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```Chapter 6
Continuous Probability Distributions
Chapter 6: Continuous Probability Distributions
6.1 Continuous Random Variables
2
 Continuous Random Variable
 A random variable that can take any value over some continuous range
of values is called a continuous random variable.
 Example 6.1.1: Height, weight, length, length of time, etc. They are
continuous random variables because they can assume any nonnegative
value.
 We are unable to list all possible values for continuous random variables.
 The probability of a specific value of a continuous random variable is
zero. That is, let X be a continuous random variable. Then P(X = any
value) = 0.
 Probability statements are concerned with probabilities over a range of
values. For examples:



The probability that a height is more than 6 ft., that is, P(X ≥ 6´) is not zero.
Chance that a height is between 5.5 ft. and 6 ft., P(5.5´ ≤ X ≤ 6´) is not zero.
To answer these, we need to assume that X = height has a particular
shape (i.e., distribution).
6.2 Normal Random Variables
 The normal random variable is one of the most commonly used
Relative frequency
continuous random variables.
 It has a bell-shaped probability distribution called the normal
distribution. The bell-shaped curve is called the normal curve.
3
μ
X
6.2 Normal Random Variables (cont.)
 Properties of Normal Distribution
 It is bell-shaped and thus symmetrical in its appearance.
 Its measures of central tendency (mean, median, mode and midrange)
are all identical.
 Two numbers (called parameters) are needed to describe a normal curve
(distribution).



4
The mean, μ. It indicates the center of the normal curve.
The standard deviation, σ. It indicates the width of a normal curve.
The total area under the curve is 1. The mean divides the curve into two
equal halves.
Area = .5
Area = .5
µ
Total area = 1
X
 Normal Curves with Unequal
Means and Equal Standard
Deviations
Relative frequency
6.2 Normal Random Variables (cont.)
Females
|
Average
female
height
Males
|
Average
male
height
Height
5
Means and Unequal Standard
Deviations
Relative frequency
 Normal Curves with Equal
Company A
Company B
Average age
in both
companies
Age
6.3 Finding a Probability for a Normal Random Variable
 Suppose the mean and the standard deviation of a normal random
variable are given. The probability of any given range of numbers is
represented by the area under the curve for that range.
 Example 6.3.1: The lifetime of an Everglo light bulb is a normal
random variable with μ = 400 hours and σ = 50 hours. Let X be the
burnout time of the Everglo bulb. What percentage of the time will
the burnout time be less than 360 hours?
6
 = 50
360
|
300
|
350
|
400
|
450
|
500
X
6.4 Finding Areas under the Standard Normal Curve
 Probabilities for all normal distributions are determined using the
standard normal distribution.
 The standard normal curve represents a standard normal variable Z
that has a mean μ = 0 and standard deviation σ = 1.
=1
7
|
-2
(µ - 2)
|
-1
(µ - )
µ=0
|
1
(µ + )
|
2
(µ + 2)
Z
6.4 Finding Areas under the Standard Normal Curve (cont.)
 Finding the probability of a normal random variable
 Convert the normal random variable X to a standard normal random
variable Z.
Z


8
X 

Shade the area of interest under the standard normal curve.
Use Standard Normal Distribution Table (Table A.4) to determine the
probability.
6.4 Finding Areas under the Standard Normal Curve (cont.)
 Understanding the Standard Normal Curve
Area under the curve = 1.00
Half the area = 0.5
|
1
|
2
Half the area = 0.5
|
3
|
1
0
|
2
|
3
Z
9
0 z
P(0 < Z < +z)
[Table value]
-z
0
P(-z < Z < 0) = P(0 < Z < +z)
[Table value]
0 z
P(Z > +z) = 0.5 – P(0 < Z < +z)
[0.5 – (Table value)]
6.4 Finding Areas under the Standard Normal Curve (cont.)
 Understanding the Standard Normal Curve
-z
0
z
P(Z < -z) = 0.5 – P(-z < Z < 0)
[0.5 – (Table value)]
-z
z
z
P(-z < Z < +z) = 2*P(0 < Z < +z)
[2*Table value]
P(Z < +z) = 0.5 + P(0 < Z < +z)
[0.5 + (Table value)]
10
-z
P(Z > -z) = 0.5 + P(-z < Z < 0)
= 0.5 + P(0 < Z < +z)
[0.5 + (Table value)]
0 z1
z2
P(+z2 < Z < +z1) =
P(0 < Z < +z2) - P(0 < Z < +z1)
[(TV for z2) – (TV for z1]
-z1
z2
P(-z1 < Z < +z2) =
P(-z1 < Z < 0) + P(0 < Z < z2)
[(TV for z2) + (TV for -z1]
6.4 Finding Areas under the Standard Normal Curve (cont.)
 Example 6.3.1 Solution:
=50
=1
-0.8
360
|
300
z
11
|
350
|
400
|
450
|
500
X
|
-2
|
-1
|
0
X   360  400  40


 0.80

50
50
Therefore, P(X<360)
= P(Z<-0.8)
= P(Z>0.8)
= 0.5 – P(0<Z<0.8)
= 0.5 – 0.2881 = 0.2119
|
1
|
2
z
|
1
|
2
z
=1
-0.8
|
-2
|
-1
|
0
6.5 Interpreting Z
 In Example 6.3.1 Z = - 0.8 means that the value 360 is .8 standard
deviations below the mean
 A positive value of Z designates how may standard deviations ()
X is to the right of the mean (µ)
 A negative value of Z designates how may standard deviations ()
X is to the left of the mean (µ)
12
6.5 Applications Where the Area Under a Normal Curve Is Provided
 Example 6.5.1: Gourmet Shack sells an average of 55 packages daily
of excellent English Stilton Cheese. The sales have a standard
deviation of 10 packages per day and sales follow a normal
distribution. If we want to meet demand 95% of the time, what is the
minimum inventory we should keep in hand each day to meet this
requirement?
 We know that μ = 55, σ = 10, Fraction of demand met = 0.95, X = ?
Fraction of demand met
= 0.95
Fraction of demand met
= 0.95
13
|
μ = 55
Z
X 

|
X=?
, That is, 1.65 
X
|
0
|
z=?
X  55
, X  55  1.65(10)  71.5
10
Z
```
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