Chapter 4 Probability Distributions (Section on Feb 10th)

Bernoulli Distribution

Definition 4.1 (Bernoulli Distribution) The Bernoulli distribution, is the discrete probability distribution of a random variable which takes the value 1 with probability $p$ and the value 0 with probability $q=1-p$.

• Probability Mass Function: $$\begin{equation} P(X=x)=p^x(1-p)^{1-x},\quad x=0,1 \tag{4.1} \end{equation}$$

• Mean: $p$

• Variance: $p(1-p)$.

Bernoulli distribution is the most basic probability distribution.

When is it appropriate to use the Bernoulli Distribution?

• random variable has two mutually exclusive events

• there is a known probability associated with a success

Binominal Distribution

Definition 4.2 (Binomial Distribution) The binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q = 1 − p).

• Probability Mass Function: $$\begin{equation} P(X=x)={n \choose x}p^x(1-p)^{n-x},\quad x=0,1,\cdots,n \tag{4.2} \end{equation}$$

• Mean: $np$

• Variance: $np(1-p)$

Binominal distribution can be viewed as sum of independent Bernoulli random variables.

When is it appropriate to use the Binomial Distribution?

• Suppose a fixed number of trials (e.g., Flip a coin a n times)

• The trials must be independent (e.g., flips do not affect each other)

• Each trial must have all outcomes classified into two mutually exclusive categories (e.g., Tail or Head, disjoint)

• The probabilities must remain constant for each trial (e.g., $P(head)= \frac{1}{2}$, and this does not change)

Exercise 4.1 An experiment was conducted in which 56 people were driving on highway 17 and probability of a tire popping was 0.28. 23 people had 1 tire popped. Define the random variable X = number of people having 1 tire popped.

1. What distribution X follows?

2. What is the probability of the event that 23 people had 1 tire popped?

3. Find the mean, variance, standard deviation of the random variable X.

Proof. (a) X follows a Binomial distribution with parameters n = 56 and p = 0.28

2. By probability mass function, the probability is $$\begin{equation} P(X=23)={56 \choose 23}0.28^{23}(1-0.28)^{56-23}\approx 0.0119 \tag{4.3} \end{equation}$$

3. Mean is $np=56\times 0.28=15.68$.

Variance is $np(1-p)=56\times 0.28\times (1-0.28)=11.2896$

Standard deviation is $\sqrt{np(1-p)}=\sqrt{56\times 0.28\times (1-0.28)}=3.36$

Poisson Distribution

Definition 4.3 (Poisson Distribution) The Poisson distribution, is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

• Probability Mass Function: $$\begin{equation} P(X=x)=\frac{\lambda^xe^{-\lambda}}{x!},\quad x=0,1,\cdots, \tag{4.3} \end{equation}$$

• Mean: $\lambda$

• Variance: $\lambda$

Poisson distribution is often used in cases like counting number of some event occuring.

When is it appropriate to use the Poisson Distribution?

• Discrete probability distribution.

• The random variable is the number of occurrences (counts) of an event in an interval

• The interval can be: time, distance, area, volume, or some similar unit.

Exercise 4.2 Use either a Poisson or a Binomial Distribution for the following:

1. In the old days, there was a probability of 0.8 of success in any attempt to make a telephone call. Calculate the probability of having 7 successes in 10 attempts.

2. If electricity power failures occur according to a poll with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Proof. Define Random Variable $\rightarrow$ What distribution dose it follow? $\rightarrow$ Calculate

1. X = random variable quantifying the number of successes.

X follows a Binomial distribution with n = 10, p = 0.8.

$$\begin{equation} P(X=7)={10 \choose 7}0.8^{7}(1-0.8)^{10-7}\approx 0.2013 \tag{4.4} \end{equation}$$

2. X = random variable quantifying the no. of failures in a week.

X follows a Poisson distribution with parameter $\lambda=\frac{3}{20}$.

$$\begin{equation} \begin{split} P(X\leq 1)&=P(X=0)+P(X=1)\\ &=\frac{(\frac{3}{20})^0e^{-3/20}}{0!}+\frac{(\frac{3}{20})^1e^{-3/20}}{1!}\\ &=e^{-3/10}(1+\frac{3}{20})\approx 0.9898 \end{split} \tag{4.5} \end{equation}$$

Normal Distribution

Definition 4.4 (Normal Distribution) The normal distribution, is a continuous probability distribution that is often used in the natural and social sciences to represent real-valued random variables whose distributions are not known.

• Probability Distribution Function (Not Required): $$\begin{equation} f(X=x)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}} \tag{4.6} \end{equation}$$

• Mean: $\mu$

• Variance: $\sigma^2$

You only need to knwo normal distribution is determined by mean and variance parmater.

Exercise 4.3 Which of the following statements describes the normal distribution accurately:

1. Bell shaped curved and mean is equal to $\sigma$

2. The random variable $X$ is continuous and the standard deviation is $\mu$

3. The random variable $X$ is continuous and the standard deviation is $\sigma$, and mean is equal to $\mu$

4. The shape of the normal distribution is uniform.

The correct answer is (c).

Exercise 4.4 Let X be a random variable following N($\mu$, $\sigma^2$). Let $\mu = 25.36, \sigma = 2.6$.

1. Find $P(X\leq 26.3)$

2. Find $P(X\geq 22.3)$

3. Find $P(21.2 \leq X \leq 26.3)$

Proof. Transfer to Standard Normal Random Variable $\rightarrow$ Refer to a distribution Table or Calculator

1. $$\begin{equation} P(X\leq 26.3) = P(Z\leq \frac{26.3-25.36}{2.6}) = P(Z\leq 0.3615) \approx 0.6412 \tag{4.7} \end{equation}$$

2. $$\begin{equation} \begin{split} P(X\geq 22.3) &= P(Z\geq \frac{22.3-25.36}{2.6}) = P(Z\geq -1.18)\\ &=1-P(Z\leq -1.18)\approx 0.8803 \end{split} \tag{4.8} \end{equation}$$

3. $$\begin{equation} \begin{split} P(21.2 \leq X \leq 26.3) &= P(X\leq 26.3) - P(X\leq 21.2)\\ &=0.6412 - P(Z\leq \frac{21.2-25.36}{2.6})\\ &=0.6412 - 0.0548\approx 0.5864 \end{split} \tag{4.9} \end{equation}$$

Reference: Standard Normal Distribution Table