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**“Teach A Level Maths” Statistics 1**

Standardizing to z © Christine Crisp

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**( available from Efofex at www.efofex.com )**

S1: Standardizing to z AQA Edexcel Normal Distribution diagrams in this presentation have been drawn using FX Draw ( available from Efofex at ) "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

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We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! ) So, we always standardise to ( Mean = 0, variance = 1 ). This is easy to do. If X is a random variable with distribution then, if Since this formula holds for X, it also holds for all the values of X, given by x. The rule is “subtract the mean and divide by the standard deviation”

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**x = 400, so e.g.1 If X is a random variable with distribution**

find (a) (b) Solution: (a) x = 400, so Tables only give 2 d.p. for z so this is all we need. So,

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**e.g.1 If X is a random variable with distribution**

find (a) (b) Solution: (b) There are 2 values to convert so we use subscripts for z. N.B. This is left of the mean so the z value will be negative. So,

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**e.g.1 If X is a random variable with distribution**

find (a) (b) Solution: (b)

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Tip: The diagrams for X and Z show the same areas so I don’t always draw both. If the question is straightforward I draw only the Z diagram but if I’m not sure what to do I’ll draw the X diagram ( and maybe the Z one as well ). SUMMARY To use tables to solve problems, we convert the values of the random variable X to values of the standardised normal variable using We need to be careful not to confuse standard deviation and variance. e.g means s = 4.

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**e. g. 2 A batch of batteries is claimed to last for 24 hours**

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time? Solution: Let X be the random variable “ life of battery ( hours )”

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**e. g. 2 A batch of batteries is claimed to last for 24 hours**

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time? Solution: Let X be the random variable “ life of battery ( hours )” We want to find

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**e. g. 2 A batch of batteries is claimed to last for 24 hours**

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time? Solution: Let X be the random variable “ life of battery ( hours )” We want to find So, Approximately 20% do not last for 24 hours.

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Exercise 1. If X is a random variable with distribution find (a) (b) (c) 2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ?

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Solutions: (a) 1. ( This is 1 standard deviation above the mean. )

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Solutions: 1. (b)

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Solutions: 1. (c)

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**2. A shop sells curtain rails labelled 90 cm**

2. A shop sells curtain rails labelled 90 cm. In fact the lengths are normally distributed with mean 90·2 cm. and standard deviation 0·4 cm. What percentage of the rails are shorter than 90 cm ? Solution: Let X be the random variable “length of rail (cm)” We want to find Approximately 31% are shorter than 90 cm.

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The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

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We can’t have a table of probabilities for every possible mean and variance ( as we would need an infinite number of tables! ) So, we always standardise to ( Mean = 0, variance = 1 ). This is easy to do. If X is a random variable with distribution then, if The rule is “subtract the mean and divide by the standard deviation” Since this formula holds for X, it also holds for all the values of X, given by x.

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SUMMARY We need to be careful not to confuse standard deviation and variance. e.g means s = 4. To use tables to solve problems, we convert the values of the random variable X to values of the standardised normal variable using

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**x = 400, so e.g.1 If X is a random variable with distribution**

Tables only give 2 d.p. for z so this is all we need. Solution: (a) x = 400, so e.g.1 If X is a random variable with distribution find (a) (b)

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Solution: (b)

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**e. g. 2 A batch of batteries is claimed to last for 24 hours**

e.g.2 A batch of batteries is claimed to last for 24 hours. In fact their running time has a normal distribution with mean time of 29 hours and standard deviation 6 hours. What proportion of batteries do not last for the claimed time? Solution: Let X be the random variable “ life of battery ( hours )” We want to find So, Approximately 20% do not last for 24 hours.

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