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