Probabaility, Statistics and Linear Programming

4th SEMESTER

BS-202

Department of Applied Science, BVCOE New Delhi Subject: PSLP

Lecture Notes

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Note:

·      f(x)=f(x;k)=P(X=x)

k is called the parameter of the distribution.

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Example 1 :�

Experiment: Tossing a balanced die.

• Sample space: S={1,2,3,4,5,6}
• Each sample point of S occurs with the same probability 1/6.
• Let X= the number observed when tossing a balanced die.
• The probability distribution of X is,

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Uniform Distribution

Also E(X)= (k+1)/2 (why?)

Hint: (1/k).k(k+1)/2 = (k+1)/2

Similarly, Var (X)=(k+1)(k-1)/12

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Proof:

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Example 5.3:

Find E(X) and Var(X) in Example 1.

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Binomial Distribution:

Bernoulli Trial:

·      Bernoulli trial is an experiment with only two possible outcomes.

·      The two possible outcomes are labeled:

success (s) and failure (f)

·      The probability of success is P(s)=p and the probability of failure is P(f)= q = 1−p.

·      Examples:

1.    Tossing a coin (success=H, failure=T, and p=P(H))

2.    Inspecting an item (success=defective, failure=non- defective, and p=P(defective))

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Bernoulli Process:

Bernoulli process is an experiment that must satisfy the following properties:

1.    The experiment consists of n repeated Bernoulli trials.

2.    The probability of success, P(s)=p, remains constant from trial to trial.

3.    The repeated trials are independent; that is the outcome of one trial has no effect on the outcome of any other trial.

Binomial Random Variable:

Consider the random variable :

X = The number of successes in n trials in a Bernoulli

process.

The random variable X has a binomial distribution with parameters n (number of trials) and p (probability of success), and we write:

X ~ Binomial(n,p) or X~b(x;n,p)

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The probability distribution of X is given by:

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We can write the probability distribution of X in table as follows.

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Example 2:

Suppose that 25% of the products of a manufacturing process are defective. Three items are selected at random, inspected, and classified as defective (D) or non-defective (N). Find the probability distribution of the number of defective items.

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Solution:

·      Experiment: selecting 3 items at random, inspected, and classified as (D) or (N).

·      The sample space is

S={DDD,DDN,DND,DNN,NDD,NDN,NND,NNN}

·      Let X = the number of defective items in the sample

·      We need to find the probability distribution of X.

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(1) First Solution:

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The probability distribution

of X is,

(2) Second Solution:

Bernoulli trial is the process of inspecting the item. The results are success=D or failure=N, with probability of success P(s)=25/100=1/4=0.25.

The experiments is a Bernoulli process with:

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 ·      number of trials: n=3

·      Probability of success: p=1/4=0.25

·      X ~ Binomial(n,p)=Binomial(3,1/4)

·      The probability distribution of X is given by:

The probability distribution of X is,

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Result:

The mean and the variance of the binomial distribution b(x;n,p) are:

μ = n p

σ² = n p (1 −p)

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Example:

In the previous example, find the expected value (mean) and the variance of the number of defective items.

Solution:

·      X = number of defective items

·      We need to find E(X)=μ and Var(X)=σ²

·      We found that X ~ Binomial(n,p)=Binomial(3,1/4)

·      .n=3 and p=1/4

The expected number of defective items is

E(X)=μ = n p = (3) (1/4) = 3/4 = 0.75

The variance of the number of defective items is

Var(X)=σ² = n p (1 −p) = (3) (1/4) (3/4) = 9/16 = 0.5625

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Example:

In the previous example, find the following probabilities:

(1) The probability of getting at least two defective items.

(2) The probability of getting at most two defective items.

Solution:

X ~ Binomial(3,1/4)

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�� Problems of Binomial Distribution.

• Ten coins are rolled simultaneously. Find the probability of getting at least 7 heads? Ans. 11/64
• In a lot of 200 articles, 10 are defective, find prob. of, no defective, one defective, atleast one defective in a random sample of 20 articles.
• The sum of mean and variance of binomial distribution of 5 trials is 4.8, find the distribution. Ans. p=4/5, q=1/5.

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solutions

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Hypergeometric Distribution :

·      Suppose there is a population with 2 types of elements:

1-st Type = success

2-nd Type = failure

·      N= population size

·      K= number of elements of the 1-st type

·      N −K = number of elements of the 2-nd type

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· We select a sample of n elements at random from the population.

· Let X = number of elements of 1-st type (number of successes) in the sample.

·      We need to find the probability distribution of X.

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(2) Now, suppose we select the elements of the sample at random and without replacement. When the selection is made without replacement, the random variable X has a hyper geometric distribution with parameters N, n, and K. and we write X~h(x;N,n,K).

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Note that the values of X must satisfy:

0≤x≤K and 0≤n−x≤ N−K

⇔

0≤x≤K and n−N+K≤ x≤ n

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Hypergeometric Distribution

Also E(X)= (nK)/N

Hint: (1/k).k(k+1)/2 = (k+1)/2

Similarly, Var (X)=NK(N-K)(N-n)/N^2(N-1)

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Example:

Lots of 40 components each are called acceptable if they contain no more than 3 defectives. The procedure for sampling the lot is to select 5 components at random (without replacement) and to reject the lot if a defective is found. What is the probability that exactly one defective is found in the sample if there are 3 defectives in the entire lot.

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Solution:

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·      Let X= number of defectives in the sample

·      N=40, K=3, and n=5

·      X has a hypergeometric distribution with parameters N=40, n=5, and K=3.

·      X~h(x;N,n,K)=h(x;40,5,3).

·      The probability distribution of X is given by:

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But the values of X must satisfy:

0≤x≤K and n−N+K≤ x≤ n ⇔ 0≤x≤3 and −42≤ x≤ 5

Therefore, the probability distribution of X is given by:

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Question?

• A bag contains 50 light bulbs of which 5 are defective and 45 are not. A Quality Control Inspector randomly samples 4 bulbs without replacement.  Let X = the number of defective bulbs selected. Find the probability mass function, f(x), of the discrete random variable X.

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Result.

• Note: One of the key features of the hypergeometric distribution is that it is associated with sampling without replacement. When the samples are drawn with replacement, the discrete random variable follows what is called the binomial distribution.
• Note: In cases where the sample size is relatively large compared to the population, a discrete distribution called hypergeometric may be useful.

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Question 1. A lake contains 600 fish, eighty (80) of which have been tagged by scientists. A researcher randomly catches 15 fish from the lake. Find a formula for the probability mass function of the number of fish in the researcher's sample which are tagged.

Question 2. Let the random variable X denote the number of aces in a five-card hand dealt from a standard 52-card deck. Find a formula for the probability mass function of X.

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�Question 3: Determine whether the given scenario describes a binomial setting. Justify your answer.�(a) Genetics says that the genes children receive from their parents are independent from one child to another. Each child of a particular set of parents has probability 0.25 of having type O blood. Suppose these parents have 5 children. Count the number of children with type O blood.�

(b). Shuffle a standard deck of 52 playing cards. Turn over the first 10 cards, one at a time. Record the number of aces you observe.

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�Question 4: Each child of a particular set of parents has probability 0.25 of having type O blood. Suppose these parents have 5 children. What’s the probability that exactly one of the five children has type O blood?��

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Solution 3 (a)

• Binary? “Success” 5 has type O blood. “Failure” 5 doesn’t have type O blood.
• Independent? Knowing one child’s blood type tells you nothing about another child’s because they inherit genes independently from their parents.
• Number? n = 5
• Same probability? p = 0.25

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Solution 3 (b)

• Binary? “Success” get an ace. “Failure” don’t get an ace.
• Independent? No. If the first card you turn over is an ace, then the next card is less likely to be an ace because you’re not replacing the top card in the deck. If the first card isn’t an ace, the second card is more likely to be an ace.

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• This is not a binomial setting because the independent condition is not met.

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Poisson Distribution:

• It is discrete distribution.
• The discrete r. v. X is said to have a Poisson distribution with parameter (average) λ if the probability distribution of X is given by

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where e = 2.71828.

We write:

X ~ Poisson ( λ )

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• The Poisson distribution is used to model a discrete r. v. which is a count of how many times a specified random event occurred in an interval of time or space.

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Example:

• No. of patients in a waiting room in an hour.
• No. of serious injuries in a particular factory in a month.
• No. of calls received by a telephone operator in a day.
• No. of rates in each house in a particular city.

Note:

• is the average (mean) of the distribution.

If X = The number of calls received in a month and

X ~ Poisson (λ)

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then:

(i) Y = The no. calls received in a year.

Y ~ Poisson (λ^*), where λ^*=12λ

Y ~ Poisson (12λ)

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2. W = The no. calls received in a day.

W ~ Poisson (λ^*), where λ^*=λ/30

W ~ Poisson (λ/30)

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Example

Suppose that the number of snake bites cases seen in a year has a Poisson distribution with average 6 bite cases.

1. What is the probability that in a year:

(i) The no. of snake bite cases will be 7?

(ii) The no. of snake bite cases will be less than 2?

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2- What is the probability that in 2 years there will be 10 bite cases?

3- What is the probability that in a month there will be no snake bite cases?

Solution:

(1) X = no. of snake bite cases in a year.

X ~ Poisson (6) (λ=6)

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Y = no of snake bite cases in 2 years

Y ~ Poisson(12)

3- W = no. of snake bite cases in a month.

W ~ Poisson (0.5)

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Note: Poisson distribution is for counts—if events happen at a constant rate over time, the Poisson distribution gives the probability of X number of events occurring in time T.

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�Continuous Random Variable

A continuous random variable can take any value in an interval, open or closed, so it has innumerable values

Examples: the height or weight of a chair

For such a variable X, the probability assigned to an exact value P(X = a) is always 0, though the probability for it to fall into interval [a, b], that is,

P(a ≤ X ≤ b), can be a positive number

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Probability Distributions and Probability Density Functions�

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Figure 4-1 Density function of a loading on a long, thin beam.

4.1 Probability Distributions Probability Density Functions�

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Definition

Probability Distributions and Probability Density Functions�

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Probability determined from the area under f(x).

Probability Distributions and Probability Density Functions�

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Probability Distributions and Probability Density Functions�

Example:

Cumulative Distribution Functions�

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Definition

Mean and Variance of a Continuous Random Variable�

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Definition

Mean and Variance of a Continuous Random Variable�

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Expected Value of a Function of a Continuous Random Variable

Question?

Solution.

4.3 Continuous Uniform Random Variable�

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Definition

Continuous Uniform Random Variable�

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Continuous uniform probability density function.

Continuous Uniform Random Variable�

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Mean and Variance

Continuous Uniform Random Variable�

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Example

Continuous Uniform Random Variable�

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Figure 4-9 Probability for Example 4-9.

4.4Continuous Uniform Random Variable�

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4.5 Normal Distribution�

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Definition

Normal Distribution�

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Figure : Normal probability density functions for selected values of the parameters μ and σ².

Normal Distribution�

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Some useful results concerning the normal distribution

4.6 Normal Distribution�

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Definition : Standard Normal

Normal Distribution�

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Example:

Figure 4-13 Standard normal probability density function.

Normal Distribution�

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Standardizing

Normal Distribution�

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Example:

Normal Distribution�

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Figure 4-15 Standardizing a normal random variable.

Normal Distribution�

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To Calculate Probability

Normal Distribution�

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Example:

Normal Distribution�

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Example (When prob. is given)

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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• Under certain conditions, the normal distribution can be used to approximate the binomial distribution and the Poisson distribution.

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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Figure 4-19 Normal approximation to the binomial.

Normal Approximation to the Binomial and Poisson Distributions�

Example (Normal Distri. can be used to compute the longer calculations)

Normal Approximation to the Binomial and Poisson Distributions�

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Normal Approximation to the Binomial Distribution

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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Example 4-18

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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Figure 4-21 Conditions for approximating hypergeometric and binomial probabilities.

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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Normal Approximation to the Poisson Distribution

4-7 Normal Approximation to the Binomial and Poisson Distributions�

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Example 4-20

4-8 Exponential Distribution�

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Definition

Poisson or not?��Which of the following is most likely to be well modelled by a Poisson distribution?��

1. Number of trains arriving at Falmer every hour
2. Number of lottery winners each year that live in Brighton
3. Number of days between solar eclipses
4. Number of days until a component fails

Are they Poisson? Answers:

1. Number of trains arriving at Falmer every hour��NO, (supposed to) arrive regularly on a timetable not at random

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• Number of lottery winners each year that live in Brighton��Yes, is number of random events in fixed interval�
• Number of days between solar eclipses��NO, solar eclipses are not random events and this is a time between random events, not the number in some fixed interval�
• Number of days until a component fails�NO, random events, but this is time until a random event, not the number of random events

What is the probability distribution for the time to the first event?

Poisson - Discrete distribution: P(number of events)

Exponential - Continuous distribution: P(time till first event)

Time between random events / time till first random event ?

Occurrence

1) Time until the failure of a part.

2) Separation between randomly happening events

Relation to Poisson distribution�

Example

E.g. Probability the time till the next death is less than one year?

Exponential distribution��A certain type of component can be purchased new or used. 50% of all new components last more than five years, but only 30% of used components last more than five years. Is it possible that the lifetimes of new components are exponentially distributed?

1. YES
2. NO

Question from Derek Bruff

4-8 Exponential Distribution�

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Mean and Variance

4-8 Exponential Distribution�

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Example 4-21

4-8 Exponential Distribution�

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Figure 4-23 Probability for the exponential distribution in Example 4-21.

4-8 Exponential Distribution�

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Example 4-21 (continued)

4-8 Exponential Distribution�

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Example 4-21 (continued)

4-8 Exponential Distribution�

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Example 4-21 (continued)

4-8 Exponential Distribution�

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Our starting point for observing the system does not matter.

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• An even more interesting property of an exponential random variable is the lack of memory property.

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In Example 4-21, suppose that there are no log-ons from 12:00 to 12:15; the probability that there are no log-ons from 12:15 to 12:21 is still 0.082. Because we have already been waiting for 15 minutes, we feel that we are “due.” That is, the probability of a log-on in the next 6 minutes should be greater than 0.082. However, for an exponential distribution this is not true.

4-8 Exponential Distribution�

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Example 4-22

4-8 Exponential Distribution�

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Example 4-22 (continued)

4-8 Exponential Distribution�

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Example 4-22 (continued)

Example: Reliability

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The time till failure of an electronic component has an Exponential distribution and it is known that 10% of components have failed by 1000 hours.

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(a) What is the probability that a component is still working after 5000 hours?

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(b) Find the mean and standard deviation of the time till failure.

(b) Find the mean and standard deviation of the time till failure.

Answer:

4-8 Exponential Distribution�

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Lack of Memory Property

4-8 Exponential Distribution�

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Figure 4-24 Lack of memory property of an Exponential distribution.

4-9 Erlang and Gamma Distributions�

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Erlang Distribution

The random variable X that equals the interval length until r counts occur in a Poisson process with mean λ > 0 has and Erlang random variable with parameters λ and r. The probability density function of X is

for x > 0 and r =1, 2, 3, ….

4-9 Erlang and Gamma Distributions�

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Gamma Distribution

4-9 Erlang and Gamma Distributions�

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Gamma Distribution

4-9 Erlang and Gamma Distributions�

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Gamma Distribution

Figure 4-25 Gamma probability density functions for selected values of r and λ.

4-9 Erlang and Gamma Distributions�

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Gamma Distribution

4-10 Weibull Distribution�

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Definition

4-10 Weibull Distribution�

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Figure 4-26 Weibull probability density functions for selected values of α and β.

4-10 Weibull Distribution�

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4-10 Weibull Distribution�

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Example 4-25

4-11 Lognormal Distribution�

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4-11 Lognormal Distribution�

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Figure 4-27 Lognormal probability density functions with θ = 0 for selected values of ω².

4-11 Lognormal Distribution�

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Example 4-26

4-11 Lognormal Distribution�

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Example 4-26 (continued)

4-11 Lognormal Distribution�

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Example 4-26 (continued)

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���������� Normal Probability � Distribution s

Properties of Normal Distributions

A continuous random variable has an infinite number of possible values that can be represented by an interval on the number line.

The probability distribution of a continuous random variable is called a continuous probability distribution.

Properties of Normal Distributions

The most important probability distribution in statistics is the normal distribution.

A normal distribution is a continuous probability distribution for a random variable, x. The graph of a normal distribution is called the normal curve.

Properties of Normal Distributions

Properties of a Normal Distribution

1. The mean, median, and mode are equal.
2. The normal curve is bell-shaped and symmetric about the mean.
3. The total area under the curve is equal to one.
4. The normal curve approaches, but never touches the x-axis as it extends farther and farther away from the mean.
5. Between μ − σ and μ + σ (in the center of the curve), the graph curves downward. The graph curves upward to the left of μ − σ and to the right of μ + σ. The points at which the curve changes from curving upward to curving downward are called the inflection points.

Properties of Normal Distributions

μ − 3σ

μ + σ

μ − 2σ

μ − σ

μ

μ + 2σ

μ + 3σ

x

Means and Standard Deviations

A normal distribution can have any mean and any positive standard deviation.

Mean: μ = 3.5

Standard deviation: σ ≈ 1.3

Mean: μ = 6,

μ -σ =7.9

Standard deviation: σ ≈ 1.9

The mean gives the location of the line of symmetry.

The standard deviation describes the spread of the data.

Means and Standard Deviations

Example:

1. Which curve has the greater mean?
2. Which curve has the greater standard deviation?

The line of symmetry of curve A occurs at x = 5. The line of symmetry of curve B occurs at x = 9. Curve B has the greater mean.

Curve B is more spread out than curve A, so curve B has the greater standard deviation.

Interpreting Graphs

Example:

The heights of fully grown magnolia bushes are normally distributed. The curve represents the distribution. What is the mean height of a fully grown magnolia bush? Estimate the standard deviation.

The heights of the magnolia bushes are normally distributed with a mean height of about 8 feet and a standard deviation of about 0.7 feet.

The inflection points are one standard deviation away from the mean.

σ ≈ 0.7

The Standard Normal Distribution

The standard normal distribution is a normal distribution with a mean of 0 and a standard deviation of 1.

Any value can be transformed into a z-score by using the formula

The horizontal scale corresponds to z-scores.

The Standard Normal Distribution

If each data value of a normally distributed random variable x is transformed into a z-score, the result will be the standard normal distribution.

After the formula is used to transform an x-value into a z-score, the Standard Normal Table in Appendix B is used to find the cumulative area under the curve.

The area that falls in the interval under the nonstandard normal curve (the x-values) is the same as the area under the standard normal curve (within the corresponding z-boundaries).

The Standard Normal Table

Properties of the Standard Normal Distribution

1. The cumulative area is close to 0 for z-scores close to z = −3.49.
2. The cumulative area increases as the z-scores increase.
3. The cumulative area for z = 0 is 0.5000.
4. The cumulative area is close to 1 for z-scores close to z = 3.49

The Standard Normal Table

Example:

Find the cumulative area that corresponds to a z-score of 2.71.

Find the area by finding 2.7 in the left hand column, and then moving across the row to the column under 0.01.

The area to the left of z = 2.71 is 0.9966.

Appendix B: Standard Normal Table

The Standard Normal Table

Example:

Find the cumulative area that corresponds to a z-score of −0.25.

Find the area by finding −0.2 in the left hand column, and then moving across the row to the column under 0.05.

The area to the left of z = −0.25 is 0.4013

Appendix B: Standard Normal Table

Guidelines for Finding Areas

Finding Areas Under the Standard Normal Curve

1. Sketch the standard normal curve and shade the appropriate area under the curve.
2. Find the area by following the directions for each case shown.
1. To find the area to the left of z, find the area that corresponds to z in the Standard Normal Table.

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Guidelines for Finding Areas

Finding Areas Under the Standard Normal Curve

• To find the area to the right of z, use the Standard Normal Table to find the area that corresponds to z. Then subtract the area from 1.

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Guidelines for Finding Areas

Finding Areas Under the Standard Normal Curve

• To find the area between two z-scores, find the area corresponding to each z-score in the Standard Normal Table. Then subtract the smaller area from the larger area.

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Guidelines for Finding Areas

Example:

Find the area under the standard normal curve to the left of z = −2.33.

From the Standard Normal Table, the area is equal to 0.0099. (0.5-0.4901)

Always draw the curve!

Guidelines for Finding Areas

Example:

Find the area under the standard normal curve to the right of z = 0.94.

From the Standard Normal Table, the area is equal to 0.1736.

Always draw the curve!

Guidelines for Finding Areas

Example:

Find the area under the standard normal curve between z = −1.98 and z = 1.07.

From the Standard Normal Table, the area is equal to 0.8338.

Always draw the curve!

§ 5.2

Normal Distributions: Finding Probabilities

Probability and Normal Distributions

If a random variable, x, is normally distributed, you can find the probability that x will fall in a given interval by calculating the area under the normal curve for that interval.

Probability and Normal Distributions

Same area

P(x < 15) = P(z < 1) = Shaded area under the curve

= 0.8413

Probability and Normal Distributions

Example:

The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score less than 90.

P(x < 90) = P(z < 1.5) = 0.9332

The probability that a student receives a test score less than 90 is 0.9332.

1.5

Probability and Normal Distributions

Example:

The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score greater than than 85.

P(x > 85) = P(z > 0.88) = 1 − P(z < 0.88) = 1 − 0.8106 = 0.1894

The probability that a student receives a test score greater than 85 is 0.1894.

0.88

Probability and Normal Distributions

Example:

The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that a student receives a test score between 60 and 80.

P(60 < x < 80) = P(−2.25 < z < 0.25) = P(z < 0.25) − P(z < −2.25)

The probability that a student receives a test score between 60 and 80 is 0.5865.

0.25

−2.25

= 0.5987 − 0.0122 = 0.5865

§ 5.3

Normal Distributions: Finding Values

Finding z-Scores

Example:

Find the z-score that corresponds to a cumulative area of 0.9973.

Find the z-score by locating 0.9973 in the body of the Standard Normal Table. The values at the beginning of the corresponding row and at the top of the column give the z-score.

The z-score is 2.78.

Appendix B: Standard Normal Table

2.7

.08

Finding z-Scores

Example:

Find the z-score that corresponds to a cumulative area of 0.4170.

Find the z-score by locating 0.4170 in the body of the Standard Normal Table. Use the value closest to 0.4170.

Appendix B: Standard Normal Table

The z-score is −0.21.

−0.2

.01

Finding a z-Score Given a Percentile

Example:

Find the z-score that corresponds to P₇₅.

The z-score that corresponds to P₇₅ is the same z-score that corresponds to an area of 0.75.

The z-score is 0.67.

0.67

Transforming a z-Score to an x-Score

To transform a standard z-score to a data value, x, in a given population, use the formula

Example:

The monthly electric bills in a city are normally distributed with a mean of $120 and a standard deviation of $16. Find the x-value corresponding to a z-score of 1.60.

We can conclude that an electric bill of $145.60 is 1.6 standard deviations above the mean.

Finding a Specific Data Value

Example:

The weights of bags of chips for a vending machine are normally distributed with a mean of 1.25 ounces and a standard deviation of 0.1 ounce. Bags that have weights in the lower 8% are too light and will not work in the machine. What is the least a bag of chips can weigh and still work in the machine?

The least a bag can weigh and still work in the machine is 1.11 ounces.

P(z < ?) = 0.08

P(z < −1.41) = 0.08

1.11

§ 5.4

Sampling Distributions and the Central Limit Theorem

Sampling Distributions

Sample

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A sampling distribution is the probability distribution of a sample statistic that is formed when samples of size n are repeatedly taken from a population.

Sample

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Sample

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Sample

​

Sample

​

Sample

​

Sample

​

Sample

​

Sample

​

Sample

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Sampling Distributions

If the sample statistic is the sample mean, then the distribution is the sampling distribution of sample means.

Properties of Sampling Distributions

Properties of Sampling Distributions of Sample Means

1. The mean of the sample means, is equal to the population mean.

​

• The standard deviation of the sample means, is equal to the population standard deviation, divided by the square root of n.

​

The standard deviation of the sampling distribution of the sample means is called the standard error of the mean.

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Sampling Distribution of Sample Means

Example:

The population values {5, 10, 15, 20} are written on slips of paper and put in a hat. Two slips are randomly selected, with replacement.

• Find the mean, standard deviation, and variance of the population.

Continued.

Population

5

10

15

20

Sampling Distribution of Sample Means

Example continued:

The population values {5, 10, 15, 20} are written on slips of paper and put in a hat. Two slips are randomly selected, with replacement.

• Graph the probability histogram for the population values.

Continued.

This uniform distribution shows that all values have the same probability of being selected.

Sampling Distribution of Sample Means

Example continued:

The population values {5, 10, 15, 20} are written on slips of paper and put in a hat. Two slips are randomly selected, with replacement.

• List all the possible samples of size n = 2 and calculate the mean of each.

Continued.

These means form the sampling distribution of the sample means.

Sampling Distribution of Sample Means

Example continued:

The population values {5, 10, 15, 20} are written on slips of paper and put in a hat. Two slips are randomly selected, with replacement.

• Create the probability distribution of the sample means.

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Sampling Distribution of Sample Means

Example continued:

The population values {5, 10, 15, 20} are written on slips of paper and put in a hat. Two slips are randomly selected, with replacement.

• Graph the probability histogram for the sampling distribution.

The shape of the graph is symmetric and bell shaped. It approximates a normal distribution.

The Central Limit Theorem

the sample means will have a normal distribution.

If a sample of size n ≥ 30 is taken from a population with any type of distribution that has a mean = μ and standard deviation = σ,

The Central Limit Theorem

If the population itself is normally distributed, with mean = μ and standard deviation = σ,

the sample means will have a normal distribution for any sample size n.

The Central Limit Theorem

In either case, the sampling distribution of sample means has a mean equal to the population mean.

Mean of the sample means

Standard deviation of the sample means

The sampling distribution of sample means has a standard deviation equal to the population standard deviation divided by the square root of n.

The Mean and Standard Error

Example:

The heights of fully grown magnolia bushes have a mean height of 8 feet and a standard deviation of 0.7 feet. 38 bushes are randomly selected from the population, and the mean of each sample is determined. Find the mean and standard error of the mean of the sampling distribution.

Mean

Standard deviation (standard error)

Continued.

Interpreting the Central Limit Theorem

Example continued:

The heights of fully grown magnolia bushes have a mean height of 8 feet and a standard deviation of 0.7 feet. 38 bushes are randomly selected from the population, and the mean of each sample is determined.

From the Central Limit Theorem, because the sample size is greater than 30, the sampling distribution can be approximated by the normal distribution.

The mean of the sampling distribution is 8 feet ,and the standard error of the sampling distribution is 0.11 feet.

Finding Probabilities

Example:

The heights of fully grown magnolia bushes have a mean height of 8 feet and a standard deviation of 0.7 feet. 38 bushes are randomly selected from the population, and the mean of each sample is determined.

Find the probability that the mean height of the 38 bushes is less than 7.8 feet.

The mean of the sampling distribution is 8 feet, and the standard error of the sampling distribution is 0.11 feet.

Continued.

Finding Probabilities

−1.82

Example continued:

Find the probability that the mean height of the 38 bushes is less than 7.8 feet.

The probability that the mean height of the 38 bushes is less than 7.8 feet is 0.0344.

= 0.0344

Probability and Normal Distributions

Example:

The average on a statistics test was 78 with a standard deviation of 8. If the test scores are normally distributed, find the probability that the mean score of 25 randomly selected students is between 75 and 79.

0.63

−1.88

Continued.

Probability and Normal Distributions

Example continued:

Approximately 70.56% of the 25 students will have a mean score between 75 and 79.

= 0.7357 − 0.0301 = 0.7056

0.63

−1.88

Probabilities of x and x

Example:

The population mean salary for auto mechanics is μ = $34,000 with a standard deviation of σ = $2,500. Find the probability that the mean salary for a randomly selected sample of 50 mechanics is greater than $35,000.

2.83

= P (z > 2.83)

= 1 − P (z < 2.83)

= 1 − 0.9977

= 0.0023

The probability that the mean salary for a randomly selected sample of 50 mechanics is greater than $35,000 is 0.0023.

Probabilities of x and x

Example:

The population mean salary for auto mechanics is μ = $34,000 with a standard deviation of σ = $2,500. Find the probability that the salary for one randomly selected mechanic is greater than $35,000.

0.4

= P (z > 0.4)

= 1 − P (z < 0.4)

= 1 − 0.6554

= 0.3446

The probability that the salary for one mechanic is greater than $35,000 is 0.3446.

(Notice that the Central Limit Theorem does not apply.)

Probabilities of x and x

Example:

The probability that the salary for one randomly selected mechanic is greater than $35,000 is 0.3446. In a group of 50 mechanics, approximately how many would have a salary greater than $35,000?

P(x > 35000) = 0.3446

This also means that 34.46% of mechanics have a salary greater than $35,000.

You would expect about 17 mechanics out of the group of 50 to have a salary greater than $35,000.

34.46% of 50 = 0.3446 × 50 = 17.23

§ 5.5

Normal Approximations to Binomial Distributions

Normal Approximation

The normal distribution is used to approximate the binomial distribution when it would be impractical to use the binomial distribution to find a probability.

Normal Approximation to a Binomial Distribution

If np ≥ 5 and nq ≥ 5, then the binomial random variable x is approximately normally distributed with mean

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and standard deviation

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Normal Approximation

Example:

Decided whether the normal distribution to approximate x may be used in the following examples.

• Thirty-six percent of people in the United States own a dog. You randomly select 25 people in the United States and ask them if they own a dog.

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• Fourteen percent of people in the United States own a cat. You randomly select 20 people in the United States and ask them if they own a cat.

Because np and nq are greater than 5, the normal distribution may be used.

Because np is not greater than 5, the normal distribution may NOT be used.

Correction for Continuity

The binomial distribution is discrete and can be represented by a probability histogram.

This is called the correction for continuity.

When using the continuous normal distribution to approximate a binomial distribution, move 0.5 unit to the left and right of the midpoint to include all possible x-values in the interval.

To calculate exact binomial probabilities, the binomial formula is used for each value of x and the results are added.

Correction for Continuity

Example:

Use a correction for continuity to convert the binomial intervals to a normal distribution interval.

• The probability of getting between 125 and 145 successes, inclusive.

The discrete midpoint values are 125, 126, …, 145.

The continuous interval is 124.5 < x < 145.5.

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• The probability of getting exactly 100 successes.

The discrete midpoint value is 100.

The continuous interval is 99.5 < x < 100.5.

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• The probability of getting at least 67 successes.

The discrete midpoint values are 67, 68, ….

The continuous interval is x > 66.5.

Guidelines

Using the Normal Distribution to Approximate Binomial Probabilities

In Words In Symbols

1. Verify that the binomial distribution applies.
2. Determine if you can use the normal distribution to approximate x, the binomial variable.
3. Find the mean μ and standard deviationσ for the distribution.
4. Apply the appropriate continuity correction. Shade the corresponding area under the normal curve.
5. Find the corresponding z-value(s).
6. Find the probability.

Specify n, p, and q.

Is np ≥ 5?

Is nq ≥ 5?

Add or subtract 0.5 from endpoints.

Use the Standard Normal Table.

Approximating a Binomial Probability

Example:

Thirty-one percent of the seniors in a certain high school plan to attend college. If 50 students are randomly selected, find the probability that less than 14 students plan to attend college.

np = (50)(0.31) = 15.5

nq = (50)(0.69) = 34.5

P(x < 13.5)

= P(z < −0.61)

= 0.2709

The probability that less than 14 plan to attend college is 0.2079.

Approximating a Binomial Probability

Example:

A survey reports that forty-eight percent of US citizens own computers. 45 citizens are randomly selected and asked whether he or she owns a computer. What is the probability that exactly 10 say yes?

np = (45)(0.48) = 12

nq = (45)(0.52) = 23.4

P(9.5 < x < 10.5)

= 0.0997

The probability that exactly 10 US citizens own a computer is 0.0997.

= P(−0.75 < z − 0.45)