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: \(p(1-p)\).
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: \(np(1-p)\)
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.
What distribution X follows?
What is the probability of the event that 23 people had 1 tire popped?
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
By probability mass function, the probability is \[\begin{equation} P(X=23)={56 \choose 23}0.28^{23}(1-0.28)^{56-23}\approx 0.0119 \tag{4.3} \end{equation}\]
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: \(\lambda\)
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:
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.
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
- X = random variable quantifying the number of successes.
X follows a Binomial distribution with n = 10, p = 0.8.
\[\begin{equation} P(X=7)={10 \choose 7}0.8^{7}(1-0.8)^{10-7}\approx 0.2013 \tag{4.4} \end{equation}\]
- 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: \(\sigma^2\)
Exercise 4.3 Which of the following statements describes the normal distribution accurately:
Bell shaped curved and mean is equal to \(\sigma\)
The random variable \(X\) is continuous and the standard deviation is \(\mu\)
The random variable \(X\) is continuous and the standard deviation is \(\sigma\), and mean is equal to \(\mu\)
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 \(\mu = 25.36, \sigma = 2.6\).
Find \(P(X\leq 26.3)\)
Find \(P(X\geq 22.3)\)
Find \(P(21.2 \leq X \leq 26.3)\)
Proof. Transfer to Standard Normal Random Variable \(\rightarrow\) Refer to a distribution Table or Calculator
\[\begin{equation} P(X\leq 26.3) = P(Z\leq \frac{26.3-25.36}{2.6}) = P(Z\leq 0.3615) \approx 0.6412 \tag{4.7} \end{equation}\]
\[\begin{equation} \begin{split} P(X\geq 22.3) &= P(Z\geq \frac{22.3-25.36}{2.6}) = P(Z\geq -1.18)\\ &=1-P(Z\leq -1.18)\approx 0.8803 \end{split} \tag{4.8} \end{equation}\]
\[\begin{equation} \begin{split} P(21.2 \leq X \leq 26.3) &= P(X\leq 26.3) - P(X\leq 21.2)\\ &=0.6412 - P(Z\leq \frac{21.2-25.36}{2.6})\\ &=0.6412 - 0.0548\approx 0.5864 \end{split} \tag{4.9} \end{equation}\]
Reference: Standard Normal Distribution Table