Transcript The normal distribution §6.2 page 237

The Normal Distribution
Limiting smooth bell shaped symmetric curve
is called the Normal p.d.f. curve.
Is symmetric about
the mean.
Mean = Median
50%
50%
Mean 
If a random variable, X, has a Normal
distribution with a mean and a standard
deviation we write:
X ~ Normal ( ,  )
parameters
C6, L2, S2
The Normal Distribution
• The Normal distribution is important because:
– it fits a lot of data reasonably well;
– it can be used to approximate other distributions;
– it is important in statistical inference (see later
work).
C6, L2, S3
GPA’s of WSU Students
Let X be the GPA of randomly selected WSU
student.
Suppose data collected from a recent survey
gave the following:
A sample mean GPA of 3.065
and a sample standard deviation of .473
These are samplebased estimates of
the population
mean  and
standard deviation
.
C6, L2, S6
GPA’s of WSU Students
Approximately 68 % of WSU students will have a
GPA within 1 standard deviation of the mean.
i.e., approximately 68 % of the WSU students have
GPA’s
=
between 3.065 - .473 and 3.065 + .473
=
between
2.592
and
3.538
C6, L2, S7
GPA’s of WSU Students
Approximately 95 % of WSU students will have a
GPA within 2 standard deviation of the mean.
i.e., approximately 95 % of WSU students have
GPA’s
= between 3.065 - 2  .473 and 3.065 + 2  .473
= between
2.119
and
4.011 (4.00)
C6, L2, S8
GPA’s of WSU Students
Approximately 99.73 % of WSU students will have
GPA’s within 3 standard deviation of the mean.
i.e., approximately 99.73% of WSU students have
GPA’s
= between 3.065 - 3  .473 and 3.065 - 3  .473
= between
1.646
and
4.484
(4.00)
C6, L2, S9
The Normal Distribution
For the Normal Distribution:
A random observation has approximately:
– 68% chance of falling within 1 of  ;
– 95% chance of falling within 2 of  ;
– 99.7% chance of falling within 3 of  .
Or:
In a Normal distribution, approximately:
– 68% of observations are within 1 of  ;
– 95% of observations are within 2 of  ;
– 99.7% of observations are within 3 of  .
C6, L2, S10
The Normal Distribution
Probabilities and numbers of standard
deviations
Shaded area = 0.683
-  +
68% chance of
falling between
 -  and  + 
Shaded area = 0.954
 - 2

 + 2
95% chance of
falling between
 - 2 and  + 2
Shaded area = 0.997
 - 3

 + 3
99.7% chance of
falling between
 - 3 and  + 3
C6, L2, S11
Problem: Diagnosing Spina Bifida
For mothers with normal foetuses, the mean level of
alpha fetoprotein is 15.73 moles/litre with a
standard deviation of 0.72 moles/litre.
For mothers carrying foetuses with spina bifida, the
mean is 23.05 and the standard deviation is 4.08.
In both groups the distribution of alpha fetoprotein
appears to be approximately Normally distributed.
Given this
For example
weinformation
might like to
want>to17.8)
be able or
to
find:we P(X
find probabilities
P(19 <with
X < these
25) etc…
associated
distributions.
for either
group.
15.73
23.05
C6, L2, S12
Standard Normal Distribution
FACT: If X ~ Normal( , ) then if we consider a new
random variable (Z) representing the z-scores of X
Z will have a standard normal distribution, i.e.
Z ~ Normal (0 , 1)
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Obtaining Probabilities
Basic method for obtaining probabilities
1. Sketch a Normal curve, marking on the mean
and values of interest.
2. Shade the area under the curve corresponding
to the required probability.
3. Convert all values to their z-scores
4. Obtain the desired probability using tables in
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Original problem:
Diagnosing Spina Bifida
15.73 23.05
Recall:
• For normal foetuses  =15.73,  = 0.72 and
for foetuses with spina bifida  = 23.05 and
 = 4.08.
• Assume the threshold for detecting
spina bifida is set at 17.8.
– (A foetus would be diagnosed as not having
spina bifida if the fetoprotein level is below 17.8)
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Original problem:
Diagnosing Spina Bifida
15.73 23.05
a) What is the probability that a foetus not
suffering from spina bifida is correctly
diagnosed?
Let X be level of fetoprotein in normal foetus
X ~ Normal (15.73, 0.72) What is P(X < 17.8)?
P(X < 17.8) = P(Z < z-score for 17.8)
z-score = (17.8 – 15.73)/.72
= 2.07/.72 =2.88
P(X < 2.88) = .9980
15.73
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17.8
Original problem:
Diagnosing Spina Bifida
15.73 23.05
b) What is the probability that a foetus with
spina bifida is correctly diagnosed?
Let Y be the level of fetoprotein in a spina
bifida foetus. Y ~ Normal (23.05, 4.08)
P(Y > 17.8) = P(Z > z-score for Y = 17.8)
P(Z > -1.29)
= .4015
+ .5000
z-score
= (17.8
– 23.05)/4.08
= 0.9015
= -5.25/4.08 = -1.29
17.8
-1.29
023.05
C6, L2, S25
Original problem:
Diagnosing Spina Bifida
15.73 23.05
If they wanted to ensure that 99% of foetuses with
spina bifida were correctly diagnosed, at what
level should they set T ?
Find a value T so that if
First
theensures
z-score
associated
with T by
T = find
13.54
From
Normal
Table
we 99%
find of foetuses
finding
z so that
with
spina
bifida
will
be
identified.
Y
~
Normal
(23.05,
4.08)
P(-2.33 < Z < 0) = .9900
P(Z < P(Z
z) =<.0100
thus
-2.33) = .0100 as desired.
This
we will
probability
have
is called the
Now convert back to original scale...
sensitivity
.9900 –or4.08
P(Y x< 2.33
T) = =.0100
T = P(Y
+  >x T)
z == 23.05
13.54
C6, L2, S26