S1: Chapter 7�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.

Height in cm (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.

180cm

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190cm

170cm

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The random variable X...

...is distributed...

Example:

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

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Example

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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!

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Expand

Minimise

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:

1. The z value is positive (i.e. we’re on the right half of the graph)
2. 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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Bro Tip: Either changing the sign of changing the direction of the inequality does “1 –”. If we do both, they cancel out.

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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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Z world

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

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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Edexcel S1 May 2012

Edexcel S1 May 2013 (R)

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

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

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

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[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

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[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

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[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

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[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

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[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

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[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

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[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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Sometimes the range is two ended.

These can be one of two possible forms in an exam:

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

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Suitable Diagram

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Suitable Diagram

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We’ll come back to this later…

Understand what a Z-value means and the formula to calculate it.

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

Calculate a z-value corresponding to a probability.

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

State the problem in probabilistic terms.

Identify z value

(as we did in the previous lesson).

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0.78

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

z = 0.77

‘Standardise’.

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What IQ corresponds to the bottom 90% of the population?

‘Standardise’

Identify z value

z = 1.2816

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z = 1.2816

State the problem in probabilistic terms.

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

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‘Standardise’

State the problem in probabilistic terms.

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Identify z value

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

Convert back into an IQ.

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Identify z value

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‘Standardise’

State the problem in probabilistic terms.

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[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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i.e. 25% of people have an IQ less than the lower quartile.

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Edexcel S1 May 2012

Edexcel S1 May 2010

1. P(D > 20) = P(Z > -1.25)� = P(Z < 1.25) = 0.8944
2. P(Z < z) = 0.75 -> z = 0.67�Q₃ = 30 + (0.67 x 8) = 35.36
3. Q₁ = 30 – (0.67 x 8) = 24.64

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(On provided sheet)

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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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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, μ = (0.55 + 0.73)/2 = 0.64kg

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

0.64 + 0.67σ = 0.73

σ = 0.134

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σ = 0.114

σ = 0.124

σ = 0.134

σ = 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.

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σ = 2.77

σ = 2.78

σ = 79

σ = 2.80

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μ = 76.15

μ = 76.25

μ = 76.35

μ = 76.45

RIP

A Ingall

He loved math

Edexcel S1 May 2013 (R)

Edexcel S1 Jan 2011

Edexcel S1 Jan 2002

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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.

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P(IQ < 115) = 0.8413

We can treat quartiles and percentiles as probabilities.

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For IQ, what is the 85^th percentile?

100 + (1.04 x 15) = 115.6

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

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P(a < X < b) = P(X < b) – P(X < a)

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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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We need to use the second z-table whenever:

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

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Edexcel S1 June 2001