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Slide1
Stats for Engineers Lecture 6
Answers for Question sheet 1 are now onlinehttp://cosmologist.info/teaching/STAT/
Answers for Question sheet 2 should be available Friday eveningSlide2
Summary From Last Time
Continuous Random
Variables
Probability Density Function (PDF)
Exponential
distribution
Probability density for separation of random independent events with constant rate
Normal/Gaussian distribution
mean
standard deviation
Slide3
Normal distribution
Consider the continuous random variable X = the weight in pounds of
a randomly
selected
new-born baby.
Suppose that X can be modelled with a normal distribution with mean μ = 7.57 and standard deviation = 1.06. If the standard deviation were
= 1.26 instead, how would that change the graph of the pdf of X?
Question from Derek
Bruff
The graph would be narrower and have a greater maximum value.The graph would be narrower and have a lesser maximum value.The graph would be narrower and have the same maximum value.The graph would be wider and have a greater maximum value.The graph would be wider and have a lesser maximum value.The graph would be wider and have the same maximum value.
Slide4
BUT: for normal distribution cannot integrate analytically.
If
, then
Instead use tables for
standard Normal
distribution:
Slide5
Change of variable
The probability for X in a range around
is for a distribution
is given by
The probability should be the same if it is written in terms of another variable
. Hence
Why does this work?
i.e. change
to
N(0, 1)
-
standard
Normal distribution
Slide6
Use Normal tables for
[also called
]
Outside of exams this is probably best evaluated using a computer package (e.g. Maple,
Mathematica
,
Matlab, Excel); for historical reasons you still have to use tables.
Slide7
zSlide8
Example
:
If
Z
~ N(0, 1):
(a)
888
Slide9
(b)
=
= 0.6915.
Slide10
(c)
=
Slide11
Symmetries
If
, w
hich of the following is NOT the same as
?
Slide12
Symmetries
If
, which of the following is NOT the same as
?
-
-
-
=
=
=
Slide13
(d)
=
= 0.2417Slide14
(e)
Between
and
Using interpolation
Fraction of distance between
1.35 and 1.36:
= 0.6
= 0.4
=0.9125Slide15
(f)
What is
?
Use table in reverse:
.
Interpolating as before
between 0.84 and 0.85
8
2
Slide16
Using Normal tables
The error
(in Celsius) on a cheap digital thermometer has a normal distribution, with
What is
the probability that a given temperature measurement is too cold by more than
C?
0.0618
0.9382
0.1236
0.0735Slide17
Using Normal tables
The error
(in Celsius) on a cheap digital thermometer has a normal distribution, with
That is the probability that a given temperature measurement is too cold by more than
C
?
Answer
:
Want
Slide18
(g) Finding a range of values within which
lies with
probability
0.95:
The answer is not unique; but suppose we want an interval which is symmetric about zero i.e. between and .
0.95
0.05/2=0.025
0.025
So
is where
0.975
0.025+0.95Slide19
Use table in reverse:
95% of the probability is in
the range
Slide20
P=0.025
P=0.025
In general 95% of the probability lies within
of the mean
The range
is called a 95%
confidence interval
.
Slide21
Normal distribution
If
has a Normal distribution with mean
and standard deviation
, which of the following could be a graph of the pdf of
?
Question from Derek
Bruff
Too wide
OKWrong meanToo narrow
1.
2.
3.
4.Slide22
Normal distribution
If
has a Normal distribution with mean
and standard deviation
, which of the following could be a graph of the pdf of
?
1.
2.
3.
4.
Too wide
Correct
Wrong mean
Too narrow
i.e. Mean at
, 95% inside (5% outside) of
i.e.
Slide23
Example
: Manufacturing variability
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222).(i) Find the probability that X
exceeds 14.99 mm.(ii) Within what range will X
lie with probability 0.95?(iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X <
Y).
X
YSlide24
Example
: Manufacturing variability
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222).(i) Find the probability that
X exceeds 14.99 mm.
Answer:
Reminder:
Slide25
Example
: Manufacturing variability
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222).(ii) Within what range will
X lie with probability 0.95?
AnswerFrom previous example
i.e.
lies in
with probability 0.95
Slide26
Where is the probability
We found 95% of the probability lies within
What is the probability that
15.04mm?
0.025
0.05
0.95
0.975
P=0.025
P=0.025Slide27
Example
: Manufacturing variability
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222).(iii)
Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y
).AnswerFor
we want
).
To answer this we need to know the distribution of , where and bothhave (different) Normal distributions
Slide28
Means and variances of independent random variables just add
. Distribution of the sum of Normal variates
A
special property of the Normal distribution is that the distribution of the sum of Normal
variates is also
a Normal distribution. [stated without proof]
Etc.
If
are independent and each have a normal distribution
I
f
are constants then:
E.g.
Slide29
Example
: Manufacturing variability
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222).(iii)
Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y
).AnswerFor
we want
).
Hence
=Slide30
Which of the following would make a random pipe more likely to fit into a random fitting?
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2
) and the fittings for joining the pipe have inside diameter
Y mm, where
Y ~ N(15.07, 0.0222).
Decreasing mean of YIncreasing the variance of XDecreasing the variance of X
Increasing the variance of Y
X
YSlide31
Which of the following would make a random pipe more likely to fit into a random fitting?
The outside diameter,
X
mm, of a copper pipe is N(15.00, 0.02
2
) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.0222
).
X
Y
Common sense.
Larger probability if
-
larger (bigger average gap between pipe and fitting)
-
smaller (less fluctuation in gap size)
, so
is smaller if variance of is decreased
Answer
Or use
Slide32
Normal approximations
Central Limit Theorem:
If
are independent random variables with the same distribution, which has mean
and variance
(both finite), then the sum tends to the distribution
as .
Hence: The sample mean
is distributed approximately as
.
For the approximation to be good, n has to be bigger than 30 or more for skewed distributions, but can be quite small for simple symmetric distributions. The approximation tends to have much better fractional accuracy near the peak than in the tails:
don’t rely on the approximation to estimate the probability of very rare events.It often also works for the sum of non-independent random variables,i.e. the sum tends to a normal distribution (but the variance is harder to calculate)Slide33
Example
: Average of
n
samples from a uniform distribution: Slide34
Answer:
The total weight of passengers is the sum of
individual weights.
Assuming independent:
Example: The mean weight of people in England is μ
=72.4kg, with standard deviation 15kg.The London Eye at capacity holds 800 people at once.
What is the distribution of the weight of the passengers at any random time when the Eye is full?
by the central limit theorem
i.e. Normal with
,
[usual caveat: people visiting the Eye unlikely to actually have independent weights, e.g. families, school trips, etc.]Slide35
Course Feedback
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Too slow
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Stopped prematurely, not many answersSlide36
Course Feedback
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I
think they are a good idea, but
need longer to answer questionsSlide37
Course Feedback
How did you find the question sheets so far?
Challenging but I managed most of it OK
Mostly fairly easy
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Had difficulty and workshops were very little help
I’ve not tried themSlide38
Normal approximation to the Binomial
If
and
is large and
is not too near 0 or 1, then
is approximately
Slide39
p
=0.5Slide40
p
=0.5Slide41
Approximating a range of possible results
from a Binomial distribution
e.g
.
if
[not always so accurate
at such low
!]
Slide42
If
what is the best approximation for
?
i.e. If
,
,
, what is the best approximation for
?
Slide43
Quality control example:
The manufacturing of computer chips produces 10% defective chips. 200 chips are randomly selected from a large production batch. What is the probability that fewer than 15 are defective?
Answer
:
mean variance
. So if is the number of defective chips, approximately
Hence
This compares to the exact Binomial answer
. The Binomial answer is easy to calculate on a computer, but the Normal approximation is
much
easier if you have to do it by hand. The Normal approximation is about right, but not accurate.