Download 30a Finding Normal Mean and Variance

“Teach A Level Maths”
Statistics 1
Finding the Normal Mean
and Variance
© Christine Crisp
Finding the Normal Mean and Variance
Statistics 1
EDEXCEL
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Finding the Normal Mean and Variance
The standardising formula to change values from a
random variable X into Z values is
z
x

We can also use this formula to find either  or  ( or
both ) provided we know, or have enough information to
find, the other unknowns.
Finding the Normal Mean and Variance
e.g.1 Find the values of  in the following:
X ~ N (  , 100) and P ( X  50)  0  8
Solution:
P ( X  50)  0  8 means P ( Z  z )  0  8
where
z
x

50  
 z
10
Z
p  0 2
08
Using the Percentage Points of the
Normal Distribution table,
z  0  8416
0 z
50
find
 the z value first or use
It doesn’t matter
whether
we
So, 0  8416 
the standardising
10formula first.
 0  8416 10  50  
  50  0  8416 10

  41  6 ( 3 s. f . )

Finding the Normal Mean and Variance
Tip: It’s easy to make a mistake and add instead of
subtract or vice versa so check that your answer is
reasonable by comparing with the information in the
question.
We had P ( X  50)  0  8
so x = 50
The mean is clearly less than 50
so the answer
  41  6 ( 3 s. f . )
is reasonable.
Z
08

x
Finding the Normal Mean and Variance
e.g.2 Find the values of the unknown in the following:
X ~ N (55,  2 ) and P ( X  70)  0  05
Solution:
where
P ( X  70)  0  05 means P ( Z  z )  0  05
70  55
x
 z
z
Z


Using the Percentage Points of the
Normal Distribution table,
p  0  05  z  1  6449
So,


1  6449 
70  55

1  6449  15
  9  12 (3 s. f . )
0
z
0  05
Finding the Normal Mean and Variance
Exercise
1. Find the values of the unknowns in the following:
(a) X ~ N (  , 25) and
P ( X  95)  0  85
(b) X ~ N (100 ,  ) and P ( X
2
 110)  0  025
Finding the Normal Mean and Variance
Solution:
(a) X ~ N (  , 25) and
P ( X  95)  0  85
P ( Z  z )  0  85
where
z
x

95  
 z
5
Z
p  0  15
0  85
Using the Percentage Points of the
Normal Distribution table,
p  0  15  z  1  0364
95  
So, 1  0364 
5
 1  0364 5  95  
  95  1  0364 5

  89  8 ( 3 s. f . )

0 z
Finding the Normal Mean and Variance
Solution:
(b) X ~ N (100 ,  ) and
P ( X  110)  0  025
P ( Z  z )  0  025
2
where
z
x

 z
110  100
So,


1  96 
0  025

Using the Percentage Points of the
Normal Distribution table,
z  1  9600
Z
110  100

1  96  10
  5  10 ( 3 s. f . )
0
z
Finding the Normal Mean and Variance
In the next two examples  and  are both unknown.
The 2nd of these is set as a problem rather than a
straightforward question.
Finding the Normal Mean and Variance
e.g. 3
Find  and  if X ~ N (  ,  2 )
and P ( X  2)  0  15 and P ( X
Solution:
We have
 7)  0  3
P ( Z  z1 )  0  15 and P ( Z  z 2 )  0  3
x
2


Z
z
 z1 


0  15
0 3
7
z2 

z1 0 z 2
N.B. z 1 is negative
Using the Percentage Points of the Normal Distribution
table,
p  0  15  z   1  0364
1
p  0  3  z 2  0  5244
2
7
So,  1  0364 
and 0  5244 


Finding the Normal Mean and Variance
 1  0364 

2

 1  0364  2  
and
and
0  5244 
7

0  5244  7  
We must solve simultaneously:
 1  0364  2  
0  5244  7  
We can
change the
signsin the
Adding:
1  5608
5 1st equation and add:
   3  20 (3 s. f . )
Substitute into either equation to find :
e.g.
  7  0  5244
   5  32 ( 3 s. f . )
Finding the Normal Mean and Variance
e.g.4 The lengths of a batch of rods follows a Normal
distribution. 10% of the rods are longer than 101 cm. and
5% are shorter than 95 cm. Find the mean length and
standard deviation.
Solution: Let X be the random variable “length of rod (cm)”
X ~ N (,  2 )
X
01
P ( X  101)  0  1 and P ( X  95)  0  05 0  05
So, P ( Z  z 2 )  0  1
P ( Z  z1 )  0  05
95
So,
z 2  1  2816
1  2816 
101  

z1  1  6449
 1  6449 
95  

0 101
Z
01
0  05
z1
0 z2
Finding the Normal Mean and Variance
e.g.4 The lengths of a batch of rods follows a Normal
distribution. 10% of the rods are longer than 101 cm. and
5% are shorter than 95 cm. Find the mean length and
standard deviation.
1  2816 

101  

1  2816  101  
and
 1  6449 
95  

 1  6449  95  
Solving simultaneously:
1  2816  101  
 1  6449  95  
Adding:
We can change
the 2signs
inthe
 9265
 62nd equation and add:
   2  05 cm. ( 3 s.f.)
Substitute into either equation to find :
  98  4 cm. ( 3 s.f.)
Finding the Normal Mean and Variance
SUMMARY
 To find one parameter ( mean or standard deviation )
of a variable with a Normal Distribution we need to
know the other parameter and one “pair” of values of
the variable and corresponding percentage or
probability.
 To find both the mean and standard deviation we
need to know two “pairs” of values of the variable
and corresponding percentages or probabilities.
Finding the Normal Mean and Variance
Exercise
1.
Find the values of  and  if X ~ N (  ,  2 )
and P ( X  80)  0  4 and P ( X  90)  0  1
2.
A large sample of light bulbs from a factory were
tested and found to have a life-time which followed a
Normal distribution. 25% of the bulbs failed in less
than 2000 hours and 15% lasted more than 2200
hours. Find the mean and standard deviation of the
distribution.
Finding the Normal Mean and Variance
1.
X ~ N (  ,  2 ) and
Solution:
P ( X  80)  0  4 and
P ( X  90)  0  1
P ( Z  z1 )  0  4 and P ( Z  z 2 )  0  1
x
80


Z
z
 z1 


04
01
90  
z2 

z
z
We have
0
1
2
Using the Percentage Points of the Normal Distribution
table,
p  0  4  z1  0  2533
p  0  1  z 2  1  2816
80  
90  
So,  0  2533 
and 1  2816 


Finding the Normal Mean and Variance
 0  2533 
80  

  0  2533  80  
and
and
1  2816 
90  

1  2816  90  
Solving simultaneously:
 0  2533 
1  2816 
80  
90  
st equation and add:
We can
change 1the
signs
in
the
1
Adding:
 5349  10

  6  52 (3 s. f . )
Substitute into either equation to find :
  90  1  2816
   81  7 ( 3 s. f . )
e.g.
Finding the Normal Mean and Variance
2.
A large sample of light bulbs from a factory were
tested and found to have a life-time which followed a
Normal distribution. 20% of the bulbs failed in less
than 2000 hours and 15% lasted more than 2200
hours. Find the mean and standard deviation of the
distribution.
Solution: Let X be the random variable “life of bulb (hrs)”
So,
X ~ N (,  2 )
We know that P ( X  2000)  0  2 and P ( X  2200)  0  15
So, P ( Z  z 1 )  0  2 and P ( Z  z 2 )  0  15
z
x

 z1 
z2 
2000  

Z
0  15
0 2
2200  

z1 0
z2
Finding the Normal Mean and Variance
z1 
2000  

z2 
2200  
Z
0  15
0 2

z1 is negative
z1 0
z2
Using the Percentage Points of the Normal Distribution
table,
p  0  2  z1  0  8416
p  0  15  z 2  1  0364
So,  0  8416 
2000  


 0  8416
1  0364
Adding: 1  8780
   106,
and
1  0364 
2200  
 2000  
 2200  
 200
  2090 ( 3 s. f . )

Finding the Normal Mean and Variance
The following slides contain repeats of
information on earlier slides, shown without
colour, so that they can be printed and
photocopied.
For most purposes the slides can be printed
as “Handouts” with up to 6 slides per sheet.
Finding the Normal Mean and Variance
The standardising formula to change values from a
random variable X into Z values is
z
x

We can also use this formula to find either  or  ( or
both ) provided we know, or have enough information to
find, the other unknowns.
Finding the Normal Mean and Variance
e.g.1 Find the values of the unknowns in the following:
(a) X ~ N (  , 100) and P ( X  50)  0  8
Solution:
P ( X  50)  0  8 means P ( Z  z )  0  8
50  
x
Z

z

z

where
08

10
Using the Percentage Points of the
Normal Distribution table,
0 z
p  0 2
p  0  2  z  0  8416
It doesn’t matter whether we
50  
So, 0  8416 
find the z value first or use the
10
standardising formula first.
 0  8416 10  50  
  50  0  8416 10

  41  6 ( 3 s. f . )

Finding the Normal Mean and Variance
Tip: It’s easy to make a mistake and add instead of
subtract or vice versa so check that your answer is
reasonable by comparing with the information in the
question.
We had P ( X  50)  0  8
so x = 50
The mean is clearly less than 50
so the answer
  41  6 ( 3 s. f . )
is reasonable.
Z
08

x
Finding the Normal Mean and Variance
e.g.2 Find the values of the unknown in the following:
X ~ N (55,  2 ) and P ( X  70)  0  05
Solution:
where
P ( X  70)  0  05 means P ( Z  z )  0  05
70  55
x
 z
z
Z


Using the Percentage Points of the
Normal Distribution table,
p  0  05  z  1  6449
So, 1  6449 


70  55

1  6449  15
  9  12 (3 s. f . )
0
z
0  05
Finding the Normal Mean and Variance
e.g.3 The lengths of a batch of rods follows a Normal
distribution. 10% of the rods are longer than 101 cm. and
5% are shorter than 95 cm. Find the mean length and
standard deviation.
Solution: Let X be the random variable “length of rod (cm)”
X ~ N (,  2 )
X
01
P ( X  101)  0  1 and P ( X  95)  0  05 0  05
So, P ( Z  z 2 )  0  1
P ( Z  z1 )  0  05
95
So,
z 2  1  2816
1  2816 
101  

z1  1  6449
 1  6449 
95  

0 101
Z
01
0  05
z1
0 z2
Finding the Normal Mean and Variance

1  2816  101  
 1  6449  95  
Solving simultaneously:
1  2816  101  
 1  6449  95  
We can change the signs in the 2nd equation and add:
1  2816  101  
 1  6449   95  
Adding:
2  9265  6
   2  05 cm. ( 3 s.f.)
Substitute into either equation to find :
  98  4 cm. ( 3 s.f.)