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IB Mathematics AA Normal distribution Study Notes

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.

The Normal Distribution and Curve

The Normal Distribution and Curve

The normal distribution is a continuous probability distribution that is symmetric about its mean. It describes many natural phenomena (e.g. heights, test scores) and is often used as an approximation for binomial distributions when \( n \) is large.

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

where:

  • \( \mu \) = mean (center of the distribution)
  • \( \sigma^2 \) = variance
  • \( \sigma \) = standard deviation (measure of spread)

The probability density function is:

\( f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{ – \frac{(x – \mu)^2}{2 \sigma^2}} \)

Properties of the Normal Distribution

  • It is bell-shaped and symmetric about \( \mu \).
  • The mean, median, and mode are equal: \( \mu = \text{median} = \text{mode} \).
  • Approximately 68% of values lie within \( \mu \pm \sigma \).
  • Approximately 95% of values lie within \( \mu \pm 2\sigma \).
  • Approximately 99.7% of values lie within \( \mu \pm 3\sigma \).
  • The total area under the curve is 1 (represents total probability).

Diagrammatic Representation

A normal distribution curve is smooth and bell-shaped. The highest point is at \( x = \mu \). The curve approaches but never touches the horizontal axis (asymptotic).

Diagram description:

  • Horizontal axis labeled \( x \)
  • Curve centered at \( \mu \)
  • Vertical lines at \( \mu \), \( \mu \pm \sigma \), \( \mu \pm 2\sigma \), \( \mu \pm 3\sigma \)
  • Shading for areas representing 68%, 95%, and 99.7%

Note: In practice, values of normal probabilities are found using technology (GDC, calculator) or standard normal tables.

Example:

The heights of adult men in a city are normally distributed with mean 175 cm and standard deviation 6 cm. Find the probability that a randomly selected man is between 169 cm and 181 cm.

▶️ Answer/Explanation

 Standardize values

\( z_1 = \frac{169 – 175}{6} = -1 \)
\( z_2 = \frac{181 – 175}{6} = 1 \)

 Use technology or tables

\( P(-1 \le Z \le 1) \approx 0.6826 \)

Conclusion: The probability is about 0.683, or 68.3%.

Example:

The time taken to complete a standardized math exam is normally distributed with mean 120 minutes and standard deviation 10 minutes. Find the probability that a student finishes the exam:

  • (a) in less than 135 minutes
  • (b) between 110 and 130 minutes
▶️ Answer/Explanation

\( \mu = 120 \), \( \sigma = 10 \)

 (a): Find \( P(X < 135) \)

Standardize: \( z = \frac{135 – 120}{10} = 1.5 \)

Using GDC: normalcdf(-1E99, 135, 120, 10)
Result: \( P(X < 135) \approx 0.9332 \)

 (b): Find \( P(110 < X < 130) \)

Standardize: \( z_1 = \frac{110 – 120}{10} = -1 \), \( z_2 = \frac{130 – 120}{10} = 1 \)

Using GDC: normalcdf(110, 130, 120, 10)
Result: \( P(110 < X < 130) \approx 0.6826 \)

Normal Probability Calculations

Normal Probability Calculations

When a variable \( X \) follows a normal distribution \( N(\mu, \sigma^2) \), probabilities and specific values of \( X \) are generally found using technology (such as a GDC, TI-84, or Casio fx-CG50).

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

Probabilities correspond to the area under the normal curve between specified values of \( X \).

Technology functions for normal probabilities:

  • normalcdf(lower, upper, μ, σ): Finds the probability \( P(\text{lower} < X < \text{upper}) \).
  • invNorm(p, μ, σ): Finds the value \( X \) such that \( P(X < x) = p \).

Why use technology?

  • The normal distribution does not have a simple formula for cumulative probabilities.

Example:

The weights of apples from a farm are normally distributed with mean 150 g and standard deviation 12 g. Find:

  • (a) the probability that a randomly chosen apple weighs less than 160 g
  • (b) the weight which separates the lightest 10% of apples
▶️ Answer/Explanation

\( \mu = 150 \), \( \sigma = 12 \)

(a): Find \( P(X < 160) \)

Use GDC: normalcdf(-1E99, 160, 150, 12)
Result: \( P(X < 160) \approx 0.7977 \)

 (b): Find \( x \) where \( P(X < x) = 0.10 \)

Use GDC: invNorm(0.10, 150, 12)
Result: \( x \approx 134.6 \) g

Conclusion:

  • Probability an apple weighs less than 160 g ≈ 0.798
  • The weight that separates the lightest 10% ≈ 134.6 g

Inverse Normal Calculations

Inverse Normal Calculations

Inverse normal calculations are used to find a value of \( X \) given a cumulative probability. This means we are given a probability \( P(X < x) = p \), and we need to find \( x \).

We use technology (GDC / calculator) for this because there is no simple algebraic method.

invNorm(p, \mu, \sigma)

Where:

  • \( p \) = cumulative probability \( P(X < x) \)
  • \( \mu \) = mean
  • \( \sigma \) = standard deviation

Key point: Mean and standard deviation will always be given in inverse normal calculations.

Example:

The scores on a standardized test are normally distributed with a mean of 500 and standard deviation of 100. Find the score that separates the top 5% of students.

▶️ Answer/Explanation

\( \mu = 500 \), \( \sigma = 100 \)

Cumulative probability

Top 5% means \( P(X > x) = 0.05 \)
So, \( P(X < x) = 0.95 \)

Use inverse normal

On GDC: invNorm(0.95, 500, 100)

Result: \( x \approx 664.5 \)

Conclusion: The score that separates the top 5% is approximately 664.5.

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