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Binomial test (a) p = 0.07 n = 17 X ~ B(17, 0.07) (i) P(X = 2) = 0.22437 = 0.224 using Bpd on calculator Or 17 C 2 x 0.07 2 x 0.93 15 (ii) n = 50 X ~ B(50, 0.07) P(X 5) = 0.8649 = 0.865 using Bcd on calculator or use tables (b) X ~ B(50, 0.55) P(X 30) = 1 P(X 29) = 1 0.71383 using Bcd on calculator* = 0.28617 = 0.286 to 3 dp Using tables involves working with Y ~ B(50, 0.45) and P(Y 20)

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Normal Distribution lesson 3 Standardising a Normal Distribution Exam qs Inverse normal Examples Textbook work calculator

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Eg Find a such that P( Z < a) = 0.66276

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Eg : Find a such that P( Z < a) = 0.95 Eg : Find a such that P( Z < a) = 0.954

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Eg : Find a such that P( Z > a) = 0.04 Inverse normal Eg : Find a such that P( Z < a) = 0.35

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Mini whiteboards

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Inverse normal Z ~ N( 0, 1 ) find a such that: (a) P(Z a) = 0.95 (c) P(Z < a) = 0.001 Ans (a) 1.881 (b) 1.645 (c) -3.090

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More inverse normal Given that X ~ N(20,25 ) find a such that: (a) P(X a) = 0.05 (c) P(X < a) = 0.004 Ans: (a) 29.4 (b) 28.2 (c) 6.74

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Oranges have weights that are normally distributed with mean of 90 grams and standard deviation of 5 grams. Determine, in grams to 1dp, the weight that is exceeded by 1% of the oranges Ans: 101.6 grams

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