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

is said to have a Normal Distribution with

parameters Q and W 2.

Symbolically, X ~ N( Q , W 2 ).

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.

!¹ º

¸©ª

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.

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.

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 .

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

Standardizing Process

This can be done by means of the transformation.

The mean of Z is zero and the variance is respectively,

¸©ª

X E Z E

¸©ª

X Var Z Var

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:

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

Example 2: P(Z < -1.28) = ?

Solution: P(Z < -1.28) = P(Z > 1.28)

= 0.5 ± 0.3997

=0.1003

Example 3: P(Z > -1.28) = ?

Solution: P(Z > -1.28) = P(Z < 1.28)

= 0.5 + 0.3997

= 0.8997

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

Example 5: P(-1.28 < Z < 2.28) = ?

Solution: P(-1.28 < Z < 2.28) = 0.3997 + 0.4887

= 0.8884

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

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

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

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) ?

Solution: (a) For the distribution of X with Q=10,

1587.0

3413.05.0

Solution: ( b) For the distribution of X with Q=10,

P(9.5 e X e 11)

= P(- 0.25e Z e 0.5)= 0.0987 + 0.1915

= 0.2902

Solution: (c) For the distribution of X with Q=10,

P(8.5 e X e 9)

= P(- 0.75e Z e - 0.5)= 0.2734 ± 0.1915

= 0.0819

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?

Solution:

Here x denoted the length of life

of dry battery cells.

Also12

(i) When x = 15, z = 1

0 .1587 15 .87%

P x P z

P z P z

(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

(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

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.

31% 45.

45 .31

0 .5 .31 .19

0 .5 .08 0.42

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

F r om t able z

when x let z z

f r om t able z

45 640.5 1.4

45 0.5 64 1.4

50 , 10

xS ince z

Q W Q W

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?

7% 35.

0 .5 .07 .43

1.47 0

0 .89 .5 .39

Let and bethemeanand S D r esp

of theitemsar eund er

W hen x let z z

f r om t able z z

W hen x let z z

f r om t able z

! ! ! !

35 631.47 1.23

1.47 35 1.23 63

50.24 10.33

xS ince z

W Q W Q

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.?

352 31, 400

400 3521.55

1.55 0 0 1.55

.5 .4394 .0606

H er e given and x

P z P z P z

Q W ! ! !

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% ?

( ) 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 ! ! !

( ) minimum350candidates .

ii H er egiven ar eshoul dbe pass

60 4060, 210

2 0 0 2

.5 .4772 .0228

. 60% 500*.022811.4 11

P z P Z P Z

no of cand id ateshavescor ed mark s above

Normal Distribution Eng

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