Title: Normal distribution Chapter 7 Sullivan
1Normal distribution Chapter 7 Sullivan
 Prof. Felix Apfaltrer
 fapfaltrer_at_bmcc.cuny.edu
 OfficeN518
 Phone 212220 8000 x 74 21
 Office hours
 Tue, Thu 1303 pm
2Definition
 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
3Uniform 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.
 ? P(x) 1
 0? P(x) ?1
4Uniform 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
5Standarizing 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  ?
6Standard 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 (A2).
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 A2 AREA 0.949
7Normal Distribution (z Score examples)
Example Find temperatures separating top and
bottom 2.5. Table A2 z 1.96
and z 1.96
Probability area under the curve integral

P(a? X ? b) P(X ? b)
 P(Xlt a)
853 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 (X36)/1.4 (38.836)/1.4 2 ( 38.8
is 2 standard deviations above the mean of 36)
Look up table P(zlt2) 0.9772
953 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
zscore (140 143)/29  0.1035 Upper
limit 211 gtgtgt zscore (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
1053 Applications
 Q10 53 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?
11Sampling 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 11, 12, 15, 2odd,
1odd, 2odd, 2/2,1/2, 2/2  21, 22, 25, 1odd, 0odd, 1odd
 51, 52, 55, 2odd, 1odd, 2odd
 Population variance / N
 Sample variance / n1
12Sampling 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.
13Sampling 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
14The 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!
15The Central Limit theorem (example)
 Example SSN (09) 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
16The 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(Zgt0.17) 10.43250.57  ? x ?172
 ? x ?/? n 29/? 12 8.37
17Normal 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.
18Normal 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.
z (119.5  100) / 7.07 10678 2.76 P(Z gt
2.76) 1 P(Z 2.76 ) 1  0.9971 0.0029
19Continuity Corrections
20Normal 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!
21End of chapter