NormalDistribution

If you observed the heights of people at a busy airport you willnotice that some are short, some are tall, but most would bearound a central value (which would probably depend on theirGender). If you displayed a distribution of heights for men and women,you would get something like:

172cm 178cm

FemaleHeights

MaleHeights

Height is an example of a CONTINUOUS RANDOM VARIABLE.Height is a particular type of random variable whose distributionIs symmetrical in shape – a sort of bell shape.This sort of distribution is called the NORMAL DISTRIBUTION

The characteristics of the normal distribution are: Symmetrical about the mean, μ Mode, median and mean are equal due to the symmetry of the distribution The range of X is from -∞ to +∞ The horizontal axis is asymptotic The total area under the curve is 1 2/3 of the values lie within 1σ of μ 95% lie within 2σ of μ 99.7% lie within 3σ of μμ

How does changing μ and σ affect the normal distribution?

μ1μ2 μ3

μ1μ2 < < μ3

Changing the mean translates the curve along the x-axis.It DOES NOT change its shape.

How does changing μ and σ affect the normal distribution?

μ

σ1σ2 < < σ3Increasing σ makes the curve fatter and shorterDecreasing σ makes the curve thinner and taller.It DOES NOT MOVE.

If a particular random variable, X, has a normal distribution,We define the distribution using the notation:

X ~ N(μ, σ2)If we are finding P(a < X < b), you arefinding an area underneath the curve

a b

mean variance

STANDARD NORMAL DISTRIBUTION

Situations that demonstrate a Normal distribution will demonstratedifferent means and variances and so we need a standardisedmodel that we can transform each situation to, to allow us toperform calculations with ease.

For this, we use the STANDARD NORMAL DISTRIBUTION.It has the properties of any other normal distribution but Has a mean of 0 and a standard deviation (and variance) of 1.We give this particular normal distribution the letter Z.So: Z ~ N(0, 12)Probabilities for this distribution are provided instatistical tables.

Any normal distribution can be transformed onto the standardnormal distribution using the following transformation:

XZ

So, if we have the model X ~ N(10, 42) and want to find P(12<X<16)We can transform the values from X to values on Z by:

)( 1612 XP

41016

41012 ZP

In other words, P(12 < X < 16) on X~ N(10, 42) would be P(0.5 < Z < 1.5) on Z~N(0, 12)

= P(0.5 < Z < 1.5)

12 16 0.5 1.5

1.5 0.5

X ~ N(10, 42) Z ~ N(0, 12)

P(12 < X < 16) P(0.5 < Z < 1.5)

P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5)

).( 51 ).( 50

).().( 5051 = 0.9332 - 0.6915

= 0.2417

If X ~ N(10, 42), what is P(X < 8)?

41088 ZPXP )( = P(Z < - 0.5)

- 0.5

).( 50

0.5

).( 50).().( 50501

0.6915P(X < 8) = 1 – 0.6915

= 0.3085

0.5

Ex 9A Qu 6The random variable, A is defined by A ~ N(28, 32). Find:(a) P(A < 32) (b) P(A > 36) (c) P(28 < A < 35)(d) P(22 < A < 26) (e) P(25 < A < 33)

(a) P(A < 32) =

3

2832ZP = P(Z < 1.33) )33.1( = 0.9082

(b) P(A > 36) =

3

2836ZP = P(Z > 2.67) )67.2(1 = 1 – 0.9962 = 0.0038

2.67

)67.2(

(c) P(28 < A < 35) =

32835

32828 ZP = P(0 < Z < 2.33)

)0()33.2( = 0.9901 – 0.5 = 0.4901

(d) P(22 < A < 26) =

32826

32822 ZP = P(-2 < Z < -0.67)

)]2(1[)]67.0(1[ = (1-0.7486) – (1-0.9772) = 0.2286

(e) P(25 < A < 33) =

32833

32825 ZP = P(-1 < Z < 1.67)

)]1(1[)67.1( = 0.9525 – (1-0.8413) = 0.7938

Ex 9A Qu 13A machine dispenses liquid into cartons in such a way that theamount of liquid dispensed on each occasion is normally distributedwith standard deviation 20ml and mean 266ml. Cartons that weighless than 260ml have to be recycled. What proportion of cartons arerecycled?Cartons weighing more than 300ml produce no profit.What percentage of cartons will this be?

X ~ N(266, 202)

X ~ N(266, 202)

Cartons that weigh less than 260ml have to be recycled.What proportion have to be recycled?

P(X < 260) =

20266260ZP = P(Z < -0.3) )3.0(1

= 1 – 0.6179 = 0.3821 = 38.21%

Cartons weighing more than 300ml produce no profit.What percentage of cartons will this be?

P(X > 300) =

20266300ZP = P(Z > 1.7) )7.1(1

= 1 – 0.9554 = 0.0446 = 4.46%

Ex 9A Qu 15Over a long period it has been found that the breaking strains ofcables produced by a factory are normally distributed with mean6000N and standard deviation 150N. Find, to 3 decimal places, theprobability that a cable chosen at random from the production willhave a breaking strain of more than 6200N.

X ~ N(6000, 1502)P(X > 6200) =

150

60006200ZP = P(Z > 1.33) )33.1(1 = 1 – 0.9082 = 0.0918

= 0.092 to 3dp

Ex 9A Qu 18The thickness of some sheets of wood follows a normal distributionwith mean μ and standard deviation σ. 96% of the sheets will gothrough an 8mm gauge while only 1.7% will go through a 7mmgauge. Find μ and σ.

8

96.0

μ

96.08

ZP

75.18

75.18 -(1)

7

017.0

μ

017.07

ZP

12.27

12.27 -(2)

983.07

ZP

75.18 -(1) 12.27 -(2)

(1)-(2)1 = 3.87σ

σ = 0.258mm

Sub σ = 0.258 into (1)8 – μ = 1.75 x 0.258

μ = 7.5485mm

Packets of breakfast cereal are said to contain 550g. Themanufacturer knows that the weights are normally distributed withmean 551.2g and standard deviation 15g. What proportion of packetswill contain more than the stated weight?

X ~ N(551.2, 152)

P(X > 550) =

152551550 .ZP

= P(Z > -0.08) ).( 0801 )].([ 08011

= 0.5319 = 53.19%

A biologist has studied a particular tropical insect and she has discovered that its life span is normally distributed. The mean lifespan of this insect is 72 days and the standard deviation of its lifespan is eight days. Find the probability that the next insect studied lives:(a) Fewer than 70 days (b) More than 76 days(c) Between 68 and 78 days.

X ~ N(72, 82)

(a) P(X < 70) =

87270ZP = P(Z < -0.25) ).( 2501

= 1 - 0.5987 = 0.4013

X ~ N(72, 82)

(b) P(X > 76) =

87276ZP = P(Z > 0.5) ).( 501

= 1 - 0.6915 = 0.3085

(c) P(68 < X < 78) =

87278

87268 ZP

= P(-0.5 < Z < 0.75)

)].([).( 501750

= 0.7734 – (1 – 0.6915)= 0.4649

The lengths of metal bolts are normally distributed with mean μ and standard deviation 7cm. If 2.5% of the bolts measure more than 68cm, find the value of μ.

X ~ N(μ, 72)

P(X > 68) = 0.025 02507

68 .

ZP

9617

68 . 721368 . cm2854.P(Z > 1.96) = 0.025

Jam is packed in tins of nominal net weight 1kg. The actual weight of jam delivered to a tin by the filling machine is normally distributed about the mean weight set on the machine with a standard deviation of 12g. The average filling of jam is 1kg.(a) Calculate the probability that a tin chosen at random contains less than 985g.

X ~ N(1000, 122)

(a) P(X < 985) =

121000985ZP = P(Z < -1.25)

).( 2511

= 1 - 0.8944 = 0.1056

It is a legal requirement that no more than 1% of tins contain less than the nominal weight.(b) Calculate the minimum setting of the filling machine which will

meet this requirement.

X ~ N(μ, 122)

(b) P(X < 1000) = 0.01 01012

1000 .

ZP

3263212

1000 .

9156271000 .

P(Z > 2.3263) = 0.01 P(Z < -2.3263) = 0.01

g1028