The Normal Model
Chapter 6, continued
We use the NORMAL MODEL to approximate real-life distributions that
are UNIMODAL and ROUGHLY SYMMETRIC. (it’s NOT exact!)
the Empirical
Rule
or, as I like to call it, the “68-95-99.7” rule
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus weights is …
a) between 745 and 879 kg?
b) less than 800 kg?
But, what do we
do when we
CAN’T use the
68-95-99.7 rule?!?
We could use calculus...
...to take the integral under the normal curve.
So … instead, we’ll use
the z-table.
(Basically, someone else did the
calculus for a bunch of different
numbers along the entire Normal
Model, and then put all those
numbers in a table for us. Yay!)
z-scores
So … instead, we’ll use
the z-table.
proportions
(represents the
area/proportion of
the graph that is
BELOW the z-score)
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus weights is …
a) less than 879 kg?
z = 879 - 812 = 1 = 1.00
67
LOOK ON Z-TABLE! Find 1.00 in the
row/column headings, write down
the proportion you see…
-> .8413
About 84.13% of female walrus
weights are less than 879 kg.
You MUST
draw and
shade normal
model graph!!
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus
weights is …
*Hmmm, does it look like 31.56% of
our graph is shaded? No! Because the
b) more than 780 kg?
z-table automatically gives you the
z = 780 - 812 = -0.48
67
From z-table:
-> .3156
1 - .3156 = .6844
About 68.44% of female walrus
weights are more than 780 kg.
proportion LESS than, and this time we
want MORE than! No worries, it’s an
easy fix … just subtract from 1 (100%)!
780
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus weights is …
c) less than 600 kg?
Such a tiiiiiiiiiny little
z = 600 - 812 = -3.16
67
part of the graph is
shaded! We better get
a pretty small answer!
-> .0008
About 0.08% of female walrus
weights are less than 600 kg.
600
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus weights is …
d) between 700 and 800 kg?
z = 700 - 812 = -1.67
67
-> .0475
z = 800 - 812 = -0.18
67
-> .5714
To find the area
BETWEEN two values,
subtract the 2
proportions you get
from the 2 z-scores.
700
.5714 - .0475 = .5239
About 52.39% of female walrus weights
are between 700 and 800 kg.
800
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
What proportion of female walrus weights is …
e) between 720 and 785 kg?
z = 720 - 812 = -1.37
67
-> .0853
z = 785 - 812 = -0.40
67
-> .3446
720
.3446 - .0853 = .2593
About 25.93% of female walrus weights
are between 720 and 785 kg.
785
using the z-table in reverse!
In other words,
we KNOW the proportion/percent,
and we want to find the value (for the
walrus example, the weight in kg) that would give
us that proportion.
What z-score corresponds with the 60th percentile?
What about the 10th percentile?
A ‘percentile’ indicates
the percent of the
population that has a
value LESS THAN the
indicated value.
For example… a newborn
baby having a weight in
the 60th percentile means
that roughly 60% of all
newborn babies weigh less
than that baby.
What z-score corresponds with the 60th percentile?
What about the 10th percentile?
around -1.28
around 0.25
To find this … we’re using
the z-table backwards.
Since we want a 60th
percentile (which means
60% of the data is BELOW
that value), we need to find a
proportion of about 0.6000
IN the z-table (where the
proportions are, not on the
edges where the z-scores
are).
Once we find a proportion
that is as close to .6000 as
we can get, we note what
z-score gave us that
proportion.
(This is hard to explain via
typing. I made a video
explaining all things z-table.)
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
f) Approximately what weight represents the cut-off
for the BOTTOM 20% of adult female walrus weights?
First, we need to find a z-score that gives us a
proportion of .2000 (or as close as we can get).
FROM THE Z-TABLE
z = -0.84
20%
??
812
Now, use the z-score formula to solve for x (the observed value):
-0.84 = x - 812
67
So, weights of 755.72 kg and under are the
x = 755.72
bottom 20% of female walrus weights.
or, “What weight
would represent
the 20th
percentile?”
While a drawing may
be helpful for this
type of question, it is
not required. :)
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
g) Approximately what weight represents the cut-off
for the TOP 5% of adult female walrus weights?
The top 5% means that 95% of the weights are below this
particular weight - so we’ll look for .9500 on the z-table.
FROM THE Z-TABLE
z = 1.64
(or 1.65 … they’re equally close - you could even use 1.645)
Now, use the z-score formula to solve for x (the observed value):
1.64 = x - 812
67
So, weights of 921.88 kg and above are the
x = 921.88
top 5% of female walrus weights.
5%
812
??
Adult FEMALE walruses weights are approximately normally
distributed, with a mean of 812 kg and a standard deviation of 67 kg.
g) What is the IQR for adult female walrus weights?
IQR = Q3 - Q1 …
Q3 (or the third quartile) is the 75th percentile. Similarly, Q1 (the first quartile) is the 25th percentile.
So I simply need to find 2 z-scores for proportions of about .2500 and .7500, then use the z-score formula to solve for the 2 weights.
Q3 (looking for .7500)
Q1 (looking for .2500)
0.67 = x - 812
67
-0.67 = x - 812
67
IQR = Q3 - Q1
= 856.89 - 767.11
= 89.78
x = 856.89
x = 767.11
The IQR is 89.78 kg.
Q3 is 856.89 kg
Q1 is 767.11 kg
From z-table: z = 0.67
From z-table: z = -0.67
Would it be appropriate to use the Normal Model for this distribution?
No! To use the Normal
Model, the distribution
should be unimodal and
roughly symmetric.
standardizing (by finding z-scores) does not change the shape of the histogram
If a distribution is NOT approximately normal...
● We CAN calculate a z-score
(and it still represents a number of standard deviations from the mean).
● We CANNOT use the Normal Model to find a
proportion/percent with that z-score.
That means:
○ We CANNOT use 68-95-99.7 rule.
○ We CANNOT use the z-table.
California condors have a mean wingspan of of 9.1 feet, with a standard
deviation of 0.63 feet.
If the distributions of these wingspans is approximately
normal, what is the probability that a randomly selected
condor has a wingspan of …
a)
b)
c)
d)
less than 8 feet?
at least 9.9 feet?
between 8 feet and 10 feet?
Find the cut-off (in feet) for the
largest 25% of wingspans.
California condors have a mean wingspan of of 9.1 feet, with a standard
deviation of 0.63 feet.
If the distributions of these wingspans is approximately
normal, what is the probability that a randomly selected
condor has a wingspan of …
a)
b)
c)
d)
less than 8 feet?
at least 9.9 feet?
between 8 feet and 10 feet?
Find the cut-off (in feet) for the
largest 25% of wingspans.
a)
b)
c)
d)
0.0404
0.1021
0.8830
about 9.53
feet