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AS-Level Maths: Statistics 1 for Edexcel S1.6 The normal distribution This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 11 of of 33 33 © Boardworks Ltd 2005 Contents Introduction: Normal distribution Introduction: Normal distribution The standard normal distribution More general normal distributions Solving problems by working backwards 22 of of 33 33 © Boardworks Ltd 2005 Introduction: Normal distribution A sample of heights of 10,000 adult males gave rise to the following histogram: Histogram showing the heights of 10000 males 1400 Frequency 1200 1000 800 600 400 200 0 140 148 156 164 172 180 188 More Height (cm ) Notice that this histogram is symmetrical and bell-shaped. This is the characteristic shape of a normal distribution. 3 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution If we were to draw a smooth curve through the mid-points of the bars in the histogram of these heights, it would have the following shape: This is called the normal curve. The normal distribution is an appropriate model for many common continuous distributions, for example: The masses of new-born babies; The IQs of school students; The hand span of adult females; The heights of plants growing in a field; etc. 4 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution All normal curves are symmetrical and bell-shaped but the exact shape is governed by 2 parameters – the mean, μ, and the standard deviation, σ. 5 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution If X has a normal distribution with mean μ, and variance σ2, we write X ~ N[μ, σ2] y x μ–σ μ+σ 68% of the distribution lies within 1 standard deviation of the mean. 6 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution If X has a normal distribution with mean μ, and variance σ2, we write X ~ N[μ, σ2] y x μ – 2σ μ + 2σ 95% of the distribution lies within 2 standard deviations of the mean. 7 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution If X has a normal distribution with mean μ, and variance σ2, we write X ~ N[μ, σ2] y x μ – 3σ μ + 3σ 99.7% of the distribution lies within 3 standard deviations of the mean. 8 of 33 © Boardworks Ltd 2005 Introduction: Normal distribution As normal distributions always represent continuous data, it only makes sense to find the probability that X takes a value in a particular interval. For example, we could find: P(X ≥ 20); P(–5 < X < 9); P(X = 19 to the nearest whole number), i.e. P(18.5 ≤ X < 19.5). y Probabilities correspond to areas underneath the normal curve. x There is no simple formula that can be used to find the probabilities. Instead, the probabilities are found from tables. 9 of 33 © Boardworks Ltd 2005 Contents The standard normal distribution Introduction: Normal distribution The standard normal distribution More general normal distributions Solving problems by working backwards 10 of 33 © Boardworks Ltd 2005 The standard normal distribution y x -3 -2 -1 1 2 3 The normal distribution with mean 0 and standard deviation 1 is called the standard normal distribution – it is denoted Z. So, Z ~ N[0, 1] Probabilities for this distribution are given in tables. 11 of 33 © Boardworks Ltd 2005 The standard normal distribution Here is an extract from a standard normal distribution table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z This column gives part ofplace the zofvalue. row gives the the nextfirst decimal the z value. The tables are cumulative, i.e. they give P(Z ≤ z). 12 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(Z ≤ 0.54) = 0.7054. So, 13 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(Z > 0.6) = 1 – P(Z ≤ 0.6) = 1 – 0.7257 = 0.2743 14 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(0.25 ≤ Z < 0.78) = P(Z ≤ 0.78) – P(Z ≤ 0.25) = 0.7823 – 0.5987 = 0.1836 15 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(Z > –0.3) = P(Z < 0.3) = 0.6179 16 of 33 Remember that the standard normal distribution is symmetrical around 0. © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(Z ≤ –0.28) = 1 – P(Z ≤ 0.28) = 1 – 0.6103 = 0.3897 17 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z P(–0.08 < Z ≤ 0.85) = P(Z ≤ 0.85) – P(Z ≤ –0.08) = 0.8023 – (1 – 0.5319) = 0.3342 18 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z Find a such that P(Z < a) = 0.6950. We search in the table to find the probability 0.6950. We see that a = 0.51. 19 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z Find b such that P(Z > b) = 0.242. i.e. such that P(Z ≤ b) = 1 – 0.242 = 0.758. We see that b = 0.7. 20 of 33 © Boardworks Ltd 2005 The standard normal distribution Extract from table: 0 1 2 3 4 5 6 7 8 9 0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359 0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753 0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141 0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517 0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224 0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549 0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852 0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133 z Find c such that P(Z < c) = 0.352. c must be negative because P(Z < c) is less than 0.5000. By symmetry, P(Z > |c|) = 0.352 and P(Z ≤ |c|) = 0.648. Therefore c = –0.38. 21 of 33 © Boardworks Ltd 2005 Contents More general normal distributions Introduction: Normal distribution The standard normal distribution More general normal distributions Solving problems by working backwards 22 of 33 © Boardworks Ltd 2005 More general normal distributions It would of course be impractical to publish tables of probabilities for every possible normal distribution. Fortunately, it is possible and easy to transform any normal distribution to a standard normal: If X ~ N[ , ] then Z 2 N[ , 2 ] X ~ N[0,1]. y N[0, 1] y Standardize x x 23 of 33 -3 -2 -1 1 2 © Boardworks Ltd 2005 3 More general normal distributions Example: If X ~ N[20, 16] , find a) P(X < 23); b) P(X > 14); c) P(16 < X < 24.8). a) If σ2 = 16, then σ = 4. Standardize y y 23 20 0.75 4 x 20 23 x 0 0.75 P( X 23) P( Z 0.75) = 0.7734 24 of 33 © Boardworks Ltd 2005 More general normal distributions Example: If X ~ N[20, 16] , find a) P(X < 23); b) P(X > 14); c) P(16 < X < 24.8). b) Standardize y 14 20 y 14 20 1.5 4 x x –1.5 0 P( X 14) P( Z 1.5) P( Z 1.5) = 0.9332 25 of 33 © Boardworks Ltd 2005 More general normal distributions Example: If X ~ N[20, 16] , find a) P(X < 23); b) P(X > 14); c) P(16 < X < 24.8). c) P 16 X 24.8 P 16 20 Z 24.8 20 P(1 Z 1.2) 4 4 y y Standardize x 16 20 24.8 x -1 0 1.2 P(Z < 1.2) = 0.8849 and P(Z < –1) = 1 – P(Z < 1) = 1 – 0.8413 = 0.1587. So, P(–1 < Z < 1.2) = 0.8849 – 0.1587 = 0.7262 26 of 33 © Boardworks Ltd 2005 More general normal distributions Examination-style question: IQs are normally distributed with mean 100 and standard deviation 15. What proportion of the population have an IQ of at least 124? Let X be the random variable for the IQ of an individual. X ~ N[100, 225]. Standardize y y 124 100 1 .6 15 x 100 124 x 0 1.6 So, we want P(X > 124) = P(Z > 1.6) = 1 – P(Z ≤ 1.6) = 1 – 0.9452 = 0.0548 27 of 33 © Boardworks Ltd 2005 Contents Working backwards Introduction: Normal distribution The standard normal distribution More general normal distributions Solving problems by working backwards 28 of 33 © Boardworks Ltd 2005 Working backwards Example: If X ~ N[4, 0.25], find the value of x if P(X < x) = 0.67. To find x, we start by finding the standardized value z such that P(Z < z) = 0.67. From tables we see that z = 0.44. We therefore need to find the value that standardizes to make 0.44 by rearranging the formula. Standardize N[4, 0.25] y y x 4 0.44 0.5 x 4.22 N[0, 1] x 4 29 of 33 x x 0 0.44 © Boardworks Ltd 2005 Working backwards Example: Marks in an examination can be assumed to follow a normal distribution with mean 62 and standard deviation 16. The pass mark is to be chosen so that 86% of candidates pass. Find the pass mark. Let X represent the marks in the examination. X ~ N[62, 256]. We need to find x such that P(X ≥ x) = 0.86. x 62 We need to solve: 1.08. Therefore x = 44.72. 16 y Standardize y x x x –1.08 So, the pass mark is 44. 30 of 33 © Boardworks Ltd 2005 Working backwards Examination-style question: A machine is designed to fill jars of coffee so that the contents, X, follow a normal distribution with mean μ grams and standard deviation σ grams. If P(X > 210) = 0.025 and P(X < 198) = 0.04, find μ and σ correct to 3 significant figures. We are given 2 pieces of information which we can use to form equations: 210 Firstly: 1.96 210 1.96 0.975 μ 31 of 33 0.975 0.025 210 0 1.96 © Boardworks Ltd 2005 Working backwards Secondly, we are told that P(X < 198) = 0.04. This gives the equation: 198 0.04 1.75 198 1.75 0.96 0.96 198 32 of 33 μ –1.75 0 © Boardworks Ltd 2005 Working backwards The two equations are: 210 1.96 198 1.75 Subtracting to eliminate μ: 12 3.71 3.2345 g This gives μ = 210 – 1.96 × 3.2345 = 203.66 g So the solutions to 3 s.f. are μ = 204 g and σ = 3.23 g. 33 of 33 © Boardworks Ltd 2005

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