Transcript Continuous Distributions

Chapter 7
Continuous
Distributions
Continuous random variables
• Are numerical variables
whose values fall within a
range or interval
• Are measurements
• Can be described by
density curves
Density curves
• Is always on or above the
horizontal axis
• Has an area exactly equal to one
underneath it
• Often describes an overall
distribution
• Describe what proportions of the
observations fall within each range
of values
Unusual density curves
• Can be any shape
• Are generic continuous
distributions
• Probabilities are calculated
by finding the area under
the curve
.5
How do you find
the area of a
triangle?
.25
1
2
3
4
2.25 
 .25
P(X < 2) =
2
5
What is the
area of a line
segment?
.5
.25
1
2
P(X = 2) = 0
P(X < 2) = .25
3
4
5
In continuous
distributions,
P(X < 2) & P(X < 2)
areHmmmm…
the same
answer.
Is this different
than discrete
distributions?
Shape is a
trapezoid –
.5
b1How
= .5long are the
bases?
.25
b2 = .375
1
2
4
h = 1
3

b1  b2 h
Area 
5
2
P(X > 3) = .5(.375+.5)(1)=.4375
P(1 < X < 3) =.5(.125+.375)(2) =.5
P(X > 1) = .75
0.50
.5(2)(.25) = .25
0.25
(2)(.25) = .5
1
2
3
4
0.50
P(0.5 < X < 1.5) = .28125
.5(.25+.375)(.5) =
.15625
0.25
(.5)(.25) = .125
1
2
3
4
Special Continuous
Distributions
Uniform Distribution
• Is a continuous distribution that is
evenly (or uniformly) distributed
• Has a density curve in the shape of a
rectangle
• Probabilities are calculated by finding
the area under the curve
a b
x 
2
 x2

b a

12
2
How
do ayou
the
Where:
& bfind
are the
area
endpoints
ofof
thea
rectangle?
uniform
distribution
The Citrus Sugar Company packs sugar in
bags labeled 5 pounds. However, the
packaging isn’t perfect and the actual
What
shape
does a uniform
weights
are
uniformly
distributed
with a
What is the height of this
distribution
have?
mean of 4.98
pounds
and
a
range
of .12
rectangle?
pounds.
How long is this rectangle?
a)Construct the uniform distribution above.
1/.12
4.92
4.98
5.04
•
What is the probability that a
randomly selected bag will weigh
more than 4.97 pounds?
P(X > 4.97) =
.07(1/.12)
= .5833
What is the
length of
the shaded region?
1/.12
4.92
4.98
5.04
• Find the probability that a
randomly selected bag weighs
between 4.93 and 5.03 pounds.
What is the
length of
P(4.93<X<5.03) = .1(1/.12)
= .8333
the shaded region?
1/.12
4.92
4.98
5.04
The time it takes for students to drive
to school is evenly distributed with a
minimum of 5 minutes and a range of 35
minutes.
What is the height of the
rectangle?
a)Draw the distribution
Where should the
rectangle end?
1/35
5
40
b) What is the probability that it takes
less than 20 minutes to drive to
school?
P(X < 20) = (15)(1/35) = .4286
1/35
5
40
c) What is the mean and standard
deviation of this distribution?
 = (5 + 40)/2 = 22.5
2 = (40 - 5)2/12 = 102.083
 = 10.104
Normal Distributions
•
•
•
•
•
Symmetrical bell-shaped (unimodal) density curve
How is this done
Above the horizontal axis
mathematically?
N(, )
The transition points occur at  + 
Probability is calculated by finding the area under
the curve
• As  increases, the curve flattens &
spreads out
• As  decreases, the curve gets
taller and thinner
A
6
B


Do these two normal curves have the same mean?
If so, what is it?
YES
Which normal curve has a standard deviation of 3?
B
Which normal curve has a standard deviation of 1?
A
Empirical Rule
• Approximately 68% of the
observations fall within  of 
• Approximately 95% of the
observations fall within 2 of 
• Approximately 99.7% of the
observations fall within 3 of 
Suppose that the height of male
students at PWSH is normally
distributed with a mean of 71 inches
and standard deviation of 2.5 inches.
What is the probability that the
height of a randomly selected male
student is more than 73.5 inches?
1 - .68 = .32
P(X > 73.5) = 0.16
68%
71
Standard Normal Density
Curves
Always has  = 0 &  = 1
To standardize:
x 
z 

Must have
this
memorized!
Strategies for finding probabilities
or proportions in normal
distributions
1. State the probability
statement
2. Draw a picture
3. Calculate the z-score
4. Look up the probability
(proportion) in the table
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours
and
a standardDraw
deviation
of 15
& shade
Write
the
hours. What
proportion of these
the curve
probability
batteries
can be expected to last less
statement
than 220 hours?
P(X < 220) = .9082
Look up z220
 200
score
in
z 
 1.33
table
15
Calculate z-score
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours and a standard deviation of 15
hours. What proportion of these
batteries can be expected to last more
than 220 hours?
P(X>220) = 1 - .9082
= .0918
220  200
z 
 1.33
15
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours and a standard deviation of 15
Look
up in
table 0.95
hours. How long
must
a battery
last to be
in the top 5%? to find z- score
P(X > ?) = .05
x  200
1.645 
15
x  224.675
.95
.05
1.645
The heights of the female students at
PWSH are normally distributed with a
What
is the zmean of 65 inches. What
is the
for the
standard deviation of this score
distribution
63?
if 18.5% of the female students are
shorter than 63 inches?
P(X < 63) = .185
63  65
 .9 

2
 
 2.22
 .9
-0.9
63
Will my calculator do any
of this normal stuff?
• Normalpdf – use for graphing ONLY
• Normalcdf – will find probability of
area from lower bound to upper
bound
• Invnorm (inverse normal) – will find
z-score for probability
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours and a standard deviation of 15
hours. What proportion of these
batteries can be expected to last less
than 220 hours?
N(200,15)
P(X < 220) =
Normalcdf(-∞,220,200,15)=.9082
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours and a standard deviation of 15
hours. What proportion of these
batteries can be expected to last more
than 220 hours?
N(200,15)
P(X>220) =
Normalcdf(220,∞,200,15) = .0918
The lifetime of a certain type of battery
is normally distributed with a mean of
200 hours and a standard deviation of 15
hours. How long must a battery last to be
in the top 5%?
P(X > ?) = .05
.95
Invnorm(.95,200,15)=224.675
.05
The heights of female teachers at
PWSH are normally distributed with
mean of 65.5 inches and standard
deviation of 2.25 inches. The heights
of male teachers are normally
distributed with mean of 70 inches and
standard deviation of 2.5 inches.
•Describe the distribution of differences
of heights (male – female) teachers.
Normal distribution with
 = 4.5 &  = 3.3634
• What is the probability that a
randomly selected male teacher is
shorter than a randomly selected
female teacher?
P(X<0) =
4.5
Normalcdf(-∞,0,4.5,3.3634 = .0901
Ways to Assess Normality
• Use graphs (dotplots,
boxplots, or histograms)
• Normal probability
(quantile) plot
Normal Probability (Quantile) plots
• The observation (x) is plotted against known
normal z-scores
• If the points on the quantile plot lie close
to a straight line, then the data is normally
distributed
• Deviations on the quantile plot indicate
nonnormal data
• Points far away from the plot indicate
outliers
• Vertical stacks of points (repeated
observations of the same number) is called
granularity
Consider a random sample with
are these
nWhy
= 5.
regions not
To find the appropriate z-scores for a
the same
sample
of
size
5,
divide
the
standard
width?
normal curve into 5 equal-area regions.
These would
be the z-scores
(from
the
Consider
a random
sample
with
standard normal curve) that we would
theto plot our data against.
n Why
= 5.isuse
median not
Next – find the median z-score for
in the
each
region.
“middle” of
each region?
-1.28
0
-.524
1.28
.524
Normal Scores
Let’s
construct
a normal
probability
Suppose
we
have
the
following
Sketch a scatterplot by pairing the
plot.
The
values
of
the
normal
scores
observations
of
widths
of
contact
smallest normal score with the
What
should
depend
oninthe
sample size
n. The
normal
windows
integrated
circuit
chips:
smallest
observation
from
the
data
1
happen
if n = set
scores
when
10 are
below:
& so
on
our data
is
3.21set2.49
2.94 4.38
normally 1
2
3
4
3.62 3.30 2.85 3.34
distributed?
4.02
5
3.81
-1.539-1 -1.001 -0.656 -0.376 -0.123
0.123 0.376 0.656 1.001 1.539
Widths of Contact Windows
Notice that the boxplot
is approximately
symmetrical and that
the normal probability
plot is approximately
Notice that linear.
the boxplot is
approximately
symmetrical except for
the outlier and that
the normal probability
plot shows the outlier.
Notice that the boxplot
is skewed left and
that the normal
probability plot shows
this skewness.
Are these approximately normally
distributed?
50 48 54 47 51 52 46 53
What
52 51 48 48 54 55
57is this
45
53 50 47 49 50 56 called?
53 52
Both the histogram & boxplot
are approximately
symmetrical, so these data
are approximately normal.
The normal probability
plot is approximately
linear, so these data are
approximately normal.