Normal Approximation to the Binomial Distribution and Inverse Exponential Functions

Conceptual Overview
- In Chapter 5, the binomial distribution is defined as a discrete distribution with parameters $n$ (finite number of trials) and $\pi$ (constant probability of success).
- Calculating cumulative probabilities for discrete distributions requires the summation of individual probabilities. For example, to find $P(X \leq 50)$ when $n = 100$ and $\pi = 0.3$ , one must calculate: $P(X = 0) + P(X = 1) + P(X = 2) + \dots + P(X = 50)$
- This is described as a "very long and tedious calculation" when performed manually.
The Symmetrical Nature of Large Trials
- Plotting a binomial distribution with a large $n$ reveals that the discrete bars begin to resemble a bell-shaped curve.
- Because the distribution becomes symmetrical and bell-shaped as $n$ increases, the normal distribution (a continuous distribution) can be used to approximate these probabilities, significantly shortening the calculation process.

Minimum Threshold Checks
- The normal approximation is not always valid. To use it, the binomial distribution must pass two tests based on a "rule of thumb" to ensure accuracy. If these values are below 5, the approximation breakdown and loses accuracy.
- The criteria are:
  1. $n \times \pi > 5$
  2. $n \times (1 - \pi) > 5$
Assessment Importance
- In exams offering calculation marks (such as second-semester assessments), explicitly writing down these checks and verifying they hold can be worth two marks.

Variable notation
- To distinguish the binomial random variable from its normal replacement, the approximated variable is denoted as $x_a$ (x approximate).
Equating the Mean and Standard Deviation
- For the approximation to be valid, the normal distribution must share the same center and spread as the original binomial distribution.
- Mean ( $\mu$ ): The expected value of the binomial distribution is calculated as: $\mu = n \times \pi$
- Standard Deviation ( $\sigma$ ): Calculated as: $\sigma = \sqrt{n \times \pi \times (1 - \pi)}$
Standardization to Z-Values
- To find the probability using the standard normal distribution table or a calculator, the approximated variable must be standardized: $Z = \frac{x_a - (n \times \pi)}{\sqrt{n \times \pi \times (1 - \pi)}}$
- The resulting $Z$ variable is normal with a mean of 0 and a standard deviation of 1: $Z \sim N(0, 1)$ .

Theory of Continuity Correction
- Discrete variables represent points, while continuous variables represent the area under a curve. Moving from one to the other is not a direct "one-to-one" transition.
- To calculate the area under the curve effectively, we "pad" the discrete value by adding or subtracting $0.5$ . This creates a small slice of area representing the discrete point.
Correction Scenarios
- Exact Value: $P(X = x)$ becomes $P(x - 0.5 < x_a < x + 0.5)$ .
- Less Than or Equal To: $P(X \leq x)$ becomes $P(x_a < x + 0.5)$ .
- Greater Than or Equal To: $P(X \geq x)$ becomes $P(x_a > x - 0.5)$ .

Scenario Details
- A flight accommodates $300 \text{ passengers}$ .
- The airline accepts $n = 350 \text{ bookings}$ , hoping no more than 300 people show up.
- The probability that a passenger shows up for their booking is $\pi = 0.8$ .
- Goal: Find the probability that more than 300 passengers show up: $P(X > 300)$ .
Step 1: Verification Checks
- $n \times \pi = 350 \times 0.8 = 280$ (Greater than 5)
- $n \times (1 - \pi) = 350 \times 0.2 = 70$ (Greater than 5)
- Conclusion: Normal approximation is valid.
Step 2: Calculating Parameters
- $\mu = 280$
- $\sigma = \sqrt{350 \times 0.8 \times 0.2} = \sqrt{56}$
Step 3: Probability Transformation and Continuity Correction
- Original binomial target: $P(X > 300)$ .
- Rewrite as cumulative: $1 - P(X \leq 300)$ .
- Apply continuity correction for the normal variable: $1 - P(x_a < 300.5)$ .
Step 4: Standardization and Result
- Standardize to Z: $Z = \frac{300.5 - 280}{\sqrt{56}} = \frac{20.5}{\sqrt{56}} \approx 2.74$
- Using an HP calculator (Shift $Z \rightarrow P$ ): $P(Z < 2.74) = 0.9969$
- Final Probability: $1 - 0.9969 = 0.0031$ .
Accuracy Comparison
- Excel exact binomial calculation ( $1 - \text{BINOM.DIST}(300, 350, 0.8, \text{TRUE})$ ): $0.00224$
- The approximation result ( $0.0031$ ) is close to the exact value ( $0.00224$ ). Accuracy typically increases with higher $n$ .

Standard Probability Problem
- In the standard exponential distribution, we know the time $t$ and solve for the probability $\alpha$ : $\alpha = P(X < t) = 1 - e^{-\lambda \times t}$
The Inverse Problem
- One may be given a specific probability (e.g., the $80^{th}$ percentile, where $\alpha = 0.8$ ) and asked to find the unknown point $p$ on the curve associated with that probability.
Mathematical Derivation for p
1. Start with the identity: $\alpha = 1 - e^{-\lambda \times p}$
2. Rearrange: $e^{-\lambda \times p} = 1 - \alpha$
3. Take the natural logarithm (ln) of both sides to cancel $e$ : $-\lambda \times p = \ln(1 - \alpha)$
4. Solve for the point $p$ : $p = -\frac{1}{\lambda} \times \ln(1 - \alpha)$
Example Calculation
- Given $\lambda = 2$ and $\alpha = 0.8$ : $p = -\frac{1}{2} \times \ln(1 - 0.8)$ $p = -0.5 \times \ln(0.2) \approx 0.8047$

Question: Where does the value of 5 come from in the verification checks?
Response: It is a standard rule of thumb developed through observation. When values are lower than 5, the model's accuracy deteriorates significantly.
Question: On the HP calculator, how do you perform the natural log ( $\ln$ )?
Response: On the HP calculator, the process is often the reverse of the Casio. You must input the number first and then press the function button. (The instructor noted they are still working on mastering the HP calculator interface for this specific function).
Question: In the Excel function BINOM.DIST, what does "True" mean?
Response: "True" is used for cumulative probabilities. If set to "False," it would calculate the probability of the variable being exactly equal to a specific value.