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,

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Presentation transcript:

Binomial test (a) p = 0.07 n = 17 X ~ B(17, 0.07) (i) P(X = 2) = = using Bpd on calculator Or 17 C 2 x x (ii) n = 50 X ~ B(50, 0.07) P(X ≤ 5) = … = using Bcd on calculator or use tables (b) X ~ B(50, 0.55) P(X ≥ 30) = 1 – P(X ≤ 29) = 1 – using Bcd on calculator* = = to 3 dp Using tables involves working with Y ~ B(50, 0.45) and P(Y ≤ 20)

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

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

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

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

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

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: grams