Normal Distribution Eng
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Transcript of Normal Distribution Eng
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Normal Distribution
A random variable X having a probability density
function given by the formula
gg!¹ º
¸©ª
¨
xe x f
x
,2
1)(
2
2
1
W
Q
T W
is said to have a Normal Distribution with
parameters Q and W 2.
Symbolically, X ~ N( Q , W 2 ).
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Properties of Normal Distribution
1. The curve extends indefinitely to the left and to
the right, approaching the x-axis as x increases
in magnitude, i.e. as xp sw, f (x) p 0.
2. The mode occurs at x= Q.
3. The curve is symmetric about a vertical axis
through the mean Q
4. The total area under the curve and above thehorizontal axis is equal to 1.
i.e.1
2
12
2
1
!¹ º
¸©ª
¨ g
gd x
e
x
W
Q
T W
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Empirical Rule (Golden Rule)
The following diagram illustrates relevant areasand associated probabilities of the NormalDistribution. Approximate 68.3% of the area
lies within Q±W, 95.5% of the area lies withinQ±2W, and 99.7% of the area lies within Q±3W.
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For normal curves with the same W, they are
identical in shapes but the means Q are centered
at different positions along the horizontal axis.
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For normal curves with the same mean Q, the
curves are centered at exactly the same positionon the horizontal axis, but with different standarddeviations W, the curves are in different shapes, i.e. the curve with the larger standard deviation islower and spreads out farther , and the curve with
lower standard deviation and the dispersion issmaller .
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Normal Table
If the random variable X ~ N(Q, W2), then we can
transform all the values of X to the standardized
values Z with the mean 0 and variance 1, i.e. Z ~
N(0, 1), on letting
WQ
! X Z
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Standardizing Process
W
Q!
x z
This can be done by means of the transformation.
The mean of Z is zero and the variance is respectively,
0
])([1
1
)(
!
!
!
¹ º
¸©ª
¨ !
QW
QW
W
Q
X E
X E
X E Z E
1
1
)(1
)(1
)(
2
2
2
2
!
�!
!
!
¹ º
¸©ª
¨ !
W W
W
QW
W
Q
X Var
X Var
X Var Z Var
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Diagrammatic of the Standardizing
Process
Transforms X ~ N(Q, W2) to Z ~ N(0, 1). Whenever X is between the values x=x1 and x=x2, Z will fall
between the corresponding values z=z1 and
z=z2, we have P( x1 < X < x2 ) = P( z1 < Z < z2 ). Itillustrates by the following diagram:
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The normal table can be used to find values like
P(Z > a), P(Z < b) and P(a e Z e b). We illustrate
with the following examples.
Example 1: P(-1.28 < Z < 0) = ?
Solution: P(-1.28 < Z < 0) = P(0 < Z < 1.28)
= 0.3997
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Example 2: P(Z < -1.28) = ?
Solution: P(Z < -1.28) = P(Z > 1.28)
= 0.5 ± 0.3997
=0.1003
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Example 3: P(Z > -1.28) = ?
Solution: P(Z > -1.28) = P(Z < 1.28)
= 0.5 + 0.3997
= 0.8997
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Example 4: P(-2.28 < Z < -1.28) = ?
Solution: P(-2.28 < Z < -1.28) = P(1.28 < Z < 2.28)
= 0.4887 ± 0.3997
= 0.0890
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Example 5: P(-1.28 < Z < 2.28) = ?
Solution: P(-1.28 < Z < 2.28) = 0.3997 + 0.4887
= 0.8884
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Example 6: If P(Z > a) = 0.8, find the value of a?
Solution: From the Normal Table
A(0.84) } 0.3
@ a } - 0.84
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Example 7: If P(Z < b) = 0.32, find the value of b?
Solution: P(Z < b) = 0.32
P( b < Z < 0) = 0.5 ± 0.32
= 0.18
From table, A(0.47) } 0.18
@ b } -0.47
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Example 8: If P(|Z` > c) = 0.1, fin the values of c?
Solution: P(|Z` > c) = 0.1
5 P(Z > c) = 0.05
@ P( c > Z > 0) = 0.5 ± 0.05
= 0.45From table, A(1.645) } 0.45
@ c } 1.645
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Transformation
Example 9: If X ~ N(10, 4), find
a) P(X u 12);
b) P(9.5 e X e 11);c) P(8.5 e X e 9) ?
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Solution: (a) For the distribution of X with Q=10,
W=2
1587.0
3413.05.0
2
1012
)12(
!
!
¹ º
¸
©ª
¨ u
!
u
Z P
X P
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Solution: ( b) For the distribution of X with Q=10,
W=2
P(9.5 e X e 11)
= P(- 0.25e Z e 0.5)= 0.0987 + 0.1915
= 0.2902
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Solution: (c) For the distribution of X with Q=10,
W=2
P(8.5 e X e 9)
= P(- 0.75e Z e - 0.5)= 0.2734 ± 0.1915
= 0.0819
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Example
A sample of 100 dry battery cells tested to find the length
of life produced the following results:
µ= 12 hours = 3 hoursAssuming the data to be normally distributed, what percentage
of battery cells are expected to have life
(i) More than 15 hours (ii) less than 6 hours
(iii) between 10 and 14 hours?
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Solution:
Here x denoted the length of life
of dry battery cells.
Also12
3
x x z
Q
W
! !
(i) When x = 15, z = 1
15 1
0 0 1
.5 .3
413
0 .1587 15 .87%
P x P z
P z P z
" ! "
! g
! ! !
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(ii) When x = 6, z = -2
6 22 0 0 2
.5 .4772 0 .0228 2.28 %
P x P z P z P z P z
!
" ! g
! ! !
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(iii) When x =10, z = 0.67
x =14, z = 0.67
10 14 0 .67 0 .672 0 0 .67
2 * 0 .2485 0 .4970
49 .70%
P x P z P z
!
!
! !
!
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ExampleIn a normal distribution, 31% of the items are under 45
and 8% are over 64. Find the mean and standard deviation
of the distribution.
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1
1
1
2
2
2
. . .
31% 45.
45 .31
45
0 .5 .31 .19
0.5
64 ,
0 .5 .08 0.42
1.4
Let and bethemeanand S D r esp
of theitems ar e und er
Ar ea to the left of the or d inate x is
when x let z z
P z z
F r om t able z
when x let z z
P z z
f r om t able z
Q W
!
! !
! !
!
! !
! !
!
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45 640.5 1.4
45 0.5 64 1.4
50 , 10
xS ince z
and
and
Q
W Q Q
W W
Q W Q W
Q W
!
! !
! !
! !
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Example:In normal distribution, 7% of the items are under 35 and 89%
are under 63. What are the mean and standard deviation of the
distribution?
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? A
1
1
1 1
2
2
2
. . .
7% 35.
35,
0 .5 .07 .43
1.47 0
63,
0 .89 .5 .39
1.23
Let and bethemeanand S D r esp
of theitemsar eund er
W hen x let z z
P z z
f r om t able z z
W hen x let z z
P z z
f r om t able z
Q W
! !
! !
!
! ! ! !
!
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35 631.47 1.23
1.47 35 1.23 63
50.24 10.33
xS ince z
and
and
Q
W
Q Q
W W
W Q W Q
Q W
!
! !
! !
! !
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Example:
It is known from the past experience that the number of telephone
calls made daily in a certain community between 3 p.m. and
4 p.m. have a mean of 352 and a standard deviation of 31.What % of the time will there be more than 400 telephone
calls made in this community between 3 p.m. and 4 p.m.?
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352 31, 400
400 3521.55
31
1.55 0 0 1.55
.5 .4394 .0606
6.06%
H er e given and x
z
P z P z P z
Q W ! ! !
! !
" ! g
! !
!
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Example:
In the examination taken by 500 candidates, the average and the
standard deviation of marks obtained are 40% and 10%. Find
approximately(i) How many will pass, if 50% is fixed as a minimum?
(ii) What should be the minimum if 350 candidates are to pass?
(iii) How many have scored marks above 60% ?
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( ) 40 , 10 50
1 0 0 1
.5 .3413 .1587
. passing .1587 *500 79.35 79
i her e x
P z P Z P Z
no of the cand id ates
Q W ! ! !
" ! g
! !
! ! ;
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1
1
( ) minimum350candidates .
.70
40
1040
.510
35%
ii H er egiven ar eshoul dbe pass
P z z
x z
x
x
" !
!
!
!
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60 4060, 210
2 0 0 2
.5 .4772 .0228
. 60% 500*.022811.4 11
x z
P z P Z P Z
no of cand id ateshavescor ed mark s above
! ! !
" ! g
! !
!
! ;
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