“Teach A Level Maths” Statistics 1 The Normal Distribution © Christine Crisp The Normal Distribution Statistics 1 AQA Edexcel Normal Distribution diagrams in the examples and exercises in this presentation have been drawn using FX Draw ( available from Efofex at www.efofex.com ) "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" The Normal Distribution Suppose we have a crate of apples which are to be sorted by weight into small, medium and large. If we wanted 25% to be in the large category, we would need to know the lowest weight a “large” apple could be. ( Here I am using weight in the everyday sense; the quantity measured in kilograms and grams. If you are a physicist you will refer to mass. ) To solve a problem like this we can use a statistical model. A model often used for continuous quantities such as weight, volume, length and time is the Normal Distribution. The Normal Distribution is an example of a probability model. The Normal Distribution Characteristics of the Normal Distribution If we were to show the weights of a large number of our apples in a histogram we might get this: There are not many very light . . . The distribution is fairly symmetric. or very heavy apples. The Normal Distribution Characteristics of the Normal Distribution The Normal distribution model is a symmetric bell-shaped curve. We fit it as closely as possible to the data. The Normal Distribution curve To fit the curve we use the mean, m, and variance, the data. These are the parameters of the model. 2, of If X is the random variable “ the weight of apples”, we write X ~ N (m , 2 ) Reminder: is the standard deviation. The Normal Distribution Characteristics of the Normal Distribution The Normal distribution model is a symmetric bell-shaped curve. We fit it as closely as possible to the data. The Normal Distribution curve For this curve we might have If a question gives me the 2 X ~ N ( 350, 110 ) standard deviation, I often write the variance in this form instead of simplifying. The Normal Distribution The axis of symmetry of the Normal distribution passes through the mean. e.g. X ~ N (4, 1) X ~ N (5, 1) The Normal Distribution A smaller variance “squashes” the distribution closer to the mean. e.g. X ~ N (4, 0 5 2 ) X ~ N (4, 1) The Normal Distribution Finding probabilities When we had a discrete distribution we could find a probability by using a formula. e.g. The r.v. X has probability distribution function (p.d.f.) given by x P( X x) 6 x 1 P (X = x) 1 6 2 2 6 for 3 3 6 x 1, 2, 3 05 P( X x) x 1 2 3 For a continuous distribution, a probability is given by an area under the graph of the p.d.f. The Normal Distribution For example, the probability that an apple taken at random weighs less than 200 grams is given by the area to the left of a line through 200. X ~ N ( 350, 110 2 ) P( X 200) The total area gives the sum of the probabilities so equals 1. ( x m )2 The p.d.f. of the Normal curve is f ( x) 1 2 e 2 2 Since this is a difficult function the probabilities have been worked out and listed in a table. The Normal Distribution Finding Probabilities Before we do an example, find the table of probabilities in your formulae book. The table gives probabilities for the random variable Z where Z ~ N (0, 1) Notice the diagram at the top of the page. The shading shows that the probabilities given are always less than the z value used. ( I will usually write the values to 4 d.p. ) Since we may need to use these values to find others, we will always draw a sketch. The Normal Distribution e.g.1 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 6) (c) P (1 Z 2) Solution: (a) P ( Z 1) Z ~ N (0, 1) (use the table ) 0 8413 0 1 (b) P ( Z 1 6) 1 P ( Z 1 6) P ( Z 1 6) is a lot to write so we can write (1·6) which means the area to the left of 1·6. Z ~ N (0, 1) (1 6) 0 1 6 The Normal Distribution e.g.1 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 6) (c) P (1 Z 2) Solution: (a) P ( Z 1) Z ~ N (0, 1) (use the table ) 0 8413 0 1 (b) P ( Z 1 6) 1 (1 6) 1 0 9452 0 0548 Z ~ N (0, 1) (1 6) Tip: It’s useful to always check 0 1 6 the answer: an area less than 0·5 corresponds to less than half the area and vice versa. The Normal Distribution e.g.1 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 6) (c) P (1 Z 2) Z ~ N (0, 1) (c) P (1 Z 2) Solution: ( 2) (1) 0 1 2 Z ~ N (0,1) ( 2 ) 0 1 2 The Normal Distribution e.g.1 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 6) (c) P (1 Z 2) Z ~ N (0, 1) (c) P (1 Z 2) Solution: ( 2) (1) (use the table ) 0 9773 0 8413 0 1 2 Z ~ N (0,1) (1) 0 1360 0 1 2 The Normal Distribution Special Cases • The probability of any single value is zero. e.g. P ( Z 1) 0 Z ~ N (0, 1) ( There is no area. ) 0 1 • Also, using e.g. is the same as using < P ( Z 1) P ( Z 1) Neither property holds for discrete distributions. The Normal Distribution SUMMARY The Normal Distribution is continuous and symmetric about the mean. An area under the curve gives a probability. The formula booklet gives a table of probabilities for the random variable Z where Z ~ N (0, 1) The table gives values of P ( Z z ). This is written as (z ). P( Z z) (z ) 0 z When using the table we always draw a sketch showing the required probability. The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Since we want 1 ( the standard deviation ), the variance, ,2 is also equal to 1. The mean can be any value, so let Z ~ N (0, 1) . We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1. Z Z P ( 1 Z 1) 1 0 1 1 The easiest way to find this area is to find and subtract ( 0) . . . 0 1 (1) . . . The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Since we want 1 ( the standard deviation ), the variance, ,2 is also equal to 1. The mean can be any value, so let Z ~ N (0, 1) . We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1. Z Z P ( 1 Z 1) 1 0 1 1 The easiest way to find this area is to find and subtract ( 0) . . . then multiply by 2. 0 1 (1) . . . The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Since we want 1 ( the standard deviation ), the variance, ,2 is also equal to 1. The mean can be any value, so let Z ~ N (0, 1) . We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1. Z Z P ( 1 Z 1) 1 0 Can you see what 1 (0) equals without (0) 0 5 1 0 1 using the table? The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Since we want 1 ( the standard deviation ), the variance, ,2 is also equal to 1. The mean can be any value, so let Z ~ N (0, 1) . We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1. Z Z P ( 1 Z 1) 1 0 1 1 0 1 (1) (0) 0 8413 0 5000 0 3413 P (1 Z 1) 2 0 3413 0 6826 The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Since we want 1 ( the standard deviation ), the variance, ,2 is also equal to 1. The mean can be any value, so let Z ~ N (0, 1) . We can find the percentage from the probability, so we need the probability that Z is between – 1 and + 1. Z P ( 1 Z 1) 1 P (1 Z 1) 0 6826 0 1 The percentage is approximately 68%. The Normal Distribution Exercise 1. If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1 3) (b) P ( Z 1 45) (c) P (0 8 Z 1 8) 2. Find the percentage of the Normal distribution that lies within (a) 2 standard deviations either side of the mean and (b) 3 standard deviations either side of the mean. The Normal Distribution Solutions: 1(a) P ( Z 1 3) (1 3) 0 9032 Z 0 1(b) P ( Z 1 45) 1 3 Z 1 (1 45) 1 0 9265 0 0735 0 1 45 The Normal Distribution Solutions: 1(c) P (0 8 Z 1 8) Z (1 8) (0 8) 0 9641 0 7881 0 1760 0 0 8 1 8 The Normal Distribution 2. Find the percentage of the Normal distribution that lies within (a) 2 standard deviations either side of the mean and (b) 3 standard deviations either side of the mean. Solution: Let Z ~ N (0, 1) (a) We want P (2 Z 2) ( 2) (0) 0 97725 0 5000 0 47725 P (2 Z 2) 2 0 47725 0 9545 Z 2 0 2 Approximately 95% of the Normal distribution lies within 2 standard deviations of the mean. (b) The method is the same. The answer is approx. 99·8%. The Normal Distribution SUMMARY The percentages of the Normal Distribution lying within the given number of standard deviations either side of the mean are approximately: 1 s.d. : 68% 68% m 2 s.d. : 95% 3 s.d. : 99·8% 99·8% 95% 2 m 2 3 m 3 The Normal Distribution e.g.3 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 5) (c) P (1 Z 2) Solution: (a) P ( Z 1) Z 1 0 The table only gives probabilities for positive z values so we have to find an equal area that is in the table. P ( Z 1) P ( Z 1) (1) 0 8413 Z 1 0 1 The Normal Distribution Solution: (b) P ( Z 1 5) Z 1 5 0 This area equals 1 (1 5) 1 0 9332 0 0668 Z 0 1 5 The Normal Distribution Solution: (c) P ( 1 Z 2) Z ( 2) (1) Tip: Work out (1) first or you could make a sign error. ( 1) (1) 1 0 8413 0 1587 So, 1 0 2 Z (1) (1) 1 0 1 ( 2) ( 1) 0 9773 0 1587 0 8186 The Normal Distribution SUMMARY To find ( z ) , the area to the left of a negative number, we use ( z ) 1 ( z ) N.B. The procedure for this and all other areas involving negative values can be seen from the diagram. NEVER try to do these questions without at least one diagram. The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) Z ~ N (0, 1) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) There are 2 methods of doing part (d). See if you can spot them both and use the quicker. The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) Z ~ N (0, 1) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) Solution: (a) P ( Z 1 3) Z Z = 1 3 0 0 P ( Z 1 3) (1 3) 0 9032 1 3 The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) Solution: (b) P ( Z Z ~ N (0, 1) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) 1) Z Z = 1 0 P ( Z 1) 1 (1) 1 0 8413 0 1587 0 1 The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) Z ~ N (0, 1) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) Solution: (c) P (2 Z 1 2) (1 2) Z 0 2 0 2) 1 2 P (2 Z 1 2) (1 2) ( 2) ( 2) 1 ( 2) 1 0 9773 0 0227 P (2 Z 1 2) 0 8849 0 0227 0 8622 2 0 1 2 The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) Z ~ N (0, 1) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) Solution: (d) P (1 26 Z 0 34) Method 1: Area equals (1 26) (0 34) Z 1 26 0 34 0 34 1 26 0 8962 0 6331 0 2631 The Normal Distribution Exercise 1. If Z is a random variable with distribution find (a) (c) (b) P ( Z 1) P ( Z 1 3) P (2 Z 1 2) (d) P (1 26 Z 0 34) Solution: (d) Z 1 26 0 34 Z ~ N (0, 1) Method 2: P (1 26 Z 0 34) P (1 26 Z 0 34) ( 0 34) ( 1 26) ( 0 34) 1 (0 34) 1 0 6331 0 3669 ( 1 26) 1 (1 26) 1 0 8962 0 1038 P(1 26 Z 0 34) 0 3669 0 1038 0 2631 The Normal Distribution 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. The Normal Distribution SUMMARY The Normal Distribution is continuous and symmetric about the mean. An area under the curve gives a probability. The formula booklet gives a table of probabilities for the random variable Z where Z ~ N (0, 1) The table gives values of P ( Z z ). This is written as (z ). P( Z z) (z ) 0 z When using the table we always draw a sketch showing the required probability. The Normal Distribution Special Cases • The probability of any single value is zero. e.g. P ( Z 1) 0 Z ( There is no area ) 0 1 • Also, using e.g. is the same as using < P ( Z 1) P ( Z 1) Neither property holds for discrete distributions. The Normal Distribution e.g.1 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 6) (c) P (1 Z 2) Solution: (a) Z P ( Z 1) (use the table ) 0 8413 0 1 (b) P ( Z 1 6) 1 (1 6) Z P ( Z 1 6) Z (1 6) 0 1 6 1 0 9452 0 0548 0 1 6 The Normal Distribution Solution: (c) P (1 Z 2) Z ( 2) (1) P (1 Z 2) 0 1 2 Z~ ( 2 ) 0 1 2 Z (1) 0 1 2 0 9773 0 8413 0 1360 The Normal Distribution e.g. 2 Find the percentage of the Normal distribution that lies within 1 standard deviation on either side of the mean. Solution: Let Z ~ N (0, 1) We can find the percentage from the probability, so we need the probability that Z is between m – 1 and m + 1 where m is the mean and equals zero. Z Z P ( 1 Z 1) 1 0 1 1 0 1 (1) (0) 0 8413 0 5000 0 3413 P (1 Z 1) 2 0 3413 0 6826 The percentage is approximately 68%. The Normal Distribution SUMMARY The percentages of the Normal Distribution lying within the given number of standard deviations either side of the mean are approximately: 1 s.d. : 68% 68% m 2 s.d. : 95% 3 s.d. : 99·8% 99·8% 95% 2 m 2 3 m 3 The Normal Distribution e.g.3 If Z is a random variable with distribution Z ~ N (0, 1) find (a) P ( Z 1) (b) P ( Z 1 5) (c) P (1 Z 2) Solution: (a) P ( Z 1) Z 1 0 The table only gives probabilities for positive z values so we have to find an area equal to this that is in the table. P ( Z 1) P ( Z 1) (1) 0 8413 Z 1 0 1 The Normal Distribution Solution: (b) P ( Z 1 5) Z 1 5 0 This area equals 1 (1 5) 1 0 9332 0 0668 Z 0 1 5 The Normal Distribution Solution: (c) P ( 1 Z 2) Z ( 2) (1) Tip: Work out first. (1) ( 1) (1) 1 0 8413 0 1587 So, 1 0 2 Z (1) (1) 1 0 1 ( 2) ( 1) 0 9773 0 1587 0 7186 The Normal Distribution SUMMARY To find ( z ) , the area to the left of a negative number, we use ( z ) 1 ( z ) N.B. The procedure for this and all other areas involving negative values can be seen from the diagram. NEVER try to do these questions without at least one diagram.