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Statistics and Probability - Continuous Probability Distributions (Normal Distribution)

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The Normal Distribution is a continuous probability distribution defined by the notation XN(μ,σ2)X \sim N(\mu, \sigma^2), where μ\mu is the mean and σ2\sigma^2 is the variance. Visually, it is represented by a 'bell-shaped' curve that is perfectly symmetrical about the vertical line x=μx = \mu. The peak of the curve occurs at the mean, which is also the mode and the median for this distribution.

The total area under the normal curve is exactly 11, representing a total probability of 100%100\%. Because of its symmetry, the area to the left of the mean P(X<μ)P(X < \mu) is 0.50.5, and the area to the right P(X>μ)P(X > \mu) is also 0.50.5. On a graph, this means the curve is divided into two identical halves by the mean line.

The parameters μ\mu and σ\sigma control the shape and position of the distribution. Changing μ\mu shifts the entire bell curve horizontally along the x-axis without changing its shape. Changing the standard deviation σ\sigma affects the 'spread': a small σ\sigma results in a tall, thin curve clustered tightly around the mean, while a large σ\sigma results in a shorter, wider, and flatter curve.

The Empirical Rule (or the 68-95-99.7 rule) describes the spread of data: approximately 68.27%68.27\% of the data lies within one standard deviation of the mean ([μσ,μ+σ][\mu - \sigma, \mu + \sigma]), 95.45%95.45\% lies within two standard deviations ([μ2σ,μ+2σ][\mu - 2\sigma, \mu + 2\sigma]), and 99.73%99.73\% lies within three standard deviations. Visually, the 'tails' of the curve approach the x-axis but never actually touch it (asymptotic behavior).

The Standard Normal Distribution is a specific normal distribution where μ=0\mu = 0 and σ=1\sigma = 1, denoted as ZN(0,1)Z \sim N(0, 1). Any normal variable XX can be transformed into a ZZ-score, which represents the number of standard deviations an observation is from the mean. This is visually useful for comparing different datasets on a standardized scale.

Probability for a continuous distribution is defined as the area under the curve between two points. For the Normal Distribution, the probability P(a<X<b)P(a < X < b) is the integral of the probability density function from aa to bb. Crucially, for any continuous distribution, the probability of the variable equaling an exact value is zero, P(X=k)=0P(X = k) = 0.

The Inverse Normal Distribution is the process of finding a boundary value xx given a known area (probability). For example, if you are given that the bottom 10%10\% of students failed a test, you use the inverse normal function on your GDC to find the score xx such that P(X<x)=0.10P(X < x) = 0.10. Visually, this corresponds to finding the x-coordinate that cuts off a specific area in the left tail of the curve.

📐Formulae

XN(μ,σ2)X \sim N(\mu, \sigma^2)

Z=XμσZ = \frac{X - \mu}{\sigma}

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

P(a<X<b)=P(aμσ<Z<bμσ)P(a < X < b) = P(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma})

f(x)=1σ2πe12(xμσ)2f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}(\frac{x-\mu}{\sigma})^2}

P(X<μ)=P(X>μ)=0.5P(X < \mu) = P(X > \mu) = 0.5

💡Examples

Problem 1:

The masses of apples in an orchard follow a normal distribution with a mean of 150150 g and a standard deviation of 1212 g. Find the probability that a randomly selected apple has a mass between 140140 g and 165165 g.

Solution:

  1. Define the variable: XN(150,122)X \sim N(150, 12^2).
  2. We need to find P(140<X<165)P(140 < X < 165).
  3. Standardize the values to ZZ-scores: Z1=14015012=0.833Z_1 = \frac{140 - 150}{12} = -0.833 Z2=16515012=1.25Z_2 = \frac{165 - 150}{12} = 1.25
  4. Using a GDC (Normal Cumulative Distribution function): P(0.833<Z<1.25)0.692P(-0.833 < Z < 1.25) \approx 0.692. Therefore, the probability is approximately 0.6920.692 (or 69.2%69.2\%).

Explanation:

This problem requires calculating the area under the normal curve between two points. We identify the parameters μ=150\mu = 150 and σ=12\sigma = 12, then use the cumulative distribution function (normCdf) on a calculator to find the area.

Problem 2:

The heights of a population are normally distributed with a mean of 170170 cm. If 10%10\% of the population are taller than 185185 cm, find the standard deviation σ\sigma.

Solution:

  1. Define the variable: XN(170,σ2)X \sim N(170, \sigma^2).
  2. We are given P(X>185)=0.10P(X > 185) = 0.10.
  3. This implies P(X<185)=10.10=0.90P(X < 185) = 1 - 0.10 = 0.90.
  4. Find the ZZ-score corresponding to a cumulative probability of 0.900.90 using Inverse Normal: P(Z<z)=0.90z1.2816P(Z < z) = 0.90 \Rightarrow z \approx 1.2816.
  5. Use the ZZ-score formula: 1.2816=185170σ1.2816 = \frac{185 - 170}{\sigma}.
  6. Solve for σ\sigma: 1.2816σ=151.2816\sigma = 15 σ=151.281611.7\sigma = \frac{15}{1.2816} \approx 11.7 cm.

Explanation:

When the standard deviation is unknown, we use the Standard Normal Distribution (ZZ) as a bridge. We find the ZZ-score that corresponds to the given percentile and then solve the linear equation for σ\sigma.