Introduction to Business Statistics QM 120 Chapter 6 Spring 2008 - - PDF document

QM-120, M. Zainal 1

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION SYSTEMS

Introduction to Business Statistics QM 120 Chapter 6

Spring 2008

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When a RV x is discrete, we can assign a positive probability

to each value that x can take and get the probability distribution for x. Th f ll b biliti f ll l f i 1

Chapter 6: Continuous Probability Distribution

The sum of all probabilities of all values of x is 1. Not all experiments result in RVs that are discrete. Continuous RV, such as heights, weights, lifetime of a

particular product, experimental laboratory error…etc. can assume infinitely many values corresponding to points on line interval

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Suppose we have a set of measurements on a continuous

RV, and we want to create a relative frequency histogram to describe their distribution. F ll b f t ll

Chapter 6: Continuous Probability Distribution

For a small number of measurements, we can use small

number of classes

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For a larger number of measurements, we must use a larger

number of classes and reduce the class width

Chapter 6: Continuous Probability Distribution

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As the number of measurements become very large, the

class width become very narrow, the relative frequency histogram appears more and more like a smooth curve.

Chapter 6: Continuous Probability Distribution

This smooth curve describes the probability distribution of

a continuous random variable.

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A continuous RV can take on any of an infinite number of

values on the real line

Chapter 6: Continuous Probability Distribution

The depth or density of the probability, which varies with x,

maybe described by a mathematical formula f(x), called the probability density function for the RV x.

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The probability distribution f(x); P(a< x <b) is equal to the

shaded area under the curve

Chapter 6: Continuous Probability Distribution

Several

important properties

• f

continuous probability distribution parallel their discrete part.

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Just as the sum of discrete probabilities is equal to 1, i.e.

ΣP(x) = 1, and the probability that x falls into a certain interval can be found by summing the probabilities in that interval, continuous probability distributions have the

Chapter 6: Continuous Probability Distribution

interval, continuous probability distributions have the following two characteristics:

The area under a continuous probability distribution is

equal to 1.

The probability that x will fall into a particular interval‐ say,

from a to b – is equal to the area under the curve between the two points a and b the two points a and b

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There is a very important difference between the continuous

case and the discrete one which is:

P(x = a) = 0

Chapter 6: Continuous Probability Distribution

P(a < x < b) = P(a ≤ x ≤ b) P( x < a) = P( x ≤ a)

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Among

many continuous probability distributions, the normal distribution is the most important and widely used

• ne.

A l b f l ld h ith tl

Chapter 6: The Normal Distribution

A large number of real world phenomena are either exactly or

approximately normally distributed.

The normal distribution or Gaussian distribution is given by

a bell‐shaped curve.

A continuous RV x that has a normal distribution is said to be

a normal RV with a mean μ and a standard deviation σ or simply x~N(μ,σ).

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Chapter 6: The Normal Distribution

• Note that not all the bell‐shaped curves represent a normal

distribution

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The normal probability distribution, when plotted, gives a

bell‐shaped curve such that

The total area under the curve is 1.0.

Chapter 6: The Normal Distribution

The curve is symmetric around the mean. The two tails of the curve extended indefinitely. The mean μ and the standard deviation σ are the parameters

• f the normal distribution. Given the values of these two

parameters, we can find the area under the normal curve for any interval.

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There is not just one normal distribution curve but rather a

family of normal distribution.

Chapter 6: The Normal Distribution

Each different set of values of μ and σ gives different

normal curve.

The value of μ determines the center of the curve on the Horizontal axis and the value

• f σ gives the spread of the

normal curve

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Like

all

• ther

distributions, the normal probability distribution can be expressed by a mathematical function

Chapter 6: The Normal Distribution

in which the probability that x falls between a and b is the

integral of the above function from a to b, i.e.

2

⎡ ⎤

2

1 2

1 ( ) 2

x

f x e

μ σ

σ π

− ⎡ ⎤ − ⎢ ⎥ ⎣ ⎦

=

2

1 2

1 ( ) 2

x b a

P a x b e dx

μ σ

σ π

− ⎡ ⎤ − ⎢ ⎥ ⎣ ⎦

< < = ∫

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However

we will not use the formula to find the

• probability. instead, we will use Table VII of Appendix C.

Chapter 6: The Normal Distribution

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The standard normal distribution is a special case of the

normal distribution where the μ is zero and the σ is 1.

Chapter 6: The Standard Normal Distribution

The RV that possesses the standard normal distribution is

denoted by z and it is called z values or z scores.

μ= 0 σ= 1

• 4
• 2

2 4 z

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Since μ is zero and the σ is 1 for the standard normal, a

specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.

Chapter 6: The Standard Normal Distribution

The z values to the right side of the mean are positive and

those on the left are negative BUT the area under the curve is always positive.

For a value of z = 2, we are 2 standard deviations from the

mean (to the right). mean (to the right).

Similarly, for z = ‐2, we are 2 standard deviations from the

mean (to the left)

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Chapter 6: The Standard Normal Distribution

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Table VII of Appendix C, list the area under the standard

normal curve between z = 0 to the values of z to 3.09.

To read the standard normal table we always start at z = 0

Chapter 6: The Standard Normal Distribution

To read the standard normal table, we always start at z

0, which represents the mean.

Always remember that the normal curve is symmetric, that is

the area from 0 to any positive z value is equal to the area from 0 to that value in the negative side. Example: Find the area under the standard normal curve between z = 0 to z = 1.95

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Example: Find the area under the standard normal curve between z = ‐2.17 to z = 0

Chapter 6: The Standard Normal Distribution

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Example: Find the following areas under the standard normal curve. a) Area to the right of z = 2.32

Chapter 6: The Standard Normal Distribution

b) Area to the left of z = ‐1.54

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Example: Find the following probabilities for the standard normal curve. ) ( )

Chapter 6: The Standard Normal Distribution

a) P(1.19 < z < 2.12) b) P(‐1.56 < z < 2.31) c) P(z > ‐.75)

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Remember:

When two points are on the same side of the mean first

Chapter 6: The Standard Normal Distribution

When two points are on the same side of the mean, first find the areas between the mean and each of the two

• points. Then, subtract the smaller area from the larger

area When two points are on different side of the mean, first find the areas between the mean and each of the two

• points. Then, add these two area

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Empirical rule

Chapter 6: The Standard Normal Distribution

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Example: Find the following probabilities for the standard normal curve. a) P(0 < z < 5.65)

Chapter 6: The Standard Normal Distribution

b) P( z < ‐ 5.3)

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As it was shown in the previous section, Table VII of

Appendix C can be used only to find the areas under the standard normal curve.

Chapter 6: Standardizing a Normal Distribution

However, in real‐world applications, most of continuous RVs

that are normally distributed come with mean and standard deviation different from 0 and 1, respectively.

What shall we do? Is there any way to bring μ to zero and σ

to 1. to 1.

Yes, it can be done by subtracting μ from x and dividing the

result by σ (standardizing)

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Standardizing x:

For a normal RV x with mean μ and standard deviation σ. The standardized RV z can be found using the following f l

Chapter 6: Standardizing a Normal Distribution

formula

x z μ σ − =

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Example: For the following data [3.2, 1.9, 2.7, 4.3, 3.5, 2.8, 3.9, 4.4, 1.5, 1.8], a) find the mean and standard deviation b) d d h l h l

Chapter 6: Standardizing a Normal Distribution

b) standardize each value in the sample

z Transformation x 3.2 1.9 2.7 4.3

n

X

∑

3.5 2.8 3.9 4.4 1.5 1.8

1

3 0.99

i i

X n μ σ

=

= = =

∑

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Example: Let x be a continuous RV that has a normal distribution with a mean 50 and a standard deviation of 10. Convert the following x values to z values a) x = 55 b) x = 35 c) x = 50

Chapter 6: Standardizing a Normal Distribution

a) x = 55 b) x = 35 c) x = 50

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To find the area between two values of x for a normal

distribution

Convert both values of x to their respective z values

Chapter 6: Standardizing a Normal Distribution

p

Find the area under the standard normal curve between those

two values

Example: Let x be a continuous RV that is normally distributed with a mean of 25 and a standard deviation of 4. Find the area a) between x = 25 and x = 32 b) between x = 18 and x = 34 a) bet ee 5 a d 3 b) bet ee 8 a d 3

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Example: Let x be a continuous RV that has a normal distribution with a mean 40 and a standard deviation of 5. Find the following probabilities a) P(x > 55) b) P(x < 49)

Chapter 6: Standardizing a Normal Distribution

a) P(x > 55) b) P(x < 49)

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Example: Let x be a continuous RV that has a normal distribution with a mean 50 and a standard deviation of 8. Find the probability P(30 < x < 39)

Chapter 6: Standardizing a Normal Distribution

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Example: Let x be a continuous RV that has a normal distribution with a mean 80 and a standard deviation of 12. Find the following probabilities a) P(70 <x < 135) b) P(x < 27)

Chapter 6: Standardizing a Normal Distribution

a) P(70 <x < 135) b) P(x < 27)

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Example: According to the U.S. Bureau of Census, the mean income of all U.S. families was $43,237 in 1991. Assume that the 1991 incomes of all U.S. families have a normal distribution with a mean of $43,237 and a standard deviation of $10,500. Find the

Chapter 6: Application of the Normal Distribution

a mean of $43,237 and a standard deviation of $10,500. Find the probability that the 1991 income of a randomly selected U.S. family was between $30,000 and $50,000.

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Example: A racing car is one of the many toys manufactured by Mack Corporation. The assembly time for this toy follows a normal distribution with a mean of 55 minutes and a standard deviation of 4 minutes. The company closes at 5 P.M. every day.

Chapter 6: Application of the Normal Distribution

deviation of 4 minutes. The company closes at 5 P.M. every day. If one worker starts assembling a racing car at 4 P.M., what is the probability that she will finish this job before the company closes for the day?

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Example: Hupper Corporation produces many types of soft drinks, including Orange Cola. The filling machines are adjusted to pour 12 ounces of soda in each 12‐ounce can of Orange Cola. However, the actual amount of soda poured into

Chapter 6: Application of the Normal Distribution

Orange Cola. However, the actual amount of soda poured into each can is not exactly 12 ounces; it varies from can to can. It is found that the net amount of soda in such a can has a normal distribution with a mean of 12 ounces and a standard deviation of .015 ounces. (a) What is the probability that a randomly selected can of Orange Cola contains 11.97 to 11.99 ounces of soda? (b) What percentage of the Orange Cola cans contain 12.02 to 12.07 ounces of soda?

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Example: The life span of a calculator manufactured by lntal Corporation has a normal distribution with a mean of 54 months and a standard deviation of 8 months. The company guarantees that any calculator that starts malfunctioning within

Chapter 6: Application of the Normal Distribution

guarantees that any calculator that starts malfunctioning within 36 months of the purchase will be replaced by a new one. About what percentage of such calculators made by this company are expected to be replaced?

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Determining the z and x values when the area under the normal distribution is known

Chapter 6: Determining the z and x values

We reverse the procedure of finding the area under the

normal curve for a specific value of z or x to finding a specific value of z or x for a known area under the normal curve. Example: Find a point z such that the area under the standard normal curve between 0 and z is 4251 and the value of z is normal curve between 0 and z is .4251 and the value of z is positive

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Chapter 6: Determining the z and x values

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Example: Find the value of z such that the area under the standard normal curve in the right tail is .0050.

Chapter 6: Determining the z and x values

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Example: Find the value of z such that the area under the standard normal curve in the left tail is .050.

Chapter 6: Determining the z and x values

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Finding an x value for a normal distribution:

To find an x value when an area under a normal

distribution curve is given we do the following

Chapter 6: Determining the z and x values

distribution curve is given, we do the following

• 1. Find the z value corresponding to that x value from

the standard normal curve.

• 2. Transform the z value to x by substituting the values
• f μ, σ, and z in the following formula

x z μ σ = +

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Example: Recall the calculators example, it is known that the life

• f a calculator manufactured by Intal Corporation has a normal

distribution with a mean of 54 months and a standard deviation

• f 8 months. What should the warranty period be to replace a

Chapter 6: Determining the z and x values

• f 8 months. What should the warranty period be to replace a

malfunctioning calculator if the company does not want to replace more than 1 % of all the calculators sold

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Example: Most business schools require that every applicant for admission to a graduate degree program take the GMAT. Suppose the GMAT scores of all students have a normal distribution with a mean of550 and a standard deviation of 90.

Chapter 6: Determining the z and x values

distribution with a mean of550 and a standard deviation of 90. Matt Sanger is planning to take this test soon. What should his score be on this test so that only 10% of all the examinees score higher than he does?

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Example: Find the value of z such that 0.95 of the area is within ±z standard deviations of the mean

Chapter 6: Determining the z and x values

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In many of the discrete cases, the sample points were

assigned equal probabilities (equally likely).

Tossing a die

Chapter 6: The Uniform Distribution

g

Tossing a coin For a continuous RV, there is an infinite number of values in

the sample space but with equally likely outcomes too (sometimes).

If a short exists in a 5 m of an electrical wire. If a safety inspector wants to inspect a plant over a time

interval

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Definition: Let the RV x be defined over the interval c to d (c < d) with

probability function given by:

Chapter 6: The Uniform Distribution

Then x is said to be uniform random variable.

Th b b l b f d h

[ ]

2 2 2

) ( ) ( ) ( ) ( ) ( x E dx x f x x Var dx x f x x E − = = = =

∫ ∫

∞ ∞ − ∞ ∞ −

σ μ

The probability can be found with

1 ( ) ,

• =

≤ ≤ f x c x d d c

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Definition: Let the RV x be defined over the interval c to d (c < d)

with probability function given by:

Chapter 6: The Uniform Distribution

Then x is said to be uniform random variable.

Th b bili b f d i h

⎪ ⎩ ⎪ ⎨ ⎧ ≤ ≤ − =

• therwise

d c if 1 ) ( x c d x f

The probability can be found with

c

• d

a

• b

dx x f b x a P

b a

= ≤ < ≤ = < <

∫

d b a c with , ) ( ) (

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Example: A special case of the continuous uniform distribution

• ccurs whenever c = 0 and d = 1. The distribution is called the

standardized uniform distribution. Determine the probability function, f(x), the mean, E(x), the variance, Var(x), of the

Chapter 6: The Uniform Distribution

function, f(x), the mean, E(x), the variance, Var(x), of the standard uniform.

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Example: A uniform random variable is defined over the interval 1 to 6. ) h h b b l h h l b

Chapter 6: The Uniform Distribution

a) What is the probability that the RV assumes a value between 2 and 4? b) What is the mean and the variance of that RV? )