Lecture 6: Normal distribution, central limit theorem

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Stats for Engineers Lecture 6
Answers for Question sheet 1 are now online
http://cosmologist.info/teaching/STAT/
Answers for Question sheet 2 should be available Friday evening
Summary From Last Time
(−1.5 <  < −0.7)
Continuous Random Variables
()
Probability Density Function (PDF) ()

 ≤≤ =
∞
  ′ ′

Exponential distribution
   = 1
−∞
 − ,
  =
0,
>0
<0
Probability density for separation of random independent events with constant rate 
Normal/Gaussian distribution
  =
1
2 2

− − 2
22
(−∞ <  < ∞)

: mean
: standard deviation
Normal distribution
Question from Derek Bruff
Consider the continuous random variable X = the
weight in pounds of a randomly selected newborn 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?
1.
2.
3.
4.
5.
6.
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.
∞
   = 1
−∞
82%
12%
3%
1
2
0%
0%
3
4
3%
5
6

 << =
  

BUT: for normal distribution cannot integrate analytically.
Instead use tables for standard Normal distribution:   =
If  ∼  ,  2 , then  =
−

1
2
∼ (0,1)
− 2
 2
Why does this work?
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
   =   
()
⇒  =  
i.e. change  to  =

.

−



⇒
=
⇒  =  + 

2
=
2
−

1
⇒  =  
=

2

2
=
1
2
− 2
 2
− 2
2 2
×
N(0, 1) - standard Normal distribution
=
Use Normal tables for  = ( < ) [also called Φ()]
−

∼ (0,1)

Q z = ≤ =
  
−∞


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.
0 ≤  ≤ 3.59

z
Example:
If Z ~ N(0, 1):
(a)   ≤ 1.22 = Q 1.22
= 0.8888
(b)   > −0.5
=   ≤ 0.5
= Q(0.5)
= 0.6915.
=
(c)   ≤ −1.0
= ( ≥ 1.0)
= 1 − ( < 1.0)
= 1 − Q 1.0 = 1 − 0.8413
= 0.1587
=
Symmetries
If  ∼ (0,1), which of the following is NOT
the same as ( < 0.7)?
1.
2.
3.
4.
1 − ( > 0.7)
( > −0.7)
1 − ( > −0.7)
1 − ( < −0.7)
60%
10%
1
17%
13%
2
3
4
Symmetries
If  ∼ (0,1), which of the following is NOT
the same as ( < 0.7)?
1 −   > 0.7
( > −0.7)
1 − ( > −0.7)
1 − ( < −0.7)

=
-

-
-
=
=

(d)  0.5 <  < 1.5
=   < 1.5 − ( < 0.5)
= Q 1.5 − Q(0.5)
= 0.9332 −0.6915
= 0.2417
=
−
(e)   < 1.356
Between Q 1.35 = 0.9115
and Q 1.36 = 0.9131
Using interpolation
Q 1.356 = A  1.35 + B (1.36)
Fraction of distance between
1.35 and 1.36:
=
1.356 − 1.35
1.36 − 1.35
 =1−
= 0.6
= 0.4
Q 1.356 = 0.4 1.35 + 0.6(1.36)
=0.9125
(f) 0.8 =   ≤  = Q()
What is ?
Use table in reverse:
 between 0.84 and 0.85
Interpolating as before
 =  × 0.084 +  × 0.085
0.8−0.7995
 = 0.8023−0.7995 = 0.18
 = 1 −  = 0.82
⇒  = 0.82 × 0.084 + 0.18 × 0.085
≈ 0.0842.
Using Normal tables
The error  (in Celsius) on a cheap digital thermometer
has a normal distribution, with  ∼  0,1 . What is the
probability that a given temperature measurement is too
cold by more than 1.54∘ C?
1.
2.
3.
4.
0.0618
0.9382
0.1236
0.0735
43%
36%
19%
2%
1.
2.
3.
4.
Using Normal tables
The error  (in Celsius) on a cheap digital thermometer has a normal
distribution, with  ∼  0,1 . That is the probability that a given temperature
measurement is too cold by more than 1.54∘ C?
Answer:
Want   < −1.54
=
= ( > 1.54)
= 1 − ( < 1.54)
= 1 − Q(1.54)
= 1 − 0.9382 = 0.0618
(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
So  is where Q  = 0.975
0.025+0.95
−
0.025
0.975


0.05/2=0.025
Use table in reverse:
Q  = 0.975
⇒  = 1.96
95% of the probability is in
the range
−1.96 <  < 1.96
In general 95% of the probability lies within 1.96 of the mean 
P=0.025
P=0.025
The range  ± 1.96 is called a 95% confidence interval.
Question from Derek Bruff
Normal distribution
If  has a Normal distribution with mean  = 20
and standard deviation  = 4, which of the
following could be a graph of the pdf of ?
1.
2.
3.
4.
45%
32%
20%
2%
1
2
3
4
Normal distribution
If  has a Normal distribution with mean  = 20
and standard deviation  = 4, which of the
following could be a graph of the pdf of ?
1.
2.
Too wide
3.
Correct
4.
Wrong mean
i.e. Mean at  = 20, 95% inside (5% outside) of  ± 2, i.e. 20 ± 8
Too narrow
Example: Manufacturing variability
The outside diameter, X mm, of a copper pipe is N(15.00, 0.022) 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).
Y
X
Example: Manufacturing variability
The outside diameter, X mm, of a copper pipe is N(15.00, 0.022) 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:
∼
,  2
Reminder:
=
=
(15.0, 0.022 )
  > 14.99 =   >
14.99 − 15.0
0.02
=   > −0.5
=   < 0.5 = (0.5)
≈ 0.6915
−

Example: Manufacturing variability
The outside diameter, X mm, of a copper pipe is N(15.00, 0.022) 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?
Answer
From previous example  −1.96 <  < 1.96 = 0.95
i.e.  lies in  ± 1.96 with probability 0.95
⇒  = 15 ± 1.96 × 0.02
⇒ 14.96mm <  < 15.04mm
Where is the probability
We found 95% of the probability lies within
14.96mm <  < 15.04mm
What is the probability that  > 15.04mm?
P=0.025
P=0.025
1.
2.
3.
4.
0.025
0.05
0.95
0.975
71%
14%
11%
4%
1.
2.
3.
4.
Example: Manufacturing variability
The outside diameter, X mm, of a copper pipe is N(15.00, 0.022) 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).
Answer
For  <  we want ( −  > 0).
To answer this we need to know the distribution of  − , where  and  both
have (different) Normal distributions
Distribution of the sum of Normal variates
Means and variances of independent random variables just add.
If 1 , 2 , … ,  are independent and each have a normal distribution  ∼
  , , 2
⇒ 1+2 = 1 + 2
21+2 = 21 + 22
Etc.
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]
If 1 , 2 , … ,  are constants then:
1 1 + 2 2 + ⋯   ∼ (1 1 + ⋯ +   , 12  2 + 22  2 + ⋯ + 2  2 )
E.g.
1 + 2 ∼ (1 + 2 , 12 + 22 )
1 − 2 ∼ (1 − 2 , 12 + 22 )
Example: Manufacturing variability
The outside diameter, X mm, of a copper pipe is N(15.00, 0.022) 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).
Answer
For  <  we want ( −  > 0).
 −  ∼   −  , 2 + 2
=  15.07 − 15,0.022 + 0.0222 = (0.07,0.000884)
Hence
 − >0 =  >
0−0.07
0.0.000884
=   > −2.35
=   < 2.35
=
≈ 0.991
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, 0.022) and the
fittings for joining the pipe have inside
diameter Y mm, where
Y ~ N(15.07, 0.0222).
Y
1.
2.
3.
4.
X
Decreasing mean of Y
Increasing the variance of X
Decreasing the variance of X
Increasing the variance of Y
55%
16%
1
16%
14%
2
3
4
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, 0.022) and the
fittings for joining the pipe have inside
diameter Y mm, where
Y ~ N(15.07, 0.0222).
Y
Answer
Common sense.
Or use  =  −  ∼   −  , 2 + 2
0 − 

 < = >0 = >
= <


Larger probability if
-  larger (bigger average gap between pipe and fitting)
-  smaller (less fluctuation in gap size)
2 = 2 + 2 , so  is smaller if variance of  is decreased
X
Normal approximations
Central Limit Theorem: If 1 , 2 … are independent random variables with the
same distribution, which has mean  and variance  2 (both finite), then the
sum =1  tends to the distribution (,  2 ) as  → ∞.
1
Hence: The sample mean  = 

=1 
is distributed approximately as
2
(,  ).
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)
Example: Average of n samples from a uniform distribution:
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?
Answer:
The total weight  of passengers is the sum of  = 800 individual weights.
Assuming independent:
⇒ by the central limit theorem  ∼ (,  2 )
 = 15Kg,  = 800 ⇒  ∼  800 × 72.4kg, 800 × 152 kg 2
= (58000kg, 180000kg 2 )
i.e. Normal with  = 58000Kg ,  = 180000Kg = 424Kg
[usual caveat: people visiting the Eye unlikely to actually have independent weights, e.g. families, school trips, etc.]
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Normal approximation to the Binomial
If  ∼ (, ) and  is large and  is not too near 0 or 1, then is approximately
 ,  1 −  .
1
2
 = 10
=
1
2
 = 50
=
p=0.5
p=0.5
Approximating a range of possible results from a Binomial distribution
e.g. (6 or fewer heads tossing a coin 10 times) = ( ≤ 6) if  ∼ (10,0.5)
  ≤ 6 =   = 0 +   = 1 + ⋯ +   = 6 = 0.8281
− 2
6.5 − 22

≈
−∞
≈Q
2 2
1.5
2.5
6.5 − 
=Q

= Q 0.9487 = 0.8286
 =  = 5
 2 =  1 −  = 2.5
[not always so accurate
at such low !]
If  ∼ (,  2 ) what is the best approximation for
(3 or more heads when tossing a coin 10 times)?
i.e. If  ∼ (10,0.5),  = 5,  2 =  1 −  = 2.5, what is
the best approximation for ( ≥ 3)?
1. ( > 2.5)
2. ( > 3)
3. ( > 3.5)
60%
27%
13%
1
2
3
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  = 200 × 0.1 = 20
variance  1 −  = 200 × 0.1 × 0.9 = 18.
So if  is the number of defective chips, approximately  ∼  20,18 .
Hence
  < 15 ≈   <
14.5 − 20
18
=   < −1.296 = 1 −   < 1.296
= 1 − [0.9015 + 0.6 × 0.9032 − 0.9015 ] ≈ 0.097
 
− ≈ 0.093. The Binomial
This compares to the exact Binomial answer 14
=0   1 − 
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.

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