Problem: The heights of adult males are normally distributed with a mean of 175 cm and a standard deviation of 10 cm. What proportion of males are shorter than 185 cm?
Solution:
- Given:
- \( \mu = 175 \), \( \sigma = 10 \).
- \( X = 185 \).
- Standardize \( X \) to find \( Z \):
\( Z = \frac{X – \mu}{\sigma} = \frac{185 – 175}{10} = 1 \).
- Using a standard normal table or technology:
\( P(X \leq 185) = P(Z \leq 1) \approx 0.8413 \).
Answer: Approximately 84.13% of males are shorter than 185 cm.
Example 2: Inverse Normal Calculation
Problem: A test score is normally distributed with \( \mu = 100 \) and \( \sigma = 15 \). Find the score corresponding to the top 5% of the distribution.
Solution:
- Top 5% corresponds to \( P(X > x) = 0.05 \), or \( P(X \leq x) = 0.95 \).
- Using technology, find the inverse normal value for \( P(X \leq x) = 0.95 \):
\( x \approx 124.7 \).
Answer: The score is approximately 124.7.
IB Mathematics AA SL Normal distribution Exam Style Worked Out Questions
Question
Let X be normally distributed with mean 100 cm and standard deviation 5 cm.
a.On the diagram below, shade the region representing \({\rm{P}}(X > 105)\) .[2]
b.Given that \({\rm{P}}(X < d) = {\rm{P}}(X > 105)\) , find the value of \(d\) . [2]
c.Given that \({\rm{P}}(X > 105) = 0.16\) (correct to two significant figures), find \({\rm{P}}(d < X < 105)\) . [2]
▶️Answer/Explanation
Markscheme
a. A1A1 N2
Note: Award A1 for vertical line to right of mean, A1 for shading to right of their vertical line.
evidence of recognizing symmetry (M1)
e.g. \(105\) is one standard deviation above the mean so \(d\) is one standard deviation below the mean, shading the corresponding part, \(105 – 100 = 100 – d\)
\(d = 95\) A1 N2
[2 marks]
evidence of using complement (M1)
e.g. \(1 – 0.32\) , \(1 – p\)
\({\rm{P}}(d < X < 105) = 0.68\) A1 N2
[2 marks]
Question
The random variable X is normally distributed with mean 1000 and standard deviation 50.
(a) Find (i) \(P(X < 925)\) (ii) \(P(925 < X < 1025)\) (iii) \(P(X > 1025)\)
(b) Sketch a graph representing the information in \((a)\)
(c) Find the standardised values of \(925\) and \(1025\)
(d) Sketch the corresponding graph of standardised values
(e) Find \(E(X^{2})\)
▶️Answer/Explanation
Ans
(a) (i) \(P(X<925)=0.0668\) (ii) \(P(925<X<1025)\) (iii) \(P(X>1025)=0.309\)
(b)
(c) \(-1.5\) and \(0.5\) respectively
(d) similar to the above the vertical boundaries are \(-1.5\) and \(0.5\)
(e) \(E(X^{2})=Var(X)+E(X)^{2}=\sigma ^{2}+\mu ^{2}=50^{2}+1000^{2}=1002500\)