Download Find specific data values for given percentages using the standard

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Central limit theorem wikipedia, lookup

Transcript
CHAPTER 6
The Normal Distribution
Objectives
•
•
•
•
•
•
•
Identify distributions as symmetrical or skewed.
Identify the properties of the normal distribution.
Find the area under the standard normal distribution, given various z values.
Find the probabilities for a normally distributed variable by transforming it into a standard
normal variable.
Find specific data values for given percentages using the standard normal distribution.
Use the central limit theorem to solve problems involving sample means for large samples.
Use the normal approximation to compute probabilities for a binomial variable.
6.1 Introduction
• Many continuous variables have distributions that are bell-shaped and are called
approximately normally distributed variables.
• A normal distribution is also known as the bell curve or the Gaussian distribution.
Normal and Skewed Distributions
•
•
•
The normal distribution is a continuous, bell-shaped distribution of a variable.
If the data values are evenly distributed about the mean, the distribution is said to be
symmetrical.
If the majority of the data values fall to the left or right of the mean, the distribution is said to
be skewed.
Left Skewed Distributions
•
When the majority of the data values fall to the right of the mean, the distribution is said to be
negatively or left skewed. The mean is to the left of the median, and the mean and the
median are to the left of the mode.
Right Skewed Distributions
•
When the majority of the data values fall to the left of the mean, the distribution is said to be
positively or right skewed. The mean falls to the right of the median and both the mean and
the median fall to the right of the mode.
6.2 Properties of a Normal Distribution
I. Continuous Probability Distributions
A continuous random variable is one that can theoretically take on any value on some line
interval. We use f(x) to represent a probability density function. Unfortunately, f(x) does not
give us the probability that the value x will be observed. To understand how a probability
density function for a continuous random variable enables us to find probabilities, it is
important to understand the relationship between probability and area. For the following given
histogram, what is the probability that x is in between 2.5 to 5.5?
1
Relative Frequency Histogram
Frequency Histogram
20
5
Percent
Frequency
4
3
2
10
1
0
0
0
0
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
C1
C1
Use the given frequency histogram to calculate P(2.5 < x < 5.5):
P(2.5 < x < 5.5) = (4 + 5 + 4) / (1+ 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1) = 13 / 25 = 52%
Use the corresponding relative frequency histogram to calculate P(2.5 < x < 5.5):
P(2.5 < x < 5.5) = 16% + 20% + 16% = 52% which is the same as the area of the three
middle bars of the relative frequency histogram. The width of each bar is
one and the height is the given percentage.
For a continuous probability distribution,
1) f ( x ) > 0 for all values x of the random variable;
2) the total area under the graph of f ( x ) is 1;
3) P(a < x < b) can be approximated by the area under the graph of f ( x )
for a < x < b.
Note: P(x = a) = 0 for continuous random variables.
This implies P(a  x  b) = P(a < x < b); P(x  a) = P(x > a);
and P(x  a) = P(x < a).
II. The Normal Distribution
Continuous probability distributions can assume a variety of shapes. However, the most
important distribution of continuous random variables in statistics is the normal distribution
that is approximately mound-shaped. Many naturally occurring random variables such as
IQs, height of humans, weights, times, etc. have nearly normal distributions.
• The mathematical equation for a normal distribution is
2
2
1
f ( x) 
e ( x   ) /(2 )
  x  
 2
Where e  2.718,   3.14,  = population mean
 = population standard deviation
The mean  is located at the center of distribution. The distribution is symmetric about its mean
.
2
There is a correspondence between area and probability.
Since the total area under the normal probability distribution is equal to 1, the symmetry implies
that the area to the right of  is 0.5 and the area to the left of  is also 0.5.
Large values of  reduce the height of the curve and increase the spread.
Small values of  increase the height of the curve and reduce the spread.
Almost all values of a normal random variable lie in the interval   3
III. Properties of the Normal Distribution
•
•
The shape and position of the normal distribution curve depend on two parameters, the mean
and the standard deviation.
Each normally distributed variable has its own normal distribution curve, which depends on
the values of the variable’s mean and standard deviation.
Normal Distribution Properties
•
•
•
•
•
•
The normal distribution curve is bell-shaped.
The mean, median, and mode are equal and located at the center of the distribution.
The normal distribution curve is unimodal (i.e., it has only one mode).
The curve is symmetrical about the mean, which is equivalent to saying that its shape is the
same on both sides of a vertical line passing through the center.
The curve is continuous—i.e., there are no gaps or holes. For each value of X, here is a
correspondingval3ue of Y.
The curve never touches the x axis. Theoretically, no matter how far in either direction the
curve extends, it never meets the x axis—but it gets increasingly closer.
6.3 The Standard Normal Distribution
• Since each normally distributed variable has its own mean and standard deviation, the shape
and location of these curves will vary. In practical applications, one would have to have a
table of areas under the curve for each variable. To simplify this, statisticians use the
standard normal distribution.
• The standard normal distribution is a normal distribution with a mean of 0 and a standard
deviation of 1.
f (x) 
1
 2
Note:   0 and   1 means z 
e ( x )
2
/2
X 0
X
1
Recall: z Values
•
The z value is the number of standard deviations that a particular X value is away from the
mean. The formula for finding the z value is:
3
Area Between 0 and z
 To find the area between 0 and any z value: Look up the z value in the table.
Area in Any Tail
•
•
Look up the z value to get the area.
Subtract the area from 0.5000.
Area Between Two z Values
•
•
Look up both z values to get the areas.
Subtract the smaller area from the larger area.
Area Between z Values—Opposite Sides
•
•
Look up both z values to get the areas.
Add the areas.
Area To the Left of Any z Value
•
•
Look up the z value to get the area.
Add 0.5000 to the area.
Area To the Right of Any z Value
•
•
Look up the z value in the table to get the area.
Add 0.5000 to the area.
Area Under the Curve
• The area under the curve is more important than the frequencies because the area
corresponds to the probability!
•
Note: In a continuous distribution, the probability of any exact Z value is 0 since area would
be represented by a vertical line above the value. But vertical lines in theory have no area.
So
Example 1:
(a) Find P(0 < z < 1.63)
4
(b) Find P(-2.48 < z < 0)
(c) Find P(-2.02 < z < 1.74)
(d) Find P(1.02 < z < 1.84)
(e) Find the probability that z is between -.58 and -.10.
(f) Find the probability that z is larger than 1.76.
(g) Find the probability that z is less than 2.04.
(h) Find the probability that z is within two standard deviations of the mean.
5
Example 2: Assume the standard normal distribution. Fill in the blanks.
(a) P( 0 < z < _______ ) = .4279
(b) P( 0 < z < _______ ) = .4997
(c) P( _______ < z < 0 ) = .4370
(d) P( z < _______ ) = .9846
(e) P( z < _______ ) = .1190
(f) Find the z value to the left of the mean so that 71.90% of the area
under the distribution curve lies to the right of it.
6
(g) Find two z values, one positive and one negative, so that the areas in
the two tails total to 12%.
6.4 Applications of the Normal Distribution
I. Calculating Probabilities for a Non-Standard Normal Distribution
Consider a normal variable x with mean  and standard deviation .
1. Standardize from x to z
z
x

2. Use Table E to find the central area corresponding to z
3. Adjust the area to answer the question
Example 1:
Let x be a normal random variable with mean 80 and standard deviation 12.
What percentage of values are
(a) larger than 56?
(b) less than 62?
(c) between 85 and 98
(d) outside of 1.5 standard deviations of the mean?
7
Example 2: (Ref: General Statistics by Chase/Bown, 4th Ed.)
The length of times it takes for a ferry to reach a summer resort from the mainland is
approximately normally distributed with mean 2 hours and standard deviation of 12
minutes. Over many past trips, what proportion of times has the ferry reached the island
in
(a) less than 1 hour, 45 minutes?
(b) more than 2 hours, 5 minutes?
(c) between 1 hour, 50 minutes and 2 hours, 20 minutes?
II. Calculating a Cutoff Value
Backward steps for calculating probabilities of a non-standard normal distribution.
1. Adjust to the corresponding central area.
2. Use Table E to find the corresponding z cutoff value.
8
3. Non-standardize from z to x:
z
x

Example 1:
Employees of a company are given a test that is distributed normally with mean 100 and
variance 25. The top 5% will be awarded top positions with the company. What score is
necessary to get one of the top positions?
Example 2:
Quiz scores were normally distributed with  = 14 and  = 2.8, the lower 20%
should receive tutorial service. Find the cutoff score.
Section 6 – 5 The Central Limit Theorem
•
I. Sampling Distribution of Sample Mean
Example 1: Population Distribution Table
x
P(x)
2
1/4
(a) Find the population mean
distribution table.
4
1/4

6
1/4
8
1/4
and population standard deviation

of the population
9
(b) Construct a probability histogram for x
Example 2: From the population distribution of example 1,
2 random variables are randomly selected.
(a) List out all possible combinations (sample space) and x for each combination.
10
(b) Construct a probability distribution table for x .
(c) Construct a probability histogram for x .
11
(d) Find the mean of the sampling distribution of x .
(e) Find the standard deviation of the sampling distribution of x .
(f) Compare  with  x .
(g) Compare  with  x .
12
Population parameter
Sample statistics
Mean


Standard deviation

x
Population Distribution
x
Sampling Distribution
P (x )
P (x )
4 / 10
3 / 10
2 / 10
1
4
1/ 10
2
4
6
8
1
x
2
3
4
6
5
7
8
x
II. Central Limit Theorem
If the population distribution is normally distributed, the sampling distribution of x will be
normally distributed, for any sample size n.
P( x )
P( x )
x
x
If the population distribution is not normally distributed, the sampling distribution of x will be
normally distributed for any size of n  30
P( x )
P( x )
x
x
x  
x 

n
Example 1: Population distribution
P( x )
Given:
x

= 50,

= 10
13
(a) Find
x
and
x
for n = 4
(b) Is the sampling distribution x normally distributed?
(c) If n is changed from 4 to 36, is the sampling distribution x normally distributed?
Example 2:
(Ref: General Statistics by Chase/Bown, 4th Ed.)
A population has mean 325 and variance 144. Suppose the distribution of
sample means is generated by random samples of size 36.
(a)
Find  x and  x
(b)
Find P( x  323)
(c)
Find P(321  x  327)
14
Example 3:
The average number of days spent in a North Carolina hospital for a coronary bypass in 1992
was 9 days and the standard deviation was 4 days (North Carolina Medical Database
Commission, Consumer’s Guide to Hospitalization Charges in North Carolina Hospitals, August
1994). What is the probability that a random sample of 30 patients will have an average stay
longer than 9.5 days?
Example 4:
Suppose the test scores for an exam are normally distributed with
(a) What percent of the students has a score greater than 85?

= 75,

=8
(b) What is the probability that 4 randomly selected students will have a mean score
higher than 85?
15
Section 6 - 6
Normal Approximation to the Binomial Distribution
I. When to use a Normal distribution to approximate a Binomial distribution?
Recall that a binomial distribution is determined by n and p. When p is approximately 0.5, and
as n increases, the shape of the binomial distribution becomes similar to the normal
distribution. In order to use a normal distribution to approximate a binomial distribution, n
must be sufficiently large. It is known n will be sufficiently large if np ≥ 5 and nq ≥ 5.
When using a normal distribution to approximate a binomial distribution, the mean and standard
deviation of the normal distribution is the same as the binomial distribution. Now recall the
formulas for finding the mean and standard deviation.
II.
Continuity Correction
• In addition to the condition np ≥ 5 and nq ≥ 5, a correction for continuity is used in employing
a continuous distribution (Normal distribution) to approximate a discrete distribution (Binomial
distribution).
Warning : The continuity correction should be used only when approximating the Binomial
probability with a normal probability. Don’t use the continuity correction with other
normal probability problems.
Continuity correction  x  0.5
Example 1:
Use the continuity correction to rewrite each expression:
(a)
Bi Dist. N Dist.
(d)
Bi Dist.
N Dist.
P( x > 6) 
P( 1 < x < 7) 
(b)
Bi Dist. N Dist.
P( x  3) 
(e)
Bi Dist.
N Dist.
P ( 5  x  10) 
(c)
Bi Dist. N Dist.
P( x  9) 
(f)
Bi Dist.
N Dist.
P (4 < x  6) 
16
III. Using a Normal Distribution to approximate a
Binomial Distribution
Step 1:
Check whether the normal distribution can be used.
( np  5 and nq  5 )
Step 2:
Find the mean  and standard deviation  .
  np ,   npq
Step 3:
Step 4:
Step 5:
Write the problem in probability notation, using x.
Rewrite the problem by using the continuity correction factor.
Continuity correction  x  0.5
Find the corresponding z value(s).
z
x

Step 6:
Use the z table to find the center area and adjust the center area to
answer the question.
Example 1:
(Ref: General Statistics by Chase/Bown, 4th Ed.)
Assume that the experiment is a binomial experiment.
Find the probability of 10 or more successes,
where n = 13 and p = .4.
(a)
Use the Binomial table
(b)
Using the normal approximation to the binomial.
17
Example 2:
A dealer states that 90% of all automobiles sold have air
conditioning. If the dealer sells 250 cars, find the probability
that fewer than 5 of them will not have air conditioning.
Example 3:
In a corporation, 30% of the people elect to enroll in the
financial investment program offered by the company.
Find the probability that of 800 randomly selected people,
between 260 and 300 inclusive have enrolled in the program.
18
Summary
•
•
•
•
•
•
•
•
•
The normal distribution can be used to describe a variety of variables, such as heights,
weights, and temperatures.
The normal distribution is bell-shaped, unimodal, symmetric, and continuous; its mean,
median, and mode are equal.
Mathematicians use the standard normal distribution which has a mean of 0 and a standard
deviation of 1.
The normal distribution can be used to describe a sampling distribution of sample means.
These samples must be of the same size and randomly selected with replacement from the
population.
The central limit theorem states that as the size of the samples increases, the distribution of
sample means will be approximately normal.
The normal distribution can be used to approximate other distributions, such as the binomial
distribution.
For the normal distribution to be used as an approximation to the binomial distribution, the
conditions np  5 and nq  5 must be met.
A correction for continuity may be used for more accurate results.
Conclusions
•
The normal distribution can be used to approximate other distributions to simplify the data
analysis for a variety of applications.
19