Analysing Data Data Presentation, Numerical Summaries: Measures for Location

Analysing Data Normal Distribution and its Applications

OtherStatisticalDistributions

(1) Bernoulli Distribution

This distribution best describes all situations where a "trial" is made resulting in either "success" or "failure," such as when tossing a coin, or when modeling the success or failure of a Success of medical treatment, or Interviewed person is female, or Student passes exam. Success happens with probabilityp, while failure happens with probability (1-p). Arandom variablethat takes value (1) in case of success and(0)in case of failure is called a Bernoulli random variable (alternatively, it is said to have a Bernoulli distribution).

The Bernoulli distribution is defined as:

This is the probability mass function of this distribution. Also this written as

wherePis the probability that a particular event (e.g., success) will occur.

Mean and Variance and Standard Deviation of Bernoulli Distribution:

Mean (Expected value) / Variance / Standard Deviation
p / p(1-p) /

Example (59):

In a study of 600 students aged between 15 and 19 in a country, 450 of the students were found to have passed the exam in driving.

(a) If one of students is chosen at random, calculate the probability that he has passed the exam.

(b) Define a suitable random variable X to indicate whether a randomly chosen student has passed the exam, and write down its probability mass function.

Solution:

(a) The probability that student randomly selected has passed the exam is

(b) Let the random variable X take the value 1 if a student has passed the exam and 0 otherwise. The random variable X is Bernoulli ~(0.75), so

 It is important to specify the range of the random variable.

The mean = = p= 0.75

The variance=p(1-p)= 0.75 (1-0.75) = 0.75(0.25)= 0.1875

Example (60):

In 1998, 95% of British households had a telephone. For a randomly chosen British household, what is the probability distribution of the random variable X which takes the value 1 if the household has a telephone and 0 otherwise? Write down its probability mass function, and find the expected valueand variance.

Solution:

This is a Bernoulli trial with P(X=0)= 0.05, P(X=1)= 0.95. That is Bernoulli ~(0.095).

The probability mass function of X is

Theexpected valueand variance of a Bernoulli random variable are:

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(2)BinomialDistribution:

A binomial experiment is an experiment, which satisfies these four conditions:

A fixed number of trials
Each trial is independent of the others
There are only two outcomes
The probability of each outcome remains constant from trial to trial.

These can be summarized as an experiment with a fixed number of independent trials, each of which can only have two possible outcomes.

The fact that each trial is independent actually means that the probabilities remain constant.

Examples of binomial experiments

Tossing a coin 20 times to see how many tails occur.
Asking 200 people if they watch ABC news.
Rolling a die to see if a 5 appears.

Examples, whichare not binomial experiments

Rolling a die until a 6 appears (not a fixed number of trials)
Asking 20 people how old they are (not two outcomes)
Drawing 5 cards from a deck for a poker hand (done without replacement, so not independent)

Binomial Probability Function (Model):

The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:

Where

This written as

Example (61):

If find the probability P(X=4)

Solution:

Here n= 6, p= 0.4, 1-p= 0.6, X= 4

Example (62):

What is the probability of rolling exactly two sixes in six rolls of a die?

Solution:

There are five things you need to do to work a binomial story problem:

Define Success first. Success must be for a single trial. Success = "Rolling a 6 on a single die"
Define the probability of success (p): p = 1/6
Find the probability of failure: 1-p = 5/6
Define the number of trials: n = 6
Define the number of successes out of those trials: x = 2

Anytime a six appears, it is a success and anytime something else appears, it is a failure. The ways you can get exactly 2 successes in 6 trials are given below. The probability of each is written to the right of the way it could occur. Because the trials are independent, the probability of the event (all six dice) is the product of each probability of each outcome (die).

Example (63):

A coin is tossed 10 times. What is the probability that:

(i) Exactly 6 heads will occur. (ii) More than 7 heads will occur.

Solution:

Success = "A head is flipped on a single coin"

p = 0.5 3. 1-p = 0.5 4. n = 10 5. x = 6

(i)

Mean, Variance, and Standard Deviation of binomial distribution:

The mean, variance, and standard deviation of a binomial distribution are extremely easy to find.

Mean (Expected value) / Variance / Standard Deviation

Example (64):

Find the mean, variance, and standard deviation for the number of sixes that appear when rolling 30 dice.

Solution:

Success = "a six is rolled on a single die". p = 1/6, 1-p = 5/6.

The mean is

The variance is

The standard deviation is the square root of the variance = 2.041(approx).

(3)TheGeometricDistribution:

Suppose that in a sequence of independent Bernoulli trials, the probability of success is constant from trial to trial and equal to p, where 0<p<1. Then the number of trials up to and including the first success is a random variable N with probability mass function given by

Where q=1-p.

The random variable N is said to have a geometric distribution with parameter p; this is written N ~G(p).

Example (65):

If find the probability P(N=10)

Solution:

p=0.5, q=1-p=1-0.5= 0.5, n=10

Then:

Example (66):

Suppose that the proportion of detective silicon chips leaving a factory's assembly line is 0.012. A quality inspector collects a random sample: she examines chips until shefound a detective one. What is the probability that she examines exactly ten chips?

Solution:

p=0.012, q=1-p=1-0.012= 0.988, n=10

Then:

Mean, Variance, and Standard Deviation of Geometric distribution:

The mean, variance, and standard deviation of a Geometricdistribution are extremely easy to find.

Mean (Expected value) / Variance / Standard Deviation

Example (67):

Find the mean, variance, and standard deviation for .

Solution:

p=0.6, q=1-p=1-0.6= 0.4

. The variance is

The standard deviation is the square root of the variance= 1.05(approx).

Additional Examples:

Example (68):

(1) In a survey, 60% of the students support that M248 Course provides new techniques for them. If 10 students are chosen at random, then let X = number of students who support that M248 Course provides new techniques for them.

(a) What is the distribution of X?

(b) What are the possible values of X (range)?

(d) What is the variance of X?

Solution:

(a) X has a Binomial distribution B (10, 0.6).

(b) Range of X: 0, 1, 2, …, 10

(c)

(d)

Example (69):

Public Opinion reported that 1 out of 10 students are afraid of being alone in a house at night. Let Y be a random variable representing the number of selections until the first person afraid of being alone is selected.

(a) What is the distribution of Y? Determine its range and parameters.

(b) Find the mean and variance of Y.

Solution:

(a) Y has a Geometric distribution:

Y= 1, 2, 3, …

(b)

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