381 Continuous Probability Distributions (The Normal Distribution-I) QSCI 381 – Lecture 16 (Larson...
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Transcript of 381 Continuous Probability Distributions (The Normal Distribution-I) QSCI 381 – Lecture 16 (Larson...
381
Continuous Probability Distributions(The Normal Distribution-I)
QSCI 381 – Lecture 16(Larson and Farber, Sect 5.1-5.3)
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Introduction The (or Gaussian) distribution
is probably the most used (and abused) distribution in statistics.
Normal random variables are continuous (they can take any value on the real line) so the Normal distribution is an example of a continuous probability distribution.
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The Normal Distribution-I
-4 -3 -2 -1 0 1 2 3 4--2-3 +3+2+
The graph of the normal distribution is called the normal (or bell) curve.
Total area=1
Inflection points
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The Normal Distribution-II The mean, median, and mode are
the same. The normal curve is symmetric
about its mean. The total area under the normal
curve is one. The normal curve approaches, but
never touches, the x-axis.
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The Normal Distribution-III Notation
We say that the random variable X is normally distributed with mean and standard deviation
The equation defining the normal curve is:
f (x) is referred to as a
2~ ( , )X N
2
2
( )
21( )
2
x
f x e
(pdf).
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We do not talk about the probability P[X=x] for continuous random variables. Rather we talk about the probability that the random variable falls in an interval, i.e. P[x1 X x2].
P[x1 X x2] can be determined by finding the area under the normal curve between x1 and x2.
Normal Distributions and Probability-I
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Data values
-3SD -2SD -SD SD 2SD 3SD
Normal Distributions and Probability-II
99.7%
95%
68%
34%
13.5%
Of the total area under the curve about: 68% lies between - and + 95% lies between -2 and +2 99.7% lies between -3 and +3
μ
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Example The mean length of a fish of age 5 is 40cm
and the standard deviation is 4. What is the probability that the length of a randomly selected fish of age 5 is less than 48cm?
48cm is two standarddeviations above themean so the areato the left of 48cm is0.5+0.475 = 0.975. -4 -3 -2 -1 0 1 2 3 4--2-3 +3+2+
This is an approximate method; we will develop a moreaccurate method later.
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The Standard Normal Distribution-I
The normal distribution with a mean of 0 and a standard deviation of 1 is called the
The standard normal distribution is closely related to the z-score:
value - mean
standard deviation
xz
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The Standard Normal Distribution-II
-4 -3 -2 -1 0 1 2 3 4
Area=1
The area under the standard normal distributionfrom - to x: is often listed in tables. can be computed using the EXCEL function NORMDIST.
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Example-I Find the area under the standard
normal distribution between -0.12 and 1.23. We use our previous approach:
P[-0.12 X 1.23] = P[X1.23] - P[X-0.12]
In EXCEL: NORMDIST(1.23,0,1,TRUE)-NORMDIST(-0.12,0,1,TRUE)
NORMDIST(x,,,cum)
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Example-IID
en
sity
-2 0 2
0.0
0.1
0.2
0.3
0.4
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
P[X1.23] P[X-0.12]
P[X1.23]-P[X-0.12]
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Further Examples How would you find the probability
that: X -0.95 X 1 0.1 X 0.5 OR 1.5 X 2 0.1 X 0.5 AND 1.5 X 2
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Generalizing-I We can transform any normal distribution
into a standard normal distribution by subtracting the mean and dividing by the standard deviation.
De
nsi
ty
240 260 280 300 320 340 360
0.0
0.0
05
0.0
15
De
nsi
ty
-2 0 2
0.0
0.1
0.2
0.3
0.4
=300; =20;x=330 =0; =1;z=1.5
Area=0.933 in both cases Z=(x-300)/20
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Generalizing-II To find the probability that X Y if
X is normally distributed with mean and standard deviation . Compute the z-score: z =(Y - )/ Calculate the area under the normal
curve between - and z using tables for the standard normal distribution.
We could calculate this area directly using the EXCEL function NORMDIST.
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Examples-1 The average swimming speed of a fish
population is 2m.s-1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s-1. Its swimming speed is greater than 2.5m.s-1. Its swimming speed is between 2 and 3 m.s-
1.
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Examples-1 The average swimming speed of a fish
population is 2m.s-1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s-1. Its swimming speed is greater than 2.5m.s-1. Its swimming speed is between 2 and 3 m.s-1.
= P(z < (1-2)/.5) = P(z < -2) = 0.0228
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Examples-1 The average swimming speed of a fish
population is 2m.s-1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s-1. Is swimming speed is greater than 2.5m.s-1. Its swimming speed is between 2 and 3 m.s-1.
= P(z > (2.5-2)/.5) = P(z > 1) = 1 - P(z ≤ 1) = 0.159
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Examples-1 The average swimming speed of a fish
population is 2m.s-1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s-1. Its swimming speed is greater than 2.5m.s-1. Its swimming speed is between 2 and 3 m.s-1.
= P( (2-2)/.5 ≤ z ≤ (3-2)/0.5 )= P(0 ≤ z ≤ 2) = P(z ≤ 2) – P(z ≤ 0) = 0.477
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Examples-1 The average swimming speed of a fish
population is 2m.s-1 (standard deviation 0.5). You select a fish at random. What is the probability that: Its swimming speed is less than 1m.s-1. Its swimming speed is greater than 2.5m.s-1. Its swimming speed is between 2 and 3 m.s-1.
Why is the normal distribution not entirely satisfactory for this case?
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Examples-2 In a large group of men 4% are
under 160cm tall and 52% are between 160 cm and 175 cm. Assuming that the heights of men are normally distributed, what are the mean and standard deviation of the distribution?
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In a large group of men 4% are under 160cm tall and 52% are between 160 cm and 175 cm. Assuming that the heights of men are normally distributed, what are the mean and standard deviation of the distribution?
P(z < (160-μ)/σ ) = 0.04 = P( z < -1.75)P( (160-μ)/σ ≤ z ≤ (175-μ)/σ ) = 0.52 0.56 = P(z ≤ (175-μ)/σ ) = P(z ≤ 0.15 )So: -1.75 = (160-μ)/ σ
0.15 = (175- μ)/ σ μ= 174 , σ= 8
Examples-2
We get these from a
standard normal table
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Examples-3 An anchovy net can hold 10 t of anchovy.
The mean mass of anchovy school is 8 t with a standard deviation of 2 t. What proportion of schools cannot be fully caught?
Hint: we are interested in the proportion of schools greater than 10t.
ie. 1- P(X ≤ 10) = 1- P(z ≤ (10-8)/2) OR: = 1 - NORMDIST(10,8,2,TRUE)
= 0.159