Modeling Random Events: The Normal and Binomial Models – Study Notes

Modeling Random Events: The Normal and Binomial Models

Normal Distribution

The normal distribution is a continuous probability distribution that is symmetric about its mean, often used to model real-world phenomena such as heights, test scores, and measurement errors. It is characterized by its bell-shaped curve.

• Key Properties:

  • Mean ($\mu$) and standard deviation ($\sigma$) determine the center and spread.

  • The total area under the curve is 1.

  • Empirical Rule: Approximately 68% of data falls within 1 standard deviation, 95% within 2, and 99.7% within 3.

• Standard Normal Distribution: A normal distribution with $\mu = 0$ and $\sigma = 1$.

• Z-score: The number of standard deviations a value $x$ is from the mean.

  • Formula: $z = \frac{x - \mu}{\sigma}$

Example: If $X \sim N(20, 3^2)$, then $P(X < 17)$ can be found by converting 17 to a z-score and using the standard normal table.

Calculating Probabilities Using the Normal Distribution

To find the probability that a normal random variable falls within a certain range, convert the values to z-scores and use the standard normal table.

• Steps:

  1. Compute the z-score for the value(s) of interest.

  2. Use the standard normal table to find the corresponding probability.

  3. For intervals, subtract probabilities as needed.

Example: $P(20 < X < 31)$ for $X \sim N(25, 5^2)$

• Find $z_1 = \frac{20-25}{5} = -1$

• Find $z_2 = \frac{31-25}{5} = 1.2$

• Probability = $P(z_1 < Z < z_2) = P(Z < 1.2) - P(Z < -1)$

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success.

• Key Properties:

  • Number of trials: $n$

  • Probability of success: $p$

  • Probability of failure: $q = 1 - p$

• Probability Mass Function:

  • $P(X = k) = \binom{n}{k} p^k q^{n-k}$

• Mean and Standard Deviation:

  • Mean: $\mu = np$

  • Standard deviation: $\sigma = \sqrt{npq}$

Example: If $n = 10$ and $p = 0.3$, then the probability of exactly 4 successes is $P(X = 4) = \binom{10}{4} (0.3)^4 (0.7)^6$.

Normal Approximation to the Binomial

When $n$ is large and $p$ is not too close to 0 or 1, the binomial distribution can be approximated by a normal distribution with mean $np$ and standard deviation $\sqrt{npq}$.

• Continuity Correction: When using the normal approximation, add or subtract 0.5 to the discrete x-value to improve accuracy.

• Conditions: Both $np \geq 10$ and $nq \geq 10$ should be satisfied.

Example: For $n = 100$, $p = 0.5$, to approximate $P(X \leq 60)$:

• Mean: $\mu = 50$

• Standard deviation: $\sigma = 5$

• Apply continuity correction: $P(X \leq 60.5)$

• Find $z = \frac{60.5 - 50}{5} = 2.1$

• Use standard normal table to find $P(Z < 2.1)$

Using Z-tables and Probability Calculations

Z-tables provide the area (probability) to the left of a given z-score in the standard normal distribution.

• To find $P(X > a)$, compute $z$ for $a$ and use $1 - P(Z < z)$.

• To find $P(a < X < b)$, compute $z$ for both $a$ and $b$, then subtract: $P(Z < z_b) - P(Z < z_a)$.

Example: If $z = 1.28$, $P(Z > 1.28) = 1 - P(Z < 1.28) = 1 - 0.8997 = 0.1003$.

Summary Table: Normal vs. Binomial Distribution

Property	Normal Distribution	Binomial Distribution
Type	Continuous	Discrete
Parameters	Mean ($\mu$), Std. Dev. ($\sigma$)	Number of trials ($n$), Probability of success ($p$)
Shape	Symmetric, bell-shaped	Symmetric if $p=0.5$, skewed otherwise
Probability Calculation	Area under curve	Sum of probabilities for integer values
Approximation	--	Can be approximated by normal if $np$ and $nq$ are large

Additional info: Some values and context were inferred from partial equations and notation in the original file, which included z-scores, probability statements, and references to normal and binomial models.