Modular 11Ch 7.1 to 7.2 Part I

Ch 7.1 Uniform and Normal Distribution

Recall: Discrete random variable probability distribution

For a continued random variable the probability of observing one particular value is zero.

Special case: Binomial distribution

i.e. 0)( axP

Finding the probability of obtaining success in independent trials of a binomial experiment is calculated by plugging the value of into the binomial formula as shown below :

Objective A : Uniform Distribution

anaan ppCaxP )1()(

a na

Continuous Random variable

A1. Introduction

We can only compute probability over an interval of values.

Continuous Probability Distribution

)()( bxaPbxaP

Since and for a continuous random variable,

0)( axP 0)( bxP

To find probabilities for continuous random variables, we use probability density functions.

Two common types of continuous random variable probability distribution :

• Uniform distribution

• Normal distribution.

Objective A : Uniform Distribution

Note : The area under a probability density function is 1.

ab 1

a b

Area of rectangle = Height x Width

)( ab 1 = Height x

Height = )(

1

ab for a uniform distribution

Ch 7.1 Uniform and Normal Distribution

A2. Uniform Distribution

Example 1 : A continuous random variable is uniformly distributed with .5010 x

x

(a) Draw a graph of the uniform density function.

40

1

10 50

Area of rectangle = Height x Width

1 = Height x

Height = )(

1

ab

)( ab

)1050(

1

= 40

1

(b) What is ?)3020( xP

40

1

2030

Area of rectangle = Height x Width

)2030(x40

1

10x40

1

25.04

1

(c) What is ?)15( xP

40

1

1015

)15()15( xPxP Area of rectangle = Height x Width

)1015(x40

1

5x40

1

125.08

1

)1510( xP

Ch 7.1 Uniform and Normal Distribution

Objective A : Uniform Distribution

Objective B : Normal distribution

Ch 7.2 Applications of the Normal Distribution

Objective A : Area under the Standard Normal Distribution

Objective B : Normal distribution – Bell-shaped CurveCh 7.1 Uniform and Normal Distribution

Example 1: Graph of a normal curve is given.Use the graph to identify the value of and .

530100

430330 530 630 730X

1 1 2 2

(b) Shade the region that represents the proportion of refrigerator that lasts for more than 17 years.

5.115.6 14 5.16 199 5.21

17

X

(a) Draw a normal curve and the parameters labeled.

Example 2: The lives of refrigerator are normally distributed with mean years and standard deviation years.14 5.2

5.115.6 14 5.16 199 5.21 X

1 1 2

3 3 2

(c) Suppose the area under the normal curve to the right = 17 is 0.1151. Provide two interpretations of this result.

11.51% of all refrigerators are kept for at least 17 years.

– the probability that a randomly selected individual from the population will have the characteristic described by the interval of values.

The probability that a randomly selected refrigerator will be kept for at least 17 years is 11.51%.

Notation: 1151.0)17( xP

x

The area under the normal curve for any interval of values of the random variable represent either:x

– the proportions of the population with the characteristic described by the interval of values.

Ch 7.1 Uniform and Normal Distribution

Objective A : Uniform Distribution

Objective B : Normal distribution

Ch 7.2 Applications of the Normal Distribution

Objective A : Area under the Standard Normal Distribution

Ch 7.2 Applications of the Normal Distribution

The standard normal distribution

– Bell shaped curve – and .10

The random variable for the standard normal distribution is .Z

Use the table (Table V) to find the area under the standard normal distribution. Each value in the body of the table is a cumulative area from the left up to a specific score.

Z

Objective A : Area under the Standard Normal Distribution

1 0 1 22 Z5.3 5.3

ZNegative ZPositive

0

1

Z

(a) what is the area to the right of ?

(b) what is the area to the left of ?

5.0

5.0

Z

Probability is the area under the curve over an interval.

The total area under the normal curve is 1.

Z

0

0Under the standard normal distribution,

0

From Table V

5.0

0 56.0 Z

038.1Z

Example 1 : Determine the area under the standard normal curve.(a) that lies to the left of -1.38.

(b) that lies to the right of 0.56.

From Table V08.0

0838.0

0838.0)38.1( ZP

3.1

06.0

7123.07123.0 1

0838.0

7123.01)56.0( ZP2877.0

Area under the whole standard normal distribution is 1.

Table V only provides area to the left of = 0.56.Z

0 85.1 47.2 Z

(c) that lies in between 1.85 and 2.47. From Table V

9678.085.1 9932.047.2 9932.09678.0

)47.285.1( ZP

0.02549678.09932.0