Lesson 3.4: Normal Approximation to the Binomial; Poisson and Exponential Distributions

Introduction

In this lesson, we will explore three important concepts in probability distributions: the normal approximation to the binomial distribution, the Poisson distribution, and the exponential distribution. Understanding these distributions will allow you to model and analyze various real-world situations effectively.

Learning Objectives

By the end of this lesson, you will be able to:

• Use the normal distribution to approximate a binomial distribution under specified conditions, accounting for continuity correction.
• Determine when the Poisson model is appropriate and evaluate Poisson probabilities, including mean and variance.
• Describe the exponential distribution as the model for time intervals between random Poisson events and evaluate its probabilities, mean, and variance.
• Apply appropriate approximations and models in real-world scenarios.
• Make informed decisions regarding probability distributions based on the characteristics of the data involved.

Section 1: Normal Approximation to the Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials, each having the same probability of success. However, as the number of trials increases, calculating binomial probabilities directly becomes complex.

When to Use Normal Approximation

The normal approximation to the binomial distribution can be used when:

• The number of trials $n$ is large.
• The expected number of successes $np$ and the expected number of failures $n(1-p)$ are both greater than 5, i.e., $np > 5$ and $n(1-p) > 5$.

Continuity Correction

When using the normal approximation for a binomial distribution, a continuity correction is applied. This means that when we want to find the probability of a discrete outcome, we adjust it to account for the continuous nature of the normal distribution. For example, to find $P(X = k)$ in a binomial distribution, we instead calculate:

$$\begin{align} P(k - 0.5 < X < k + 0.5).\end{align}$$

This adjustment allows us to achieve a more accurate approximation of the binomial probability using the normal curve.

Worked Example 1

Problem: A fair coin is flipped 100 times (i.e., $n = 100$, $p = 0.5$). What is the probability of getting exactly 55 heads?

Solution:

1. First, calculate the expected number of heads:

$$\begin{align} np &= 100 \times 0.5 = 50,\ n(1-p) &= 100 \times 0.5 = 50. \end{align}$$

Since both values are greater than 5, we can use the normal approximation.

1. Now, we apply the continuity correction to find:

$$\begin{align} P(X = 55) \approx P(54.5 < X < 55.5).\end{align}$$

1. Convert this to a standard normal variable $Z$:

$$\begin{align} Z &= \frac{X - \mu}{\sigma},\ \mu &= np = 50,\ \sigma &= \sqrt{np(1-p)} = \sqrt{100 \times 0.5 \times 0.5} = 5. \end{align}$$

1. Calculate the $Z$-scores:

• For $X = 54.5$:

$$\begin{align} Z_{54.5} &= \frac{54.5 - 50}{5} = 0.9,\end{align}$$

• For $X = 55.5$:

$$\begin{align} Z_{55.5} &= \frac{55.5 - 50}{5} = 1.1.\end{align}$$

1. Using a standard normal distribution table, find:

• $P(Z < 1.1) \approx 0.8643$ and $P(Z < 0.9) \approx 0.8159.$

1. Therefore:

$$\begin{align} P(54.5 < X < 55.5) &\approx 0.8643 - 0.8159 = 0.0484.\end{align}$$

Thus, the probability of getting exactly 55 heads is approximately $0.0484$ or $4.84\%$.

Section 2: The Poisson Distribution

The Poisson distribution models the number of events occurring in a fixed interval of time or space, given that these events happen with a known average rate and independently of the time since the last event.

When to Use the Poisson Model

You can use the Poisson distribution when the following conditions are met:

• The events occur independently.
• The average rate (mean) of occurrence is constant.
• Two events cannot occur at the same instant in the infinitesimal interval.

Key Parameters

The Poisson distribution has one parameter, $\lambda$, which is the average rate of occurrence.

Probability Mass Function

The probability of observing $k$ events in an interval is given by the probability mass function:

$$P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

where $e$ is the base of the natural logarithm (approximately 2.71828).

Worked Example 2

Problem: Suppose a call center receives an average of 3 calls per minute. What is the probability that they will receive exactly 5 calls in a minute?

Solution:

1. Determine $\lambda$: Here, $\lambda = 3$ (the average number of calls).
2. Find the probability using the PMF:

$$P(X = 5) = \frac{3^5 e^{-3}}{5!}$$

1. Calculate:

• First, find $3^5 = 243$.
• Next, find $5! = 120$.
• Then calculate $e^{-3} \approx 0.0498$ (using a calculator).

$P(X = 5) = \frac{243 \times 0.0498}{120} \approx \frac{12.0942}{120} \approx 0.1008.$

1. Therefore, the probability that exactly 5 calls will be received in a minute is approximately $0.1008$ or $10.08\%$.

Mean and Variance of the Poisson Distribution

For a Poisson distribution, the mean ($\mu$) and variance ($\sigma^2$) are both equal to $\lambda$:

$$\begin{align} \mu &= \lambda,\ \sigma^2 &= \lambda.\end{align}$$

This property makes the Poisson distribution unique among probability distributions.

Section 3: The Exponential Distribution

The exponential distribution is closely related to the Poisson distribution. It models the time between consecutive events in a Poisson process, where events occur continuously and independently at a constant average rate.

Key Parameter

The exponential distribution has a single parameter $\lambda$, which represents the rate (mean occurrences per unit time).

Probability Density Function (PDF)

The PDF of the exponential distribution is given by:

$$f(x; \lambda) = \lambda e^{-\lambda x} \quad (x \geq 0)$$

Cumulative Distribution Function (CDF)

The CDF, which gives the probability that a random variable $X$ is less than or equal to some value $x$, is calculated as:

$$F(x; \lambda) = 1 - e^{-\lambda x} \quad (x \geq 0)$$

Worked Example 3

Problem: If the average time between calls at a call center is 1 minute, what is the probability that the time until the next call is more than 2 minutes?

Solution:

1. Identify $\lambda$. Since the average time is 1 minute, $\lambda = 1$.
2. Use the complementary CDF to find:

$$P(X > 2) = 1 - P(X \leq 2) = 1 - F(2; 1)$$

1. Calculate:

$$F(2; 1) = 1 - e^{-1 \cdot 2} = 1 - e^{-2}.$$

1. Using $e^{-2} \approx 0.1353$, calculate:

$$P(X > 2) = 1 - (1 - 0.1353) = 0.1353.$$

Thus, the probability that the time until the next call exceeds 2 minutes is approximately $0.1353$ or $13.53\%$.

Conclusion

In this lesson, we have explored the normal approximation to the binomial distribution, identified when to use the Poisson model for counting events, and learned about the exponential model for time intervals between events. Mastering these concepts is essential for analyzing real-world situations involving random variables and probability distributions.

Study Notes

• The normal approximation to the binomial is useful when $n$ is large and both $np$ and $n(1-p)$ are greater than 5.
• Use continuity correction when approximating binomial probabilities with the normal distribution.
• The Poisson distribution is appropriate for modeling the number of events in a fixed interval when events occur independently at an average rate.
• The mean and variance of a Poisson distribution are equal to $\lambda$.
• The exponential distribution describes the time between events in a Poisson process, with a PDF given by $f(x; \lambda) = \lambda e^{-\lambda x}$ and a CDF of $F(x; \lambda) = 1 - e^{-\lambda x}$.