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381
Continuous Probability Distributions
(The Normal Distribution-I)
QSCI 381 – Lecture 16
(Larson and Farber, Sect 5.1-5.3)
Introduction
381


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.
381
The Normal Distribution-I
Inflection
points
Total area=1
-4
-3
-3
-2
-2
-
-1
0
+
1
+2
2
+3
3
The graph of the normal distribution
is called the normal (or bell) curve.
4
The Normal Distribution-II
381

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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.
The Normal Distribution-III
381


Notation
X ~ N ( ,  2 )
We say that the random variable X is
normally distributed with mean  and
standard deviation 
The equation defining the normal curve is:

1
f ( x) 
e
2 

( x   )2
2 2
f (x) is referred to as a
(pdf).
Normal Distributions and Probability-I
381
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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.
381
Normal Distributions and Probability-II
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
68%
34%
95%
13.5%
-3SD
-2SD
-SD
99.7%
μ
Data values
SD
2SD
3SD
Example
381
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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 standard
deviations above the
mean so the area
to the left of 48cm is
0.5+0.475 = 0.975.
-4
-3
-3
-2
-2
-
-1
0
This is an approximate method; we will develop a more
accurate method later.
+
1
+2
2
+3
3
4
The Standard Normal Distribution-I
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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
x
z

standard deviation

381
The Standard Normal Distribution-II
Area=1
-4
-3
-2
-1
0
1
2
3
4
The area under the standard normal distribution
from - to x:
is often listed in tables.
can be computed using the EXCEL function NORMDIST.
Example-I
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Find the area under the standard
normal distribution between -0.12 and
1.23.

We use our previous approach:
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In EXCEL:
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P[-0.12  X  1.23] = P[X1.23] - P[X-0.12]
NORMDIST(1.23,0,1,TRUE)-NORMDIST(-0.12,0,1,TRUE)
NORMDIST(x,,,cum)
0.0
0.1
0.2
Density
0.3
0.4
0.0
0.0
0.1
0.1
-2
-2
0
0
2
P[X1.23]-P[X-0.12]
2
0.2
Density
0.2
Density
0.3
0.3
0.4
0.4
381
Example-II
P[X1.23]
P[X-0.12]
-2
0
2
Further Examples
381
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How would you find the probability
that:
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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
Generalizing-I
381
We can transform any normal distribution into a
standard normal distribution by subtracting the
mean and dividing by the standard deviation.
=300; =20;x=330
0.4
=0; =1;z=1.5
0.0
0.1
0.2
Density
0.3
0.015
Density
0.005
0.0

240
260
280
300
320
340
360
-2
Area=0.933 in both cases
0
2
Z=(x-300)/20
Generalizing-II
381
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To find the probability that X Y if X is
normally distributed with mean  and
standard deviation .


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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.
Examples-1
381
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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:
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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.
Examples-1
381

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
381

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:

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

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
Examples-1
381

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
Examples-1
381

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?
Examples-2
381

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?
Examples-2
381

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)
We get these
from a standard
P( (160-μ)/σ ≤ z ≤ (175-μ)/σ ) = 0.52
normal table
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-3
381
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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