s1: chapter 9 the normal distribution

S1: Chapter 9The Normal Distribution

Dr J Frost ([email protected])www.drfrostmaths.com

Last modified: 7th February 2016

Teacher Guidance

Possible lesson structure:

Lesson 1: Introduction and using z-tables

Lesson 2: Standardising, calculating probability of ranges

Lesson 3: Reverse z-table, calculating values given probabilities

Lesson 4: More on reverse z-tables: Quartiles, dealing with and

Lesson 5 : Unknown and/or

Lesson 6: Harder questions (including conditional probabilities)

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What does it look like?

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.

Height in cm (x)

p(x)

180cm

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.

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.

What does it look like?

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

𝑝(𝑥

)

Height in cm ()180cm

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 , we could:find the area between these values.

Would we ever want to find say?Since height is continuous, the probability someone is ‘exactly’ 200cm is infinitesimally small. So not a ‘probability’ in the normal sense.

Normal Distribution Q & AQ1

Q2

Q3

190cm170cm

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𝑝 (𝑥 )

Q4

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Notation

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

𝑋 𝑁 (𝜇 ,𝜎 2 )

The random variable X...

...is distributed......using a Normal distribution with mean and variance

Example:

“ represents the height of a randomly chosen person, with mean 160cm and standard deviation 10cm.”

𝑋 𝑁 (160,102 )?

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

100

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

IQ ()

Example

IQ Z

100 0

130 2

85 -1

165 4.333

62.5 -2.5

?????

p(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!

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

𝑃 (𝑍<2 )=0.9772?

Z table

IQ ()100 130115

𝑧=0𝑧=1𝑧=2

Expand

Minimise

Use of the z-tableSuppose 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:a) The z value is

positive (i.e. we’re on the right half of the graph)

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

1

𝑧=− 0.3

This is clear by symmetry.?

𝑧

2

𝑧=− 0.3

?𝑧

3

𝑧=1

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𝑧

4

𝑧=−2

?𝑧

Bro Tip: Either changing the sign of changing the direction of the inequality does “1 –”. If we do both, they cancel out.

Test Your Understanding

𝑃 (𝑍>0.70 )=𝟏−𝑷 (𝒁<𝟎.𝟕𝟎 )=𝟎 .𝟐𝟒𝟐𝟎???

Exercise 1Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).

123456789

10

??

??

??

???

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‘Standardising’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.

!

Bro Side Note: , but why? 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!

𝑋=145 𝑍=3IQ (X) world (recall that and )

Z world

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By thinking of what calculation you did, can you therefore come up with a formula for in terms of , and ?

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Example

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

𝑋 𝑁 (160 , 102 )

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.

No marks attached with this, but good practice!

A statement of the problem.

M1 for “attempt to standardise”

Look up in z-table

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Test Your UnderstandingEdexcel S1 May 2012

Edexcel S1 May 2013 (R)𝑋 𝑁 (162,7.52 )

𝑋 𝑁 (150 ,102 )

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Probabilities for RangesAgain, let represent the IQ of a randomly chosen person, where Thinking about the graph of the normal distribution, find:

IQ ()

100

z=0

96 112

This easiest way is to find and ‘cut out’ the area corresponding to :

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! Probabilities of ranges:

Test Your UnderstandingLet 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(123 < X < 151) = P(1.53 < Z < 3.4) = P(Z < 3.4) – P(Z < 1.53) = 0.0627

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

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Exercise 2[Jan 2013 Q4a] The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. Find P(L > 127). = 0.0359

[Jan 2012 Q7a] A manufacturer fills jars with coffee. The weight of coffee, W grams, in a jar can be modelled by a normal distribution with mean 232 grams and standard deviation 5 grams. Find P(W < 224). = 0.0548

[May 2011 Q4a] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.A child from the school is selected at random. Find the probability that this child runs 100 m in less than 15 s. = 0.2420

[Jan 2011 Q8a] The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random. Find the probability that this tin contains more than 168 g. = 0.0548

[May 2010 Q7a] The distances travelled to work, D km, by the employees at a large company are normally distributed with D N( 30, 82 ). Find the probability that a randomly selected employee has a journey to work of more than 20 km. = 0.8944

[May 2009 Q8a,b] The lifetimes of bulbs used in a lamp are normally distributed. A company X sells bulbs with a mean lifetime of 850 hours and a standard deviation of 50 hours.(a) Find the probability of a bulb, from company X, having a lifetime of less than 830 hours. = 0.3446(b) In a box of 500 bulbs, from company X, find the expected number having a lifetime of less than 830 hours. = 172.3

[Jan 2009 Q6a] The random variable X has a normal distribution with mean 30 and standard deviation 5.Find P(X < 39). = 0.9641

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(On provided sheet)

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[May 2008 Q7a] A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg. Find P(X > 53). = 0.0668

The heights of raccoons in a Canadian town are normally distributed with mean 30cm and standard deviation 5cm. Determine the probability of a randomly chose raccoon having a height between:a) 28cm and 31cm

b) 33cm and 37cm

c) 25cm and 28cm

Exercise 2 (On provided sheet)

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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!

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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????

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

Dealing with two-ended inequalitiesSometimes the range is two ended.These can be one of two possible forms in an exam:

Find the value of such that

Bro Tip: The key is to use a diagram (or otherwise) to convert to a single-ended inequality.

𝑎This can be simplified to:

Thus:

0.3

?Suitable Diagram

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𝑎This can be simplified to:

Thus:

0.7

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0.15 0.15−𝑎

?Suitable Diagram



Find the value of such that ?

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We’ll come back to this later…

Where we are so farUnderstand what a Z-value means and the formula to calculate it.

“IQ has the distribution An IQ of 130 corresponds with what z-value?”

Calculate a probability of being above or below a particular value.

“Find the probability that I have an IQ above 115.”

Calculate a z-value corresponding to a probability.

Calculate a value corresponding with a probability.

Calculate a missing value of and/or .

𝑃 (𝑍<𝑧 )=0.58 𝒛=𝟎.𝟐𝟎COMING SOON!

COMING NOW!

“Find the IQ corresponding with the bottom 30% of the population.”

Solve more complex problems (e.g. involving conditional probabilities)

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Retrieving the original value

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

What IQ corresponds to the bottom 78% of the population?

State the problem in probabilistic terms.

𝑃 (𝑍<𝑧 )=0.78

Identify z value(as we did in the previous lesson).𝑧=0.77

Use z formula to find

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0.78

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

z = 0.77

𝑃 (𝑋<𝑥 )=0.78

‘Standardise’.

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‘Standardise’𝑃 (𝑍<𝑧 )=0. 9

Identify z value

z = 1.2816


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?0.9

z = 1.2816

State the problem in probabilistic terms.𝑃 (𝑋<𝑥 )=0.9?

𝑃 (𝑍<−𝑧 )=0.7




𝑃 (𝑋<𝑥 )=0.3


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𝑃 (𝑍<𝑧 )=0.3 ‘Standardise’


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Identify z value

𝑃 (𝑍<−𝑧 )=0. 8



What IQ does 80% of the population have a value more than?

𝑃 (𝑋>𝑥 )=0. 8

Convert back into an IQ.

?Identify z value

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𝑃 (𝑍>𝑧 )=0. 8

‘Standardise’


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Test Your Understanding[May 2008 Q7b] A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg.Find the weight that is exceeded by 99% of the bags. (5)

[May 2011 Q4b] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.On sports day the school awards certificates to the fastest 30% of the children in the

100 m race. Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate. (4)

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Remember:1. State your problem in probabilistic terms.2. Standardise. Manipulate if necessary so

that your probability is above 0.5 and you’re finding .

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

4. Use to find your value of .

Additional Practice (Outside of Class)

Remember:1. State your problem in probabilistic terms.2. Standardise. Manipulate if necessary so that your probability is above 0.5 and you’re finding .3. Use your z-table backwards to find the z value.4. Use to find your value of .

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

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

b) 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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Exercise 3a (On provided sheet)

[Jan 09 Q6b-d] The random variable X has a normal distribution with mean 30 and standard deviation 5.(b) Find the value of d such that P(X < d) = 0.1151. (4) (c) Find the value of e such that P(X > e) = 0.1151. (2) (d) Find P(d < X < e). (2)

[Jan 2013 Q4b] The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. Find the value of d such that P(L < d) = 0.10. (3)

[Jan 2011 Q8b] The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. The weight stated on the tin is w grams. Find w such that P(X < w) = 0.01. (3)

[Ex9C Q6] Given , find such that

[Ex9C Q7] Given , find such that

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Exercise 3a (On provided sheet)

[Jan 2007 Q7b] The measure of intelligence, IQ, of a group of students is assumed to be Normally distributed with mean 100 and standard deviation 15.The probability that a randomly selected student as an IQ of at least 100 + k is 0.2090.Find, to the nearest integer, the value of k. (6)

[June 2005 Q6b,c] A scientist found that the time taken, M minutes, to carry out an experiment can be modelled by a normal random variable with mean 155 minutes and standard deviation 3.5 minutes.(b)P(150 ≤ M ≤ 157), (4)

(c) the value of m, to 1 decimal place, such that P(M ≤ m) = 0.30. (4)

[Solomon Paper E Q3c] The random variable is normally distributed with a mean of 42 and a variance of 18. Find the value of such that

[Ex9C Q9] Given . Find the value of and such that:a)

b)

c)

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Harder reverse probability questions

Recap: Find the value such that

𝑷 ( 𝒁<𝒂)=𝟎 .𝟖𝟓Q: If , find the value of such that

Observe that the 30 at the bottom end of the inequality is the mean. Thus:

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Quartiles and Percentiles

Using IQ: Find:

𝑷 (𝑿<𝑸𝟏 )=𝟎 .𝟐𝟓i.e. 25% of people have an IQ less than the lower quartile.

The Lower Quartile The Upper Quartile

𝑷 (𝑿<𝑸𝟑 )=𝟎 .𝟕𝟓? ?

? ?

The 70th percentile 𝑷 (𝑿<𝑷 𝟕𝟎)=𝟎 .𝟕?


Edexcel S1 May 2012

Edexcel S1 May 2010

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

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

c) Q1 = 30 – (0.67 x 8) = 24.64

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Exercise 3b (On provided sheet)

Finding missing and

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

(Standardising)

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

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

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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

P(X < 35) = 1 – 0.025 = 0.975If P(Z < z) = 0.975, then z = 1.96.

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First deal with P(X > 35) = 0.025 Next deal with P(X < 15) = 0.1469

We now have two simultaneous equations. Solving gives:

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Test your understanding

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.

Due to symmetry, = (0.55 + 0.73)/2 = 0.64kgIf P(Z < z) = 0.75, then z = 0.67.0.64 + 0.67 = 0.73 = 0.134

= 0.114 = 0.124 = 0.134 = 0.144

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.

= 2.77 = 2.78 = 79 = 2.80

= 76.15 = 76.25 = 76.35 = 76.45Similarly – 0.5244 = 75

RIP

A IngallHe loved math

Exam QuestionsEdexcel S1 May 2013 (R)

Edexcel S1 Jan 2011

Edexcel S1 Jan 2002

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

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

c) Solving, , d) 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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Summary

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

A z-value is:The number of standard deviations above the mean.

A z-table is:A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1.

P(IQ < 115) = 0.8413

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.

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F P(a < X < b) = P(X < b) – P(X < a)?

A normal distribution is good for modelling data which:tails off symmetrically about some mean/ has a bell-curve like distribution.

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E We need to use the second z-table whenever:we’re looking up the z value for certain ‘round’ probabilities.?

Mixed QuestionsEdexcel S1 June 2001

s1: chapter 9 the normal distribution

Documents

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probability distribution

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