© Boardworks Ltd 20061 of 39 © Boardworks Ltd 2006 1 of 39 A2-Level Maths: Statistics 2 for Edexcel S2.3 Continuous distributions This icon indicates the.

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© Boardworks Ltd 20061 of 39 © Boardworks Ltd 2006 1 of 39 A2-Level Maths: Statistics 2 for Edexcel S2.3 Continuous distributions This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation.

© Boardworks Ltd 20062 of 39 Contents © Boardworks Ltd 2006 2 of 39 Continuous uniform distribution Approximating the binomial using a normal Approximating the Poisson using a normal

© Boardworks Ltd 20063 of 39 A random variable X is said to have a continuous uniform distribution (or rectangular distribution) over the interval [a, b] if its probability density function has the form: f(x)f(x) The graph of its probability density function is as follows: x Continuous uniform distribution

© Boardworks Ltd 20064 of 39 and Proof of E[ X ]: The result for E[ X ] follows immediately from the symmetry of the p.d.f.. Continuous uniform distribution Key result: If X has a continuous uniform distribution over the interval [ a, b ], then

© Boardworks Ltd 20065 of 39 Continuous uniform distribution Proof of Var[ X ]: So,

© Boardworks Ltd 20066 of 39 Example: A random variable Y has a continuous uniform distribution in the interval [2, 8]. Find P( Y < μ + σ). Continuous uniform distribution Using the formulae for E[ X ] and Var[ X ], we get: The required probability is P( Y < μ + σ ) = P( Y < 5 + √3). This probability is represented by the shaded area. Therefore P( Y < 5 + √3) =

© Boardworks Ltd 20067 of 39 Examination-style question: A random variable X is given by the probability density function f ( x ), where Find: a)E[ X ] and Var[ X ] b)P(7 ≤ X < 10) Examination-style question

© Boardworks Ltd 20068 of 39 Solution: X has a uniform distribution over the interval (5, 15). a) Examination-style question b) The p.d.f. for X is shown on the diagram below. The probability we require is shaded. So, P(7 ≤ X < 10) =

© Boardworks Ltd 20069 of 39 Note: If X has a uniform distribution over the interval ( a, b ) then the cumulative distribution function of X is: Continuous uniform distribution

© Boardworks Ltd 200610 of 39 Contents © Boardworks Ltd 2006 10 of 39 Approximating the binomial using a normal Continuous uniform distribution Approximating the binomial using a normal Approximating the Poisson using a normal

© Boardworks Ltd 200611 of 39 Approximating the binomial using a normal

© Boardworks Ltd 200612 of 39 Calculating probabilities using the binomial distribution can be cumbersome if the number of trials ( n ) is large. Consider this example: Approximating the binomial using a normal Let the number of left-handed people in the school be X. Then X ~ B[1200, 0.1]. Introductory example:10% of people in the United Kingdom are left-handed. A school has 1 200 students. Find the probability that more than 140 of them are left-handed.

© Boardworks Ltd 200613 of 39 The required probability is P( X > 140) = P( X = 141) + P( X = 142) + … + P( X = 1200). As no tables exist for this distribution, calculating this probability by hand would be a mammoth task. A further problem arises if you attempt to work out one of these probabilities, for example P( X = 141): One way forward is to approximate the binomial distribution using a normal distribution. Approximating the binomial using a normal Calculators cannot calculate the value of this coefficient – it is too large!

© Boardworks Ltd 200614 of 39 Key result: If X ~ B( n, p ) where n is large and p is small, then X can be reasonably approximated using a normal distribution: X ≈ N[ np, npq ] where q = 1 – p. There is a widely used rule of thumb that can be applied to tell you when the approximation will be reasonable: Approximating the binomial using a normal A binomial distribution can be approximated reasonably well by a normal distribution provided np > 5 and nq > 5.

© Boardworks Ltd 200615 of 39 Approximating the binomial using a normal

© Boardworks Ltd 200616 of 39 Continuity correction: Exact distribution: B( n, p ) P( X ≥ x ) P( X ≥ x – 0.5) P( X ≤ x ) P( X ≤ x + 0.5) Approximating the binomial using a normal Approximate distribution: N[ np, npq ] This 0.5 is called the continuity correction factor. A continuity correction must be applied when approximating a discrete distribution (such as the binomial) to a continuous distribution (such as the normal distribution).

© Boardworks Ltd 200617 of 39 Approximating the binomial using a normal

© Boardworks Ltd 200618 of 39 Introductory example (continued): 10% of people in the United Kingdom are left-handed. A school has 1 200 students. Find the probability that more than 140 of them are left-handed. Approximating the binomial using a normal Solution: Let the number of left-handed people in the school be X. Then the exact distribution for X is X ~ B[1200, 0.1]. Since np = 120 > 5 and nq = 1080 > 5 we can approximate this distribution using a normal distribution: X ≈ N[120, 108]. npnpq

© Boardworks Ltd 200619 of 39 So, P( X > 140) = P( X ≥ 141) → P( X ≥ 140.5) Approximating the binomial using a normal Standardize N[120, 108] Using continuity correction You convert 140.5 to the standard normal distribution using the formula: N[0, 1]

© Boardworks Ltd 200620 of 39 Approximating the binomial using a normal Therefore P( X ≥ 140.5) = P( Z ≥ 1.973) = 1 – Φ (1.973) = 1 – 0.9758 = 0.0242 So the probability of there being more than 140 left-handed students at the school is 0.0242.

© Boardworks Ltd 200621 of 39 Example: It has been estimated that 15% of schoolchildren are short-sighted. Find the probability that in a group of 80 schoolchildren there will be Approximating the binomial using a normal Solution: Let the number of short-sighted children in the group be X. Then the exact distribution for X is X ~ B[80, 0.15]. Since np = 12 > 5 and nq = 68 > 5 we can approximate this distribution using a normal distribution: X ≈ N[12, 10.2]. a) no more than 15 children that are short-sighted b) exactly 10 children that are short-sighted.

© Boardworks Ltd 200622 of 39 a) So P( X ≤ 15) → P( X ≤ 15.5) Therefore P( X ≤ 15.5) = P( Z ≤ 1.096) = Φ (1.096) = 0.8635 Approximating the binomial using a normal Using continuity correction Standardize N[12, 10.2] N[0, 1] So the probability that no more than 15 children will be short-sighted is 0.8635.

© Boardworks Ltd 200623 of 39 b) So P( X = 10) → P(9.5 ≤ X ≤ 10.5) Approximating the binomial using a normal Using continuity correction N[12, 10.2] Standardize N[0, 1]

© Boardworks Ltd 200624 of 39 Approximating the binomial using a normal Therefore P(9.5 ≤ X ≤ 10.5) = P(–0.783 ≤ Z ≤ –0.470) = P(0.470 ≤ Z ≤ 0.783) = 0.7832 – 0.6808 = 0.1024 The probability that 10 children will be short-sighted is 0.1024.

© Boardworks Ltd 200625 of 39 Examination-style question: A sweet manufacturer makes sweets in 5 colours. 25% of the sweets it produces are red. The company sells its sweets in tubes and in bags. There are 10 sweets in a tube and 28 sweets in a bag. It can be assumed that the sweets are of random colours. Examination-style question a) Find the probability that there are more than 4 red sweets in a tube. b) Using a suitable approximation, find the probability that a bag of sweets contains between 5 and 12 red sweets (inclusive).

© Boardworks Ltd 200626 of 39 Solution: a) Let the number of red sweets in a tube be X. Then the exact distribution for X is X ~ B[10, 0.25]. This distribution cannot be approximated by a normal but its probabilities are tabulated: P( X > 4) = 1 – P( X ≤ 4) = 1 – 0.9219 = 0.0781 So the probability that a tube contains more than 4 red sweets is 0.0781. Examination-style question

© Boardworks Ltd 200627 of 39 Solution: b) Let the number of red sweets in a bag be Y. Then the exact distribution for Y is Y ~ B[28, 0.25]. This distribution can be approximated by a normal since np = 7 and nq = 21 (both greater than 5): Y ≈ N[7, 5.25] P(5 ≤ Y ≤ 12) → P(4.5 ≤ Y ≤ 12.5) Examination style question npq Using continuity correction

© Boardworks Ltd 200628 of 39 Therefore P(4.5 ≤ Y ≤ 12.5) = P(–1.091 ≤ Z ≤ 2.400) = P( Z ≤ 2.400) – P( Z ≤ –1.091) = Φ(2.400) – (1 – Φ(1.091)) = 0.9918 – (1 – 0.8623) So the probability that a bag will contain between 5 and 12 red sweets is 0.8541. Examination style question N[7, 5.25]N[0, 1] = 0.8541 Standardize

© Boardworks Ltd 200629 of 39 Contents © Boardworks Ltd 2006 29 of 39 Approximating the Poisson using a normal Continuous uniform distribution Approximating the binomial using a normal Approximating the Poisson using a normal

© Boardworks Ltd 200630 of 39 Approximating the Poisson using a normal

© Boardworks Ltd 200631 of 39 Key result: If X ~ Po( λ ) and λ is large, then X is approximately normally distributed: X ≈ N[λ, λ] There is a widely used rule of thumb that can be applied to tell you when the approximation will be reasonable: Recall that the mean and variance of a Poisson distribution are equal. Note: A continuity correction is required because we are approximating a discrete distribution using a continuous one. Approximating the Poisson using a normal A Poisson can be approximated reasonably well by a normal distribution provided λ > 15.

© Boardworks Ltd 200632 of 39 Example: An animal rescue centre finds a new home for an average of 3.5 dogs each day. Approximating the Poisson using a normal a) What assumptions must be made for a Poisson distribution to be an appropriate distribution? b) Assuming that a Poisson distribution is appropriate: i.Find the probability that at least one dog is rehoused in a randomly chosen day. ii.Find the probability that, in a period of 20 days, fewer than 65 dogs are found new homes.

© Boardworks Ltd 200633 of 39 Solution: a)For a Poisson distribution to be appropriate we would need to assume the following: b) i) Let X represent the number of dogs rehoused on a given day. So, X ~ Po(3.5). P( X ≥ 1) = 1 – P( X = 0) = 1 – 0.0302 (from tables) = 0.9698 Approximating the Poisson using a normal 1.The dogs are rehoused independently of one another and at random; 2.The dogs are rehoused one at a time; 3.The dogs are rehoused at a constant rate.

© Boardworks Ltd 200634 of 39 b) ii) Let Y represent the number of dogs rehoused over a period of 20 days. So, Y ~ Po(3.5 × 20) i.e. Po(70). As λ is large, we can approximate this Poisson distribution by a normal distribution: Y ≈ N[70, 70]. P( Y < 65) → P( Y ≤ 64.5) Approximating the Poisson using a normal Standardize N[70, 70]N[0, 1]

© Boardworks Ltd 200635 of 39 Approximating the Poisson using a normal P( Y ≤ 64.5) = P( Z ≤ –0.657) = 1 – Φ(0.657) = 1 – 0.7445 = 0.2555 So the probability that less than 65 dogs are rehoused is 0.2555.

© Boardworks Ltd 200636 of 39 Examination-style question: An electrical retailer has estimated that he sells a mean number of 5 digital radios each week. Examination-style question a)Assuming that the number of digital radios sold on any week can be modelled by a Poisson distribution, find the probability that the retailer sells fewer than 2 digital radios on a randomly chosen week. b)Use a suitable approximation to decide how many digital radios he should have in stock in order for him to be at least 90% certain of being able to meet the demand for radios over the next 5 weeks.

© Boardworks Ltd 200637 of 39 Solution: a) Let X represent the number of digital radios sold in a week. So X ~ Po(5). P( X < 2) = P( X ≤ 1) = 0.0404 (from tables). Examination-style question b) Let Y represent the number of digital radios sold in a period of 5 weeks. So, Y ~ Po(25). We require y such that P( Y ≤ y ) = 0.9. So the probability that the retailer sells fewer than 2 digital radios in a week is 0.0404.

© Boardworks Ltd 200638 of 39 Since the parameter of the Poisson distribution is large, we can use a normal approximation: Y ≈ N[25, 25]. P( Y ≤ y ) → P( Y ≤ y + 0.5)(using a continuity correction). Examination-style question N[25, 25] N[0, 1] Standardize The 10% point of a normal is 1.282

© Boardworks Ltd 200639 of 39 Examination-style question So, So the retailer would need to keep 31 digital radios in stock.

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