...6
The NormalDistribution
Objectives
Outline
After completing this chapter, you should be able to
1
2
3
Identify distributions as symmetric or skewed.
4
Find probabilities for a normally distributed
variable by transforming it into a standardnormal variable.
Introduction
6–1
NormalDistributions
Identify the properties of anormaldistribution.
Find the area under the standardnormaldistribution, given various z values.
5
Find speciﬁc data values for given
percentages, using the standardnormaldistribution.
6
6–3 The Central Limit Theorem
6–4 The Normal Approximation to the Binomial
Distribution
Use the central limit theorem to solve
problems involving sample means for large
samples.
7
6–2 Applications of the NormalDistribution
Use the normal approximation to compute
probabilities for a binomial variable.
Summary
6–1
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Confirming Pages
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Chapter 6 The NormalDistribution
Statistics
Today
What Is Normal?
Medical researchers have determined so-called normal intervals for a person’s...

...
NormalDistributionNormaldistribution is a statistics, which have been widely applied of all mathematical concepts, among large number of statisticians. Abraham de Moivre, an 18th century statistician and consultant to gamblers, noticed that as the number of events (N) increased, the distribution approached, forming a very smooth curve.
He insisted that a new discovery of a mathematical expression for this curve could lead to an easier way to find solutions to probabilities of, “60 or more heads out of 100 coin flips.” Along with this idea, Abraham de Moivre came up with a model that has a drawn curve through the midpoints on the top of each bar in a histogram of normally distributed data, which is called, “Normal Curve.”
One of the first applications of the normaldistribution was used in astronomical observations, where they found errors of measurement. In the seventeenth century, Galileo concluded the outcomes, with relation to the measurement of distances from the star. He proposed that small errors are more likely to occur than large errors, random errors are symmetric to the final errors, and his observations usually gather around the true values. Galileo’s theory of the errors were discovered to be the characteristics of normaldistribution and the formula for...

...Deﬁnition.
Deﬁnition 1. A standard (one-dimensional) Wiener process (also called Brownian motion) is
a stochastic process {Wt }t≥0+ indexed by nonnegative real numbers t with the following
properties:
(1)
(2)
(3)
(4)
W0 = 0.
With probability 1, the function t → Wt is continuous in t.
The process {Wt }t≥0 has stationary, independent increments.
The increment Wt+s − Ws has the N ORMAL(0, t) distribution.
A Wiener process with initial value W0 = x is gotten by adding x to a standard Wiener
process. As is customary in the land of Markov processes, the initial value x is indicated
(when appropriate) by putting a superscript x on the probability and expectation operators. The term independent increments means that for every choice of nonnegative real
numbers 0 ≤ s1 < t1 ≤ s2 < t2 ≤ · · · ≤ sn < tn < ∞, the increment random variables
Wt1 − Ws1 , Wt2 − Ws2 , . . . , Wtn − Wsn
are jointly independent; the term stationary increments means that for any 0 < s, t < ∞
the distribution of the increment Wt+s − Ws has the same distribution as Wt − W0 = Wt .
In general, a stochastic process with stationary, independent increments is called a L´vy
e
process; more on these later. The Wiener process is the intersection of the class of Gaussian
processes with the L´vy processes.
e
It should not be obvious that properties (1)–(4) in the deﬁnition of a...

...1
Marks: 1
Assume that X has a normaldistribution, and find the indicated probability.
The mean is μ = 60.0 and the standard deviation is σ = 4.0.
Find the probability that X is less than 53.0.
Choose one answer.
a. 0.5589
b. 0.0401
c. 0.9599
d. 0.0802
Question2
Marks: 1
Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation. Assume that the population has a normaldistribution.
Weights of eggs: 95% confidence; n = 22, x = 1.77 oz, s = 0.47 oz
Choose one answer.
a. (0.38 oz; 0.63 oz)
b. (0.36 oz; 0.67 oz)
c. (0.36 oz; 0.65 oz)
d. (0.37 oz; 0.61 oz)
Question3
Marks: 1
The weights of the fish in a certain lake are normally distributed with a mean of 12 lb and a standard deviation of 12. If 16 fish are randomly selected, what is the probability that the mean weight will be between 9.6 and 15.6 lb?
Choose one answer.
a. 0.0968
b. 0.6730
c. 0.3270
d. 0.4032
Question4
Marks: 1
Find the appropriate minimum sample size.
You want to be 95% confident that the sample variance is within 40% of the population variance.
Choose one answer.
a. 57
b. 14
c. 224
d. 11
Question5
Marks: 1
Find the indicated probability.
The weekly salaries of teachers in one state are normally distributed with a mean of $490 and a...

...NormalDistribution:- A continuous random variable X is a normaldistribution with the parameters mean and variance then the probability function can be written as
f(x) = - < x < , - < μ < , σ > 0.
When σ2 = 1, μ = 0 is called as standardnormal.
Normaldistribution problems and solutions – Formulas:
X < μ = 0.5 – Z
X > μ = 0.5 + Z
X = μ = 0.5
where,
μ = mean
σ = standard deviation
X = normal random variable
NormalDistribution Problems and Solutions – Example Problems:
Example 1:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X<50). When mean μ = 41 and standard deviation = 6.5
Solution:
Given
Mean μ = 41
Standard deviation σ = 6.5
Using the formula
Z =
Given value for X = 50
Z =
=
= 1.38
Z = 1.38
Using the Z table, we determine the Z value = 1.38
Z = 1.38 = 0.4162
If X is greater than μ then we use this formula
X > μ = 0.5 + Z
50 > 41 = 0.5 + 0.4162
P(X) = 0.5 + 0.4162
= 0.9162
Example 2:
If X is a normal random variable with mean and standard deviation calculate the probability of P(X< 37). When mean μ = 20 and standard deviation = 15
Solution:
Given
Mean μ = 20...

...NormalDistribution
It is important because of Central Limit Theorem (CTL), the CTL said that Sum up a lot of i.i.d random variables the shape of the distribution will looks like Normal.
Normal P.D.F
Now we want to find c
This integral has been proved that it cannot have close form solution. However, someone gives an idea that looks stupid but actually very brilliant by multiply two of them.
reminds the function of circle which we can replace them to polar coordinate
Thus
Mean
By symmetry if g(x) is odd function g-x=-g(x) then -abgxdx=0
Variance
Notation
CDF is standardNormal CDF
by symmetric
,CDF , , All the odd moment of standardnormal are zero. However, even moment is not easy to calculate by integral
(Symmetry)
Then we say
Most of Statistics books will write the pdf then explain the mean and variance but it is not intuitive.
Standardization
Find PDF of
CDF:
The PDF is derivative of the CDF (using chain rule)
PDF:
Later we’ll show if independent
68-95-99.7% Rule
Because you can’t actually calculate the , somebody create a rule of thumb
The properties of variance
If you shift the variance by c, the mean also shift by c. Thus, the variance doesn’t change.
Remember to square. It is easy to validate if you think c is negative and don’t square it....

...under a StandardNormal curve
a) to the right of z is 0.3632;
b) to the left of z is 0.1131;
c) between 0 and z, with z > 0, is 0.4838;
d) between -z and z, with z > 0, is 0.9500.
Ans : a) z = + 0.35 ( find 0.5- 0.3632 = 0.1368 in the normal table)
b) z = -1.21 ( find 0.5 – 0.1131 = 0.3869 in the normal table)
c ) the area between 0 to z is 0.4838, z = 2.14
d) the area to the right of +z = ( 1-0.95)/2 = 0.025, therefore z = 1.96
3. Given the Normally distributed variable X with mean 18 and standard deviation 2.5, find
a) P(X < 15);
b) the value of k such that P(X < k) = 0.2236;
c) the value of k such that P(X > k) = 0.1814;
d) P( 17 < X < 21).
Ans : X ~ N ( 18, 2.52)
a) P ( X < 15)
P ( Z < (15-18)/2.5) = P ( Z < -1.2) = 0.1151 ( 4 decimal places)
b) P ( X < k) = 0.2236
P ( Z < ( k – 18) / 2.5 ) = 0.2236
From normal table, 0.2236 = -0.76
(k-18)/2.5 = - 0.76, solve k = 16.1
c) P (X > k) = 0.1814
P ( Z > (k-18)/2.5 ) = 0.1814
From normal table, 0.1814 = 0.91
(k-18)/ 2.5 = 0.91, solve k = 20.275
d) P ( 17 < X < 21)
P ( (17 -18)/2.5 < Z < ( 21-18)/2.5)
P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places)...

...√9 = 3), we have
z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2
From the area between z =±2 is 2(0.4772) = 0.9554
Therefore the probability that a measurement selected at random will be between 19 and 31 is about 0.95. This area (probability) is shown fir the X values and for the z values.
σ = 3 0.95 σ = 1 0.95
X
19 25 31 -2 0 +2
Normal curve showing Standardnormal curve showing
area between 19 and 31 area between -2 and +2
Entry to a certain University is determined by a national test. The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100.
Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Will he be admitted to this university?
Solution:
Let x be the random variable that represents the scores. x is normally distributed with a mean of 500 and a standard deviation of 100. The total area under the normal curve represents the total number of students who took the test. If we multiply the values of the areas under the curve by 100, we obtain percentages.
For x = 585, z = (585 - 500) / 100 = 0.85
The proportion P of students who scored below 585 is given by
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom...

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