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S1: Chapter 9 The Normal DistributionPowerPoint Presentation

S1: Chapter 9 The Normal Distribution

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The following shows what the probability distribution might look like for a random variable X, if X is the height of a randomly chosen person.

We expect this ‘bell-curve’ shape, where we’re most likely to pick someone with a height around the mean of 180cm, with the probability diminishing symmetrically either side of the mean.

p(x)

180cm

Height in cm (x)

A variable with this kind of distribution is said to have a normal distribution.

For normal distributions we tend to draw the axis at the mean for symmetry.

We can set the mean and the standard deviation of the Normal Distribution.

For a Normal Distribution to be used, the variable has to be:

continuous

With a discrete variable, all the probabilities had to add up to 1.

For a continuous variable, similarly:

the area under the probability graph has

to be 1.

To find P(170cm < X < 190cm), we could:

find the area between these values.

Would we ever want to find P(X = 200cm) say?

Since height is continuous, the probability someone is ‘exactly’ 200cm is infinitesimally small. So not a ‘probability’ in the normal sense.

Q1

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Q2

p(x)

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Q3

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Q4

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170cm

180cm

190cm

Height in cm (x)

If a variable is ‘normally distributed’ (i.e. its probability function uses a normal distribution), then we write:

...is distributed...

...using a Normal distribution with mean and variance

The random variable X...

!

?

The Z value is the number of standard deviations a value is above the mean.

Example

IQ, by definition, is normally distributed for a given population. By definition, and

i.e.

p(x)

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?

?

?

100

IQ (x)

Minimise

A z-table allows us to find the probability that the outcome will be less than a particular z value.

For IQ, P(Z < 2) would mean “the probability your IQ is less than 130”.

(You can find these values at the back of your textbook, and in a formula booklet in the exam.)

Expand

P(Z < 2) = 0.9772

?

z=0

z=1

z=2

100

130

115

IQ (x)

You may be wondering why we have to look up the values in a table, rather than being able to calculate it directly. The reason is that calculating the area under the graph involves integrating (see C2), but the probability function for the normal distribution (which you won’t see here) cannot be integrated!

Suppose we’ve already worked out the z value, i.e. the number of standard deviations above or below the mean.

Bro Tip: We can only use the z-table when:

The z value is positive (i.e. we’re on the right half of the graph)

We’re finding the probability to the left of this z value.

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This is clear by symmetry.

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Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).

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10

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We have seen that in order to look up a value in the table, we needed to first convert our IQ into a value. We call this ‘standardising’ our variable, because we’re turning our normally distributed variable into another one where the mean is 0 and the standard deviation is 1.

Standardise

e.g.

Why is ? Well consider a z value of 3 for example. We understand that to mean 3 standard deviations above the mean. But if and , the 3 is 3 lots of 1 above 0!

The heights in a population are normally distributed with mean 160cm and standard deviation 10cm. Find the probability of a randomly chosen person having a height less than 180cm.

Here’s how they’d expect you to lay out your working in an exam:

No marks attached with this, but good practice!

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A statement of the problem.

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M1 for “attempt to standardise”

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Look up in z-table

Steel rods are produced by with a mean weight of 10kg, and a standard deviation of 1.6kg. Determine the probability that a randomly chosen steel rod has a weight below 9kg.

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Edexcel S1 May 2012

P(Z > -1.6)

= P(Z < 1.6)

= 0.9452

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Edexcel S1 May 2013 (Retracted)

?

P(Z < -0.5)

= P(Z > 0.5)

= 1 – P(Z < 0.5)

= 0.3085

Edexcel S1 Jan 2011

P(Z > 1.6)

= 1 – P(Z < 1.6)

= 0.0548

?

Quickfire Probabilities

(For further practice outside class)

These are for further practice if you’re viewing these slides outside of class. You don’t need a calculator, as the values are obvious!

Let X represent the IQ of a randomly chosen person, where

Use your Z-table to find:

Example:

P(X < 115) = P(Z < 1) = 0.8413

P(X < 107.5) = P(Z < 0.5) = 0.6915

P(X > 122.5) = P(Z > 1.5) = 1 – P(Z < 1.5) = 0.0668

P(X > 85) = P(X < 115) = P(Z < 1) = 0.8413

P(X < 106) = P(Z < 0.4) = 0.6554

P(X > 95) = P(X < 105) = P(Z < 0.33) = 0.6293

P(X < 92.5) = P(Z < -0.5) = P(Z > 0.5) = 1 – P(Z < 0.5) = 0.3085

P(X < 80) = P(Z < -1.33) = P(Z > 1.33) = 1 – P(Z < 1.33) = 0.0918

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Again, let X represent the IQ of a randomly chosen person, where

Thinking about the graph of the normal distribution, find:

P(96 < X < 112)

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This easiest way is to find P(X < 112) and ‘cut out’ the area corresponding to

P(X < 96):

z=0

96

100

112

We can see that, in general:

P(a < Z < b) = P(Z < b) – P(Z < a)

IQ (x)

Quickfire Probabilities

Let X represent the IQ of a randomly chosen person, where

Use your Z-table to find:

P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915

P(145 < X < 151) = P(3 < Z < 3.4) = P(Z < 3.4) – P(Z < 3) = 0.001

P(116 < X < 120) = P(1.07 < X < 1.33) = 0.9082 – 0.8577 = 0.0505

P(80 < X < 110) = P(-1.33 < X < 0.67) = 0.7486 – (1 – 0.9082) = 0.6568

P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486) – (1 – 0.9772) = 0.2286

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Probability

z-value

(num standard deviations above mean)

Original value

or

This essentially says that our value is standard deviations above the mean.

must be positive.

Use Z-table

must be <

The reverse: Finding the z-value for a probability

Sometimes we’re given the probability, and need to find the value, so that we can determine a missing value or the standard deviation/mean.

Just use the z-table backwards!

For nice ‘round’ probabilities, we have to look in the second z-table. You’ll lose a mark otherwise.

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Bro Tip: Remember that either flipping the inequality, or changing the sign of will cause your probability to become 1 minus it.

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The reverse: Finding the z-value for a probability

Again, let X represent the IQ of a randomly chosen person, where

What IQ corresponds to the top 22% of the population?

State the problem in probabilistic terms.

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‘Standardise’.

0.78

Find the closest probability in the z-table.

z = 0.77

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Bro Tip: Draw a diagram for these types of questions if it helps.

We can use this value to find the value of .

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The reverse: Finding the z-value for a probability

Again, let X represent the IQ of a randomly chosen person, where

What IQ corresponds to the top 10% of the population?

?

State the problem in probabilistic terms.

‘Standardise’

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0.9

Find the closest probability in the z-table.

z = 1.2816

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z = 1.2816

Convert from value to IQ.

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The reverse: Finding the z-value for a probability

Again, let X represent the IQ of a randomly chosen person, where

What IQ corresponds to the bottom 30% of the population?

State the problem in probabilistic terms.

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‘Standardise’

(I haven’t written yet because it’ll make the manipulation more tedious. I’ve used a variable to represent the value)

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We can’t look up probabilities less than 0.5 in the table, so manipulate:

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Find the closest probability in the z-table.

Again, use the second table.

z = -0.5244

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Convert back into an IQ.

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On the planet Frostopolis, the mean height of a Frongal is 1.57m and the standard deviation 0.2m. Determine:

The height for which 65% of Frongals have a height less than.

The height for which 40% of Frongals have a height more than.

c) The height for which 23% of Frongals have a height less than.

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Remember:

State your problem in probabilistic terms.

Standardise. Manipulate if necessary so that your probability is above 0.5 and you’re finding .

Use your z-table backwards to find the z-table.

Use to find your value of .

Page 184

Given

a)

b)

Given , find such that

Given that , find such that

. Find the value of and such that:

a)

b)

c)

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P(X = x)

IQ (x)

Q3 = 110

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Q1 = 90

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Q2 = 100

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i.e. The upper quartile is two-thirds of a standard deviation above a mean (useful to remember!)

Edexcel S1 Jan 2011

?

z = -2.3263 (using 2nd z-table)

w = 160 – (5 x 2.3263) = 148.3685

Edexcel S1 May 2012

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P(Z < z) = 0.6

z = 0.2533 (using 2nd table)

W = 162 + (7.5 x 0.2533)

= 163.89975cm

Edexcel S1 May 2010

?

P(D > 20) = P(Z > -1.25) = P(Z < 1.25) = 0.8944

P(Z < z) = 0.75 -> z = 0.67Q3 = 30 + (0.67 x 8) = 35.36

Q1 = 30 – (0.67 x 8) = 24.64

Finding missing and

The random variable

Given that P(X > 20) = 0.20, find the value of .

The random variable

Given that P(X < 46) = 0.2119, find the value of .

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(Normalising)

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If your standard deviation is negative, you know you’ve done something wrong!

Finding missing and

The random variable

Given that P(X > 35) = 0.025 and P(X < 15) = 0.1469, find the value of and the value of

First deal with P(X > 35) = 0.025

Next deal with P(X < 15) = 0.1469

P(X < 35) = 1 – 0.025 = 0.975

If P(Z < z) = 0.975, then z = 1.96.

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We now have two simultaneous equations. Solving gives:

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For the weights of a population of squirrels, which are normally distributed, Q1 = 0.55kg and Q3 = 0.73kg.

Find the standard deviation of their weights.

= 0.114

= 0.124

= 0.134

= 0.144

Due to symmetry, = (0.55 + 0.73)/2 = 0.64kg

If P(Z < z) = 0.75, then z = 0.67.

0.64 + 0.67 = 0.73

= 0.134

Only 10% of maths teachers live more than 80 years. Triple that number live less than 75 years. Given that life expectancy of maths teachers is normally distributed, calculate the standard deviation and mean life expectancy.

RIP

Similarly

– 0.5244 = 75

= 76.15

= 76.25

= 76.35

= 76.45

A Ingall

He loved math

= 2.77

= 2.78

= 79

= 2.80

Edexcel S1 May 2013 (Retracted)

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P(Z < z) = 0.85, then z = 1.04.

So 115 minutes is 1.04 standard deviations above the mean.

Edexcel S1 Jan 2011

Using z-table, 160 is 2.32 standard deviations above the mean. So:

Similarly, z-value for 0.9 is 1.28. By symmetry, 152 has z value of -1.28. So:

Solving, and

?

Edexcel S1 Jan 2002

z-value for 0.975 is 1.96. By symmetry, 235 is 1.96 standard deviations below mean.So . The result follows.

P(Z < z) = 0.85. So z = 1.04

Solving, ,

If 0.683 in the middle, (0.683/2)+0.5=0.8415 prob below value above mean. Thus z = 1.So values are 154.8 – 2.22 and 154.8 + 2.22

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We need to use the second z-table whenever:

we’re looking up the z value for certain ‘round’ probabilities.

E

A normal distribution is good for modelling data which:

tails off symmetrically about some mean/ has a bell-curve like distribution.

A

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P(a < X < b) = P(X < b) – P(X < a)

F

A z-value is:

The number of standard deviations above the mean.

B

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We can treat quartiles and percentiles as probabilities.

For IQ, what is the 85th percentile?

100 + (1.04 x 15) = 115.6

We can form simultaneous equations to find the mean and standard deviation, given known values with their probabilities.

G

If a random variable X is normally distributed with mean 50 and standard deviation 2, we would write:

C

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A z-table is:

A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1.

P(IQ < 115) = 0.8413

D

H

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Edexcel S1 June 2001

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