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Stats for Engineers Lecture 6 Stats for Engineers Lecture 6

Stats for Engineers Lecture 6 - PowerPoint Presentation

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Stats for Engineers Lecture 6 - PPT Presentation

Answers for Question sheet 1 are now online httpcosmologistinfoteachingSTAT Answers for Question sheet 2 should be available Friday evening Summary From Last Time Continuous Random Variables ID: 250039

distribution normal pipe probability normal distribution probability pipe diameter random answer graph standard approximation 025 randomly range copper joining

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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

Which best describes your experience of the lectures so far?

Too slow

Speed OK, but struggling to understand many things

Speed OK, I can understand most things

A bit fast, I can only just keep up

Too fast, I don’t have time to take notes though I still follow most of it

Too fast, I feel completely lost most of the time

I switch off and learn on my own from the notes and doing the questions

I can’t hear the lectures well enough (e.g. speech too fast to understand or other people talking)

Stopped prematurely, not many answersSlide36

Course Feedback

What do you think of clickers?

I think they are a good thing, help me learn and make lectures more interesting

I enjoy the questions, but don’t think they help me learn

I think they are a waste of time

I think they are a good idea, but better questions would make them more useful

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

Had difficulty, but workshops helped me to understand

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.