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A2-Level Maths: Statistics 2 for Edexcel S2.2 Continuous random variables This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 61 © Boardworks Ltd 2006 Relative frequency histograms Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 2 of 61 © Boardworks Ltd 2006 Relative frequency histograms A histogram has the important property that the area of each bar is in proportion to the corresponding frequency. A histogram with unequal widths can be drawn by plotting the frequency density on the vertical axis, where: frequency density = frequency class width A relative frequency histogram is one in which the area of a bar corresponds to the proportion of the data falling into the corresponding interval. The relative frequency is defined as: relative frequency density = 3 of 61 frequency density total frequency © Boardworks Ltd 2006 Relative frequency histograms Example: The relative frequency histogram below shows the distribution of ages of the UK population. Relative frequency density Relative frequency histogram to show the age distribution in the UK 0.018 0.016 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 0 10 20 30 40 50 60 70 80 90 100 Age The area of each bar corresponds to the proportion of the population with ages in that interval. The total area of all the bars is 1. 4 of 61 © Boardworks Ltd 2006 Relative frequency histograms Relative frequency histogram to show the age distribution in the UK Relative frequency density 0.02 0.015 0.01 0.005 0 0 10 20 30 40 50 60 70 80 90 100 Age The distribution of the ages can be modelled by a curve. This curve is called a probability density function. 5 of 61 © Boardworks Ltd 2006 Probability density functions Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 6 of 61 © Boardworks Ltd 2006 Probability density functions Relative frequency density A probability density function (or p.d.f.) is a curve that models the shape of the distribution corresponding to a continuous random variable. 0.02 0.015 0.01 0.005 0 0 10 20 30 40 50 60 70 80 90 100 Age A probability density function has several important properties. 7 of 61 © Boardworks Ltd 2006 Probability density functions If f(x) is the p.d.f corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold: 0.02 1. b f ( x)dx 1 a 0.015 0.01 0.005 0 0 10 20 30 40 50 60 70 80 90 100 i.e. the total area under a p.d.f. is 1. 8 of 61 © Boardworks Ltd 2006 Probability density functions If f(x) is the p.d.f. corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold: 2. f ( x) 0 for a x b 0.02 0.015 0.01 0.005 0 0 10 20 30 40 50 60 70 80 90 100 i.e. the graph of the p.d.f. never dips below the x-axis. 9 of 61 © Boardworks Ltd 2006 Probability density functions If f(x) is the p.d.f. corresponding to a continuous random variable X and if f(x) is defined for a ≤ x ≤ b then the following properties must hold: x2 3. P( x1 X x2 ) f ( x)dx x1 i.e. probabilities correspond to areas under the curve. 10 of 61 © Boardworks Ltd 2006 Probability density functions Example: Sketch the graph of each of the following functions. Decide in each case whether it could be the equation of a probability density function: 1. 1 24 x 3 x 2 36 f ( x) 32 0 2. 1 3 f ( x) x 0 3. x2 4 x 3 0 x 4 f ( x) 0 otherwise 11 of 61 2 x6 otherwise x 1 x 1 © Boardworks Ltd 2006 Probability density functions 1 24 x 3 x 2 36 1. f ( x) 32 0 2 x6 otherwise The function is non-negative everywhere. If f represents a p.d.f., then f ( x)dx 1 all x 6 1 16 2 3 2 12 x x 36 x 2 f ( x)dx (24 x 3 x 36)dx 32 all x 32 2 1 (432 216 216) (48 8 72) 1 32 So f(x) could represent a probability density function. 12 of 61 © Boardworks Ltd 2006 Probability density functions 1 2. f ( x) x3 0 x 1 x 1 The function is non-negative everywhere. For f to represent a p.d.f. we need to check that f ( x)dx 1 all x 1 1 2 1 1 3 But: 3 dx x dx x 0 1 x 1 2 2 2 1 So f(x) could not represent a probability density function. 13 of 61 © Boardworks Ltd 2006 Probability density functions 3. x2 4 x 3 0 x 4 f ( x) 0 otherwise The function is clearly negative for some values of x. Consequently f(x) cannot represent a probability density function. 14 of 61 © Boardworks Ltd 2006 Probability density functions 15 of 61 © Boardworks Ltd 2006 Question 1 Question 1: A continuous random variable X is defined by the probability density function k (5 x) 0 x 5 f ( x) otherwise 0 a) Sketch the probability density function. b) Find the value of the constant k. c) Find P(1 ≤ X ≤ 3). 16 of 61 © Boardworks Ltd 2006 Question 1 k (5 x) 0 x 5 a) f ( x) otherwise 0 The diagram shows the probability density function. b) To find k, we can use the property that f ( x)dx 1 all x 5 1 k (5 x)dx k 5 x x2 k (25 12.5) 0 12.5k 0 2 0 5 2 Therefore, 12.5k = 1 k 25 17 of 61 © Boardworks Ltd 2006 Question 1 c) P(1 ≤ X ≤ 3) 3 2 3 2 2 x (5 x)dx 5 x 25 25 2 1 1 2 (15 4.5) (5 0.5) 25 = 0.48 18 of 61 © Boardworks Ltd 2006 Examination-style question 1 Examination-style question 1: A continuous random variable X is defined by the probability density function 1 x 3 k ( x 1) f ( x) k (5 x)( x 2) 3 x 5 0 otherwise a) Sketch the probability density function. b) Find the value of the constant k. c) Find P(X > 2). 19 of 61 © Boardworks Ltd 2006 Examination-style question 1 a) 1 x 3 k ( x 1) f ( x) k (5 x)( x 2) 3 x 5 0 otherwise 20 of 61 © Boardworks Ltd 2006 Examination-style question 1 f ( x)dx 1 b) To find k, we can use the property that Note that (5 – x)(x – 2) = 7x – 10 – So, 3 1 k ( x 1)dx 3 5 all x x2 k 7 x 10 x2 dx 1 3 5 1 3 1 2 7 2 k x x k x 10 x x 1 3 3 2 1 2 1 1 1 1 k 1 k 4 7 1 2 2 2 6 1 Therefore 5 k 1 3 21 of 61 i.e. k 3 16 © Boardworks Ltd 2006 Examination-style question 1 5 c) P(X > 2) = f ( x)dx 2 3 ( x 1)dx 2 16 3 3 7 x 10 x 2 dx 3 16 5 3 5 3 1 2 3 7 2 1 3 x x x 10 x x 16 2 3 3 2 16 2 22 of 61 3 1 1 3 1 1 0 4 7 16 2 16 6 2 29 32 An alternative method would be to utilise P(X > 2) = 1 – P(X ≤ 2) © Boardworks Ltd 2006 Examination-style question 2 Examination-style question 2: The life, T hours, of an electrical component is modelled by the probability density function ke0.001t f (t ) 0 t 1000 otherwise a) Sketch the probability density function. b) Find the value of the constant k. c) Find P(1500 ≤ T ≤ 2000). 23 of 61 © Boardworks Ltd 2006 Examination-style question 2 Solution: a) 24 of 61 © Boardworks Ltd 2006 Examination-style question 2 ke0.001t f (t ) 0 t 1000 otherwise f (t )dt 1 b) To find k, we use the fact that So: all t ke0.001t dt 1 1000 k 1000e 0.001t 1000 1 k 0 (1000e1) 1 Therefore 25 of 61 k 1 1000e1 e 0.00272 1000 © Boardworks Ltd 2006 Examination-style question 2 ke0.001t f (t ) 0 t 1000 otherwise 2000 c) P(1500 ≤ T ≤ 2000) = 2000 f (t )dt 1500 k e0.001t dt 1500 k 1000e 0.001t 2000 1500 e 1000e2 1000e1.5 1000 = 0.239 (3 s.f.) 26 of 61 © Boardworks Ltd 2006 Mode Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 27 of 61 © Boardworks Ltd 2006 Mode Suppose that a random variable X is defined by the probability density function f(x) for a ≤ x ≤ b. The mode of X is the value of x that produces the largest value for f(x) in the interval a ≤ x ≤ b. A sketch of the probability density function can be very helpful when determining the mode. Differentiation could be used to find the mode here. 28 of 61 © Boardworks Ltd 2006 Mode Example: A random variable X has p.d.f. f(x), where x 2 (2 x ) 0 x 2 f ( x) 0 otherwise Find the mode. Sketch of f(x): 29 of 61 © Boardworks Ltd 2006 Mode The mode can be found using differentiation: f ( x) 2x2 x3 f ( x) 4x 3x2 To find a turning point, we solve f ( x) 0 4 x 3 x2 0 Factorize: x(4 3 x) 0 x 0 or x 4 3 Check that x 4 3 gives the maximum value: f ( x) 4 6 x f ( 4 3) 4 8 4 0 So the mode is x 4 3 . 30 of 61 © Boardworks Ltd 2006 Cumulative distribution functions Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 31 of 61 © Boardworks Ltd 2006 Cumulative distribution functions The cumulative distribution function (c.d.f.) F(x) for a continuous random variable X is defined as F(x) = P(X ≤ x). Therefore, the c.d.f. is found by integrating the p.d.f.. Example: A random variable X has p.d.f. f(x), where 1 3 x 1 f ( x) 6 0 0 x2 otherwise Find the c.d.f. and find P(X < 1). 32 of 61 © Boardworks Ltd 2006 Cumulative distribution functions Solution: The c.d.f., F(x) is given by: 1 3 1 F( x) f ( x)dx x dx 6 6 1 4 1 x xc 24 6 To find the constant c we can use the fact that P(X ≤ 0) = 0 (because the random variable X is only nonzero between 0 and 2) Therefore F(0) = 0, i.e. c = 0. 33 of 61 © Boardworks Ltd 2006 Cumulative distribution functions So the c.d.f. is 0 x0 1 4 1 F( x) 24 x 6 x 0 x2 1 x2 1 1 5 P(X < 1) = F(1) = 24 6 24 34 of 61 © Boardworks Ltd 2006 Median and quartiles Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 35 of 61 © Boardworks Ltd 2006 Median and quartiles The median, m, of a random variable X is defined to be the value such that F(m) = P(X ≤ m) = 0.5 where F is the cumulative distribution function of X. Likewise the lower quartile is the solution to the equation F(x) = 0.25 and the upper quartile is the solution to F(x) = 0.75. 36 of 61 © Boardworks Ltd 2006 Median and quartiles 37 of 61 © Boardworks Ltd 2006 Median and quartiles Example: A random variable X is defined by the cumulative distribution function: 0 1 2 F( x) 24 x x 6 1 x2 2 x5 x5 a) Calculate and sketch the probability density function. b) Find the median value. c) Work out P(3 ≤ X ≤ 4). 38 of 61 © Boardworks Ltd 2006 Median and quartiles a) We can get the p.d.f. by differentiating the c.d.f. 1 x 1 f ( x) F( x) 12 24 So the p.d.f. is 1 x 1 12 2 x5 24 f ( x) otherwise 0 Sketch of f(x): 39 of 61 © Boardworks Ltd 2006 Median and quartiles b) The median, m, satisfies F(m) = 0.5. Therefore 1 24 m2 m 6 0.5 m2 m 6 12 m2 m 18 0 1 1 4 1 (18) m 2 m 4.77 or m 3.77 The median must be 3.77 (as the p.d.f. is only non-zero for values in the interval [2, 5]). 40 of 61 © Boardworks Ltd 2006 Median and quartiles c) P(3 ≤ X ≤ 4) = F(4) – F(3) 1 42 4 6 1 32 3 6 24 24 7 1 12 4 1 3 41 of 61 © Boardworks Ltd 2006 Median and quartiles Example: A random variable X has p.d.f. f(x), where 3 x2 3 x 3 1 x 2 2 4 4 f ( x) 3 3 x 2 x4 2 8 0 otherwise a) Calculate the cumulative distribution function and verify that the lower quartile is at x = 2. b) Work out the median value of X. 42 of 61 © Boardworks Ltd 2006 Median and quartiles a) For f ( x) 34 x2 32 x 34 , F( x) 41 x3 34 x2 34 x c We know that P(X ≤ 1) = 0, i.e., that F(1) = 0. So, 1 4 34 34 c 0 c 41 Therefore F( x) 41 x3 34 x2 34 x 41 3 x2 c For f ( x) 32 38 x, F( x) 32 x 16 We know that P(X ≤ 4) = 1, i.e. that F(4) = 1. 3 42 c 1 c 2 So 32 4 16 3 x2 2 Therefore F( x) 32 x 16 43 of 61 © Boardworks Ltd 2006 Median and quartiles 0 x 1 1 x3 3 x 2 3 x 1 1 x 2 4 F( x) 4 3 4 3 2 4 2 x4 2 x 16 x 2 1 x4 So To verify that the lower quartile is 2, we simply need to check that F(2) = 0.25: 1 3 3 2 3 1 F(2) 2 2 2 0.25 4 4 4 4 Therefore the lower quartile is 2. 44 of 61 © Boardworks Ltd 2006 Median and quartiles b) The median, m, must lie in the interval [2, 4] because F(2) = 0.25. To find the median we must solve F(m) = 0.5: 3 3 2 1 i.e. F(m) m m 2 2 16 2 This can be rearranged to give the quadratic equation: 3m2 24m 40 0 Using the quadratic formula, m 24 576 4 3 40 6 m = 5.63 or m = 2.37 As 5.63 does not lie in the interval [2, 4], the median must be 2.37. 45 of 61 © Boardworks Ltd 2006 Expectation Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 46 of 61 © Boardworks Ltd 2006 Expectation If X is a continuous random variable defined by the probability density function f(x) over the domain a ≤ x ≤ b, then the mean or expectation of X is given by b E[ X ] xf ( x)dx a E[X] is the value you would expect to get, on average. This mean value of X is also sometimes denoted μ. [Note: if the p.d.f. is symmetrical, then the expected value of X will be the value corresponding to the line of symmetry]. We can also find the expected value of g(X), i.e. any function of X: b E[g( X )] g( x)f ( x) dx a 47 of 61 © Boardworks Ltd 2006 Expectation Example: A random variable X is defined by the probability density function 2 x 1 3 f ( x) x 0 otherwise Calculate the value of E[X] and E[1/X]. 2 2 E[ X ] xf ( x)dx x. 3 dx 2 dx x x all x 1 1 2x 2 1 dx 2 x (0) (2) 2 1 1 48 of 61 © Boardworks Ltd 2006 Expectation 1 1 2 1 E f ( x)dx . 3 dx X x x x all x 1 2x 4 dx 1 2 3 x 3 1 2 (0) 3 2 3 2 1 So, E[ X ] 3 49 of 61 © Boardworks Ltd 2006 Variance Relative frequency histograms Contents Probability density functions Mode Cumulative distribution functions Median and quartiles Expectation Variance 50 of 61 © Boardworks Ltd 2006 Variance If X is a continuous random variable defined by the probability density function f(x) over the domain a ≤ x ≤ b, then the variance of X is given by Var[ X ] E X 2 E[ X ] 2 b or Var[ X ] x 2f ( x)dx μ 2 a The standard deviation of X is the square root of the variance. The standard deviation is sometimes denoted by the symbol σ. 51 of 61 © Boardworks Ltd 2006 Variance Example: A continuous random variable Y has a probability density function f(y) where 3 y (4 y ) 0 y 4 32 f ( y) 0 otherwise Calculate the value of Var[Y]. Sketch of f(y): The p.d.f. is symmetrical. Therefore E[Y] = 2. 52 of 61 © Boardworks Ltd 2006 Variance 4 4 3 E[Y ] y f ( y )dy y . y ( 4 y )dy 32 0 0 2 2 2 4 3 3 4 ( 4 y y )dy 32 0 4 3 4 1 5 y y 32 5 0 3 4 1 5 ( 4 4 ) 0 32 5 4 4 5 Therefore Var[Y] = 4 53 of 61 4 4 22 5 5 © Boardworks Ltd 2006 Variance Example: A continuous random variable x has a probability density function f(x) where 41 0 x2 3 x f ( x) 8 16 2 x 6 otherwise 0 Calculate a) the mean value, μ. b) the standard deviation, σ. 54 of 61 © Boardworks Ltd 2006 Variance 41 0 x2 x f ( x) 38 16 2 x6 0 otherwise 2 6 3 x x2 1 x 3 x dx a) E[ X ] x. dx x. dx dx 4 4 8 16 8 16 0 2 0 2 2 6 2 6 x 3x x 8 0 16 48 2 2 2 3 4 1 7 0 2 8 4 12 2 55 of 61 1 6 © Boardworks Ltd 2006 Variance 41 0 x2 x f ( x) 38 16 2 x6 0 otherwise 6 2 2 6 2 3 1 3 x x 3 x x 2 2 2 dx b) E[ X ] x . dx x . dx dx 4 4 8 16 8 16 0 2 0 2 2 2 6 x x x 12 0 8 64 2 3 3 4 2 3 3 0 6 3 4 4 2 6 3 56 of 61 2 2 1 35 So, Var[ X ] 6 2 1 3 6 36 Therefore σ = 71 36 = 1.40 (3 s.f.) © Boardworks Ltd 2006 Examination-style question Examination-style question: The mass, X kg, of luggage taken on board an aircraft by a passenger can be modelled by the probability density function kx3 (30 x) 0 x 30 f ( x) 0 otherwise a) Sketch the probability density function and find the value of k. b) Verify that the median weight of luggage is about 20.586 kg. c) Find the mean and the variance of X. 57 of 61 © Boardworks Ltd 2006 Examination-style question a) kx3 (30 x) 0 x 30 f ( x) 0 otherwise 30 To find k we use 3 kx (30 x)dx 1 0 30 k (30 x 3 x 4 )dx 1 0 30 30 4 1 5 k x x 1 5 0 4 k 1215000 0 1 58 of 61 k 1 1215000 © Boardworks Ltd 2006 Examination-style question b) kx3 (30 x) 0 x 30 f ( x) 0 otherwise To verify that the median is about 20.586, we need to check that P(X ≤ 20.586) = 0.5 20.586 P(X ≤ 20.586) = kx3 (30 x)dx 0 20.586 1 30 4 1 5 x x 1215000 4 5 0 1 607525 0 1215000 0.500 59 of 61 © Boardworks Ltd 2006 Examination-style question c) kx3 (30 x) 0 x 30 f ( x) 0 otherwise 30 E[ X ] 0 30 x.kx3 (30 x)dx k (30 x 4 x 5 )dx 0 30 1 5 1 6 6x x 1215000 6 0 1 24300000 0 1215000 20 60 of 61 © Boardworks Ltd 2006 Examination-style question c) kx3 (30 x) 0 x 30 f ( x) 0 otherwise E[ X 2 ] 30 30 0 0 2 3 5 6 x . kx ( 30 x ) dx k ( 30 x x )dx 30 1 6 1 7 5x x 1215000 7 0 428.5714 Therefore, Var[ X ] 428.5714 202 28.57 (to 4 s.f.). 61 of 61 © Boardworks Ltd 2006