Working With Normal Models - math-b
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Transcript Working With Normal Models - math-b
Driving
It takes you 20 minutes, on
average, to drive to school with
a standard deviation of 2
minutes
Suppose a Normal Model is
appropriate for the distribution
of driving times
A) How often will you arrive at
school in less than 22 minutes?
Answer:
68% of the time we’ll be within 1
SD, or two minutes, of the
average 20 minutes.
So 32% of the time we’ll arrive in
less than 18 minutes or in more
than 22 minutes.
Half of those times (16%) will be
greater than 22 minutes, so 84%
will be less than 22 minutes
Driving
It takes you 20 minutes,
on average, to drive to
school with a standard
deviation of 2 minutes
B) How often will it take
you more than 24
minutes?
Answer: 24 minutes is 2
Suppose a Normal Model
is appropriate for the
distribution of driving
times
SD above the mean. By
the 95% rule, we know
2.5% of the times will be
more than 24 minutes
Driving
It takes you 20 minutes, on
C) Do you think the
average, to drive to school
with a standard deviation of
2 minutes
distribution of your driving
times is unimodal and
symmetric?
Suppose a Normal Model is
Answer: “Good” traffic will
appropriate for the
distribution of driving times
speed up your time by a bit
but traffic incidents may
occasionally increase the
time it takes so times may
be skewed to the right and
there may be outliers.
Driving
It takes you 20 minutes,
D) What does the shape of
on average, to drive to
school with a standard
deviation of 2 minutes
the distribution then say
about the accuracy of
your predictions?
Suppose a Normal Model
Answer: If this is the case
is appropriate for the
distribution of driving
times
the Normal Model is not
appropriate and the
percentages we predict
would not be accurate.
SAT Scores Example
Each 800 point section of
the SAT exam is
constructed to have a
mean of 500 and a
Standard Deviation of 100
The distribution of scores
is unimodal and
symmetric
Suppose you earned a 600
on one part of the SAT,
where do you fall among
all students who took the
test?
SAT Scores Example
Does the z-Score tell you
anything?
𝑧=
600−500
100
= 1.00
We need to employ our
Normal Model!
SAT Scores Example
68% of students had scores
that fell no more than 1 SD
from the mean.
100% - 68% = 32% had scores
more than 1 standard deviation
from the mean
Only half of those were on the
high side, so 16% of students
were better than my score of
600.
My score is higher than about
84% of students taking the
exam.
i.e. what if your score was 680?
We Have Two Options:
If the value doesn’t fall at
exactly 1, 2, or 3 SDs from
the mean, we have two
options:
Technology
Tables
Either way, start by
standardizing into z-scores
Table
𝑧=
680−500
100
= 1.80
To Find In Table:
Look down left column
for the first two digits, 1.8
Then across the row for
the third digit, 0
Z
.00
.01
⋮
⋮
⋮
1.7
.9554
.9564
1.8
.9641
.9649
1.9
.9713
.9719
⋮
⋮
⋮
The table gives .9641, this means
96.41% of the z-scores are less than
1.80. Only 3.6% of people, then,
scored better than 680 on the SATs
Calculator!
Look under
2nd DISTR
There are three “norm”
functions:
normalpdf(
normalcdf(
invNorm(
normalpdf( calculatesy-values
for graphing a Normal Curve
We won’t use this one much
but lets try it now,
Y1=normalpdf(X) in a graphing
WINDOW with Xmin=-4, Xmax=4,
Ymin=-0.1, and Ymax=0.5
Calculator!
Normalcdf( finds the
proportion of area under
the curve between two zscore cut points, by
specifying normalcdf(zLeft,
zRight)
You will use this function
often!
Calculator! Example
Example 1
Let’s find the shaded area:
Under 2nd DISTR:
Select normalcdf(
Hit ENTER
Specify the cut points
Normalcdf(-.5,1) and hit ENTER
You get 0.533, and that is the
area between those two
points, 53.3%
Approximately 53% of a Normal Model falls
between half a standard deviation and below
1 standard deviation above the mean
Example 2
Previous SAT example:
We determined the
fraction of scores above
your score of 680
SAT Example
First we need to find z-scores
for the cut points:
680 is 1 SD above the mean,
your z-score is 1.8
This is the left cut-point
The standard Normal extend
rightward forever, but
remember how little of the
data lies beyond +/- 3 SD
Take the upper cut point of
say, 99. Use +/-99 as your
general rule of thumb
Use the command:
Normalcdf(1.8,99)
Answer: .0359302655
So 3.6% of SAT scores are higher
than 680.
Typical Question
What z-score cuts off the
top 10% in a Normal
Model?
Before – find areas
(percentages of total)
from the z-score
Now – Find z-score from
percentages
By Table
From the table, we need an
area of .900
This exact area is not there,
but .8997 is pretty close
This shows up in table with
1.2 in left margin and .08 in
top margin
Z-score for 90th percentile is
1.28
By Calculator
The function we will use is
under 2nd DISTR -> invNorm(
For the top 10%, we take
invNorm(.90) and get:
1.281551567
Only 10% of the area in a
Normal Model is more than
about 1.28 SDs above the
mean
Pesky Word Problems!
(pg 121)
A cereal manufacturer has a
machine that fills cereal boxes
Boxes are labeled 16oz
The machine is never going to
be perfect so there will be
minor variations
If the machine is set to 16oz
and the Normal model applies,
about half the boxes will be
underweight, leading to angry
customers
Pesky Cereal Problem!
THINK
Question: What
portion of cereal boxes
will be under 16 oz?
Variable: Let y =
weight of each cereal box
Use a N(16.3,0.2) model
Pesky Word Problems!
To avoid possible lawsuits
from angry customers, the
company sets the mean to
be a little higher than
16oz
The company believes the
packaging machine fills in
an amount of cereal that
fits a Normal model with a
SD = .2oz
The company decides to
put an average of 16.3oz
of cereal in each box
Question 1: What fraction
of cereal boxes will be
underweight?
Cereal!!!
Convert cut-off value into a z-
score:
𝑧=
𝑦−𝜇
𝜎
=
16−16.3
0.2
= −1.5
Find the area with your
calculator:
Area(y<16)=Area(z<-1.5)=.0668
Cereal…
Question 1: What percent
of cereal boxes will be
underweight?
Answer 1: Approximately
6.7% of the boxes will
contain less than 16oz
Cereal Question 2
The company’s lawyers
say that 6.7% is too high.
They insist that no more
than 4% of cereal boxes
be underweight.
What new mean setting
does the company need to
insure this happens?
Let y=weight of cereal box
I don’t know 𝜇 the mean
amount of cereal. The
standard deviation is still
0.2 oz, so the model is
𝑁(𝜇, 0.2)
Cereal Q2
Solution:
Find a z-score that cuts off
the lowest 4%
Use this information to find
𝜇.
The z-score with .04 area to
the left is z= -1.75
What function finds this?
Cereal Q2
For 16 to be -1.75 SD
below the mean, the
mean must be:
16+1.75(.02)=16.35oz
The company must set the
machine to average
16.35oz of cereal per box
to have less than 4% be
underweight
Cereal……….Question 3!
The President of the cereal
company isn’t happy. He thinks
they should give away less free
cereal, not more.
His goal is to set the machine
no higher than 16.2oz and still
have only 4% underweight
boxes
The only way to accomplish
this is to reduce the standard
deviation. What SD must the
company achieve and what
does this mean about the
machine?
Cereal 3
Question:
What SD will allow
the mean to be 16.2oz
and still have only 4% of
boxes underweight?
Let y = weight cereal in
box
I know the mean, but not
the SD, so use 𝑁(16.2, 𝜎)
Cereal 3
I know the z-score with
4% below it is -1.75 from
previous problem
Sole for 𝜎:
𝑧=
𝑦−𝜇
𝜎
−1.75 =
16−16.2
𝜎
−1.75𝜎 = 0.2
𝜎 = 0.114
Cereal 3
The company must get the
machine to box cereal
with a standard deviation
of only 0.114 ounces. This
means the machine must
be more consistent (by
nearly a factor of 2) at
filling the boxes.
Pg 130, # 17, 19, 21, 37, 41