Continuous
Distributions

The distributions that we have looked
at so far have involved DISCRETE
Data
• The outcome was a natural number or a
count
Many Data Sources however do not
always have a discrete outcome
You may have data that is fractional or
has decimal values which is
classified under CONTINUOUS
DATA.
Continuous data also has Probability
Distributions
Ex
PG 414 Investigation
a)
b)
c)
Create a Scatter plot
Sketch a smooth
curve
Describe the
probability
Distribution
Age of
Printer
0-6
6-12
12-18
18-24
24-30
30-36
36-42
42-48
Failure
Rate
4.5
2.4
1.2
1.5
2.2
3.1
4.0
5.8
Failure Rate(%)
9
8
Failure Rate %
7
6
5
Failure Rate(%)
4
3
2
1
0
0
10
20
30
AGE of Computer
40
50
These Probability Distribution graphs are
smooth curves
Positively
Skewed
Negatively
Skewed
Positively Skewed




Tail pulled to the right…the Mean is to the
right of the hump
Mode at hump
# children in families
Growth rate of plants
Negatively Skewed


Tail to the left…mean is to the left of
the hump
Mode at hump
Bimodal




A distribution with 2 groups with
different attributes
Mean in the middle of humps
2 modes
Adult Shoe size or height
The curves represent the probability
density, which is the probability per unit of a
continuous variable
To find the probability that a variable falls in a
particular range of values we determine the
area under the distribution curve
This can be quite difficult, so we have given
curves an equation that models it to allow us
to calculate the area
Uniform Distribution
Ex
The time it takes to drive from Toronto
to North Bay is found to range
evenly between 195-240 minutes.
What is the probability that the drive
will take 210 min?
Since we are told in the question that the time ranges
evenly…THIS is a Uniform Distribution and that every range is
equally likely giving us a straight line for a graph
The area under the curve must be 1
Since Uniform the area can be found using Area of a
rectangle
1 lh
1  (240  195)  h
1
 0.0222
h
45
P( X  210)  0.022(210  195)
 0.022(15)
 0.33
Normal Distribution
Normal Distribution (Bell Curve)


A continuous distribution that is distributed
symmetrical and unimodal about the mean.
The mean and standard deviation intervals are
labelled on the curve Now we are going to discuss
and develop different models for distributions that
will show ALL possible outcomes
-3 - 2 -  
 +  + 2  + 3
Normal Distributions can be completely
described their mean, , and standard
deviation, .
Mode and median can also be used in place
of mean if they are a better
representation of the middle value of the
data.
If the standard deviation of the data is small,
the data is
_________________________
and the bell shape is
___________________.
If the standard deviation of the data is larger,
the data is
_________________________
and the bell shape is
______________________.
A normal Distribution Curve has highly
predictable properties. The areas under the
normal distribution curve are always the
same for each interval of standard deviation.
It does not matter how the curve is stretched.
Example
The weights of 10000 women
athletes are normally
distributed with a mean of 55 kg
and a standard deviation of 5 kg.
  55   5
a) Draw and label the normal distribution
curve that represents the data.
40
45
50
55
60
65
70
b) Find the number of women weighing
between 50 and 60 kg.
40
45
50
55
60
68.2%
65
70
10000  0.682  6820
c) Find the number of women weighing less
than 45kg.
40
45
50
55
2.15%
60
65
70
10000  0.0215  215
d) Find the range of weight for 99.7% of the
population.
Range from 40-70kg
40
45
50
55
60
99.7%
65
70
e) Find the number of athletes that weigh
more than 68 Kg.



68kg will be ½ the distance from 65
to 70 which is 2.15% of the values
Therefore 2.15/2=1.25% will be
more than 68kg
10 000 x .0125=125 people
Example:
Giselle is 168 cm tall. In her high school,
boys’ heights are normally distributed with a
mean of 174cm and a standard deviation of
6cm. What is the probability that the first
boy Giselle meets at school will be taller
than she is?
  174cm
  6cm


You need to find the Probability that
the boys height will be greater than
Giselle's (168cm)
P(168<boys height)
168 174
Boy’s Height (cm)
168 174
Boy’s Height (cm)
P(168  boysheight  174)  34.1%
P(174  boysheight )  50%
P(boy will be taller than Giselle)  34.1%  50%
 84%
Equations and Probabilities for
Normal Distribution

The curve of a normal distribution with
mean
and standard deviation
is
given by

1
f ( x) 
e
 2

1  x 
 

2  
2
Very Scary!!!!




That equation corresponds to the area
under the normal distribution curve.
We use z-scores instead of a normal grid
with squares
Recall z-scores is the number of standard
deviations a value lies above or below the
mean
Recall that z-scores or
z
x

Standard Normal Distribution



Is a special type of normal
distribution where   0 and   1
The areas under this type of normal
curve are known to a very high
degree of accuracy and are printed in
the back of your text book in a table
Pg 606-607
Ex: The mass of a packaged
product is normally distributed with
a mean of 50.5g and a standard
deviation of 0.6g. The company
wants to ensure that each package
has at least 49.5 g of product in it.
What percent of packages do not
contain this much product?


Let X be the mass of product in
package
P(X<49.5) is what we are trying to
find
49.5
50.5
P( X  49.5)
z

x

49.5  50.5

0.6
 1.67

Use the table in
the back of text
book
Find the probability
that a standard
normal variable is
less than this zscore
P( X  49.5)  P(
x

 1.67)
 0.0475
So, 4.8% of the packages will
contain less than 49.5g of product
The mass of 49.5 lies 1.67 standard deviations below the mean
Homework!
Pg 430 #1-4,7,8,9,