Normal distribution Chapter 7 Sullivan - PowerPoint PPT Presentation

Loading...

PPT – Normal distribution Chapter 7 Sullivan PowerPoint presentation | free to download - id: d3510-ZDc1Z



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Normal distribution Chapter 7 Sullivan

Description:

A continuous random variable whose distribution has a graph that is. bell ... Example: Ski gondola safety, Vail, CO. Maximal capacity 12 people or 2004 pounds, ... – PowerPoint PPT presentation

Number of Views:138
Avg rating:3.0/5.0
Slides: 22
Provided by: socrates9
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Normal distribution Chapter 7 Sullivan


1
Normal distribution Chapter 7 Sullivan
  • Prof. Felix Apfaltrer
  • fapfaltrer_at_bmcc.cuny.edu
  • OfficeN518
  • Phone 212-220 8000 x 74 21
  • Office hours
  • Tue, Thu 130-3 pm

2
Definition
  • A continuous random variable whose distribution
    has a graph that is
  • bell shaped
  • Symmetric
  • is said to have a normal distribution
  • if the distribution is described by the formula
  • Response to the formula
  • Uh!
  • x random variable
  • e 2.718282 Euler
  • ? 3.1415929Pi
  • Two parameters
  • ? mean
  • ? standard deviation

3
Uniform distributions
  • A density curve (or probability density function
    PDF) is a graph of a continuous probability
    density function, such that
  • The total area under the graph is 1.
  • Every point in the graph has a vertical height
    larger than (or equal) to zero.

The continuous random variable has uniform
distribution if its values are spread evenly on
an interval (a, b). The probability of being in
the interval is constant, and zero elsewhere. The
distribution is specified by two parameters the
end points a and b. We denote the distribution
U(a,b). Its probability density function (PDF)
is
Example A prof. plans classes so carefully that
their length is uniformly distributed between 100
and 102 minutes.
  1. ? P(x) 1
  2. 0? P(x) ?1

4
Uniform distributions and normal distributions
(area)
  • For normal distributions
  • Not so easy to calculate area
  • Many normals with different means and standard
    deviations
  • Women ? 63.6, ? 2.5
  • Men ? 69.0, ? 2.8

Example (continued) Kim has an interview
immediately following the class. If class runs
longer than 101.5 minutes, she will be
late. Calculate the probability of Kim being late
to the interview.
P(class gt 51.5 min) area of shaded region
0.5 0.5 0.25
5
Standarizing the normal distribution
  • Properties of the normal distribution
  • Symmetric about the mean ?
  • meanmedianmode
  • Highest point there at x ?
  • Inflection points at ? -? and ? ?
  • Area to the right and left of ? is ½. Area under
    the whole curve is 1.
  • Area between
  • ? -? and ? ? is 68
  • ? -2? and ? 2? is 95
  • ? -3? and ? 3? is 99.7
  • There are many normal distributions, but one
    standard normal distribution.
  • To standarize any normal distribution X, convert
    to Z, the standard normal, by the following
    calculation

x-? z ----- ?
6
Standard Normal Distribution
  • The standard normal distribution is a normal
    probability distribution that has
  • Mean 0
  • Standard variation 1
  • ? 0, ? 1, ? 2 1,
  • we denote a random variable X that has such
    distribution by X N( 0, 1 )
  • A r.v. Y with normal distribution with mean ? and
    variance ? 2 by Y N( ?, ? 2 )
  • The total area under the density curve is always
    1.
  • z - scores
  • The area under the standard normal has been
    calculated for many different values and can be
    calculated using Excel, scientific calculators,
    or tables (A-2).

Example Scientific thermometers should show 0C
at the temperature that water freezes. Some show
positive values, some negative values at that
temp.Assume ? 1. Find the probability, that for
a thermometer the temp. shown is below
1.58C. Table A-2 AREA 0.949
7
Normal Distribution (z Score examples)
Example Find temperatures separating top and
bottom 2.5. Table A-2 z -1.96
and z 1.96
Probability area under the curve integral
-

P(a? X ? b) P(X ? b)
- P(Xlt a)
8
5-3 Normal Distribution (Applications)
x-? z ----- ?
Example When designing a seat in a car, heights
of people must be considered. Mean men sitting
height is normally distributed with a mean of
36 in, and the standard deviation of 1.4 in,
N(36,1.4). A car is designed that cannot
accommodate men higher than 38.8 in sitting. Find
the probability of a man not fitting in the car.
Answer Z (X-36)/1.4 (38.8-36)/1.4 2 ( 38.8
is 2 standard deviations above the mean of 36)
Look up table P(zlt2) 0.9772
9
5-3 Applications
Jet seats must be able to eject people between
140 lb and 211. Women have a mean weight of 143
and a standard deviation of 29lb. What
probability of women are within the right limits?
x-? z ----- ?
Solution Lower limit 140 gtgtgt
z-score (140- 143)/29 - 0.1035 Upper
limit 211 gtgtgt z-score (211- 143)/29
2.3448
  • .

Answer P( 140 ? X ? 211 )
P(-0.1035? Z? 2.3448) P(Z? 2.3448) P(Z
?-0.1035) 0.9904 0.4602
0.4302
10
5-3 Applications
  • Q10 5-3 Pregnancy length
  • ?268 days ? 15 days
  • Letter Wife says she was pregnant for 308 days,
    time since her husband was in for a short visit
    from Navy.
  • Calculate P(Xgt308 days).
  • What does this suggest?
  • Premature if below 4. Find length.
  • Q11 SAT ?998 ? 202
  • College requires 1100 minimum.
  • Find percentage satisfying requirement.
  • Find 40 percentile. Why does college not ask
    top 40?

11
Sampling distributions
  • Population 1,2, 5 pizzas sold (3 days only)
  • Sample size 2 (with replacement)
  • same results when sample small
  • Independence
  • Samples of size 2 1-1, 1-2, 1-5, 2odd,
    1odd, 2odd, 2/2,1/2, 2/2
  • 2-1, 2-2, 2-5, 1odd, 0odd, 1odd
  • 5-1, 5-2, 5-5, 2odd, 1odd, 2odd
  • Population variance / N
  • Sample variance / n-1

12
Sampling distributions, cont
  • The sampling distribution of the mean is the
    probability distribution of the sample means,
    when all samples have the same size.

The value of a statistic (like the sample mean x
) depends on the particular values in the
sample. It is called sampling variability.
13
Sampling proportions
  • The sampling distribution of the proportion is
    the probability distribution of the sample
    proportions, when all samples have size n.
  • Sample proportions tend to target the population
    proportion
  • Under certain conditions, the distribution tends
    to the normal distribution

14
The Central Limit theorem
  • The random variable x has a distribution with
    mean ? and variance ?2.
  • Simple random samples of size n are selected from
    the population.
  • Then
  • the sample means x approach a normal
    distribution, as n increases.
  • The mean of all sample means is ?
  • and the variance of the sample means is ?2/n .

? x n
N( ? ,?2/n )
x
? . ? n
? x ?
? x
  • Practical rules
  • Good approximation for n gt30, for general
    distributions.
  • For normal distributions, sample means are
    normally distributed for any n!

15
The Central Limit theorem (example)
  • Example SSN (0-9) generate 50 samples of size n
    4.
  • For example, for second set 7,3,5,6

? 5.25 ? 1.71
In general, the distribution of the mean tends to
a normal with
16
The Central Limit theorem (applications)
  • We can standarize the normal distribution of the
    sample means
  • Example Ski gondola safety, Vail, CO
  • Maximal capacity 12 people or 2004 pounds,
  • i.e., 2004/12167 lbs per person
  • Worst case scenario, all men (heavier) ? 172
    lbs, ?29 lbs
  • Find the probability that one random man is
    heavier than 167lbs.
  • Find the probability that 12 men have an average
    greater than 167.
  • z(x-? )/ ? (167- 172) / 29 - 0.17.
  • From table A2 P(Zlt -0.17) 0.4325, hence
    P(Zgt-0.17) 10.43250.57
  • ? x ?172
  • ? x ?/? n 29/? 12 8.37

17
Normal approximating binomial (applications)
  • If np ? 5 and nq ? 5, then the binomial
    distribution can be approximated by a normal
    distribution with mean ? np
  • ? ? npq
  • Airlines
  • Plane with 200 passengers has to unload cargo
    when at least 120 men.
  • Find the probability that among 200 random
    passangers, at least 120 are men.
  • Binomial, successes 120 or more, out of n200.
    Not in table, or calculator! Use normal
    approximation, npnq200 gt5!
  • ? np2001/2100
  • ? npq ? 2001/21/2 7.07
  • The discrete value 120 is represented
    continuously as the interval
  • 119.5 to 120.5, hence X120 is represented by
    the area 119.5.

18
Normal approximating binomial (applications)
  • Airlines (cont)
  • The discrete value 120 is represented
    continuously as the interval
  • 119.5 to 120.5, hence X120 is represented by
    the area X 119.5.
  • 100

z (119.5 - 100) / 7.07 10678 2.76 P(Z gt
2.76) 1- P(Z 2.76 ) 1 - 0.9971 0.0029
19
Continuity Corrections
20
Normal approximating binomial (example)
  • TV ratings
  • CBS 60 minutes share 20, i.e., 20 of tvs
  • Survey 200 households, 16 watching CBS 60
    minutes.
  • Assuming 20 share, is there enough evidence to
    conclude that 20 share is wrong? (rare event rule)

Need to calculate probability that at most 32
households are watching CBS. np2000.240 ?
5 nq2000.8160 ? 5 ? np 40 ? ? npq ?
2000.20.85.65
  • P(X ? 32.5)
  • P( (X-? ) /? ? (32.5 40)/5.65)
  • P( Z? -1.33)
  • 0.0918
  • Not enough evidence to throw away the claim!

21
End of chapter
  • Hw p. 259 1,2, and 4
About PowerShow.com