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A2-Level Maths: Statistics 2 for Edexcel S2.3 Normal approximation to Binomial 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 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 2 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal Calculating probabilities using the binomial distribution can be cumbersome if the number of trials (n) is large. Consider this example: 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. Let the number of left-handed people in the school be X. Then X ~ B[1200, 0.1]. 3 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 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): P( X 141) 1200 C141 0.1141 0.91059 Calculators cannot calculate the value of this coefficient – it is too large! One way forward is to approximate the binomial distribution using a normal distribution. 4 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 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: A binomial distribution can be approximated reasonably well by a normal distribution provided np > 5 and nq > 5. 5 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 6 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal A continuity correction must be applied when approximating a discrete distribution (such as the binomial) to a continuous distribution (such as the normal distribution). Continuity correction: Exact distribution: B(n, p) P(X ≥ x) Approximate distribution: N[np, npq] P(X ≥ x – 0.5) This 0.5 is called the continuity correction factor. P(X ≤ x) 7 of 39 P(X ≤ x + 0.5) © Boardworks Ltd 2006 Approximating the binomial using a normal 8 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 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. 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]. np 9 of 39 npq © Boardworks Ltd 2006 Approximating the binomial using a normal So, P(X > 140) = P(X ≥ 141) → P(X ≥ 140.5) Using continuity correction N[120, 108] Standardize N[0, 1] 140.5 120 1.973 108 You convert 140.5 to the standard normal distribution using the formula: Z 10 of 39 X ~ N[0,1]. © Boardworks Ltd 2006 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. 11 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal 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 a) no more than 15 children that are short-sighted b) exactly 10 children that are short-sighted. 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]. 12 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal a) So P(X ≤ 15) → P(X ≤ 15.5) N[12, 10.2] Using continuity correction Standardize N[0, 1] 15.5 12 1.096 10.2 Therefore P(X ≤ 15.5) = P(Z ≤ 1.096) = Φ(1.096) = 0.8635 So the probability that no more than 15 children will be short-sighted is 0.8635. 13 of 39 © Boardworks Ltd 2006 Approximating the binomial using a normal b) So P(X = 10) → P(9.5 ≤ X ≤ 10.5) Using continuity correction Standardize 9.5 12 0.783 10.2 N[12, 10.2] N[0, 1] 10.5 12 0.470 10.2 14 of 39 © Boardworks Ltd 2006 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. 15 of 39 © Boardworks Ltd 2006 Examination-style question 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. 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). 16 of 39 © Boardworks Ltd 2006 Examination-style question 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. 17 of 39 © Boardworks Ltd 2006 Examination style question 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) 18 of 39 npq Using continuity correction © Boardworks Ltd 2006 Examination style question Standardize N[7, 5.25] 4.5 7 1.091 5.25 12.5 7 2.400 5.25 N[0, 1] 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) = 0.8541 So the probability that a bag will contain between 5 and 12 red sweets is 0.8541. 19 of 39 © Boardworks Ltd 2006