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junior college 2 | H2 Maths
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Pls help me with (ii) onwards. Thanks
Part (ii), 4.8 is closer to 5 than 4, so I would think that the most likely number of completed crossword puzzles is 5. A binomial distribution testing for X = 4, X = 5 and X = 6 should tell the difference. X = 0, 1, 2, 3 is just too far off.
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For part ii, we simply model the binomial distribution by n = 6, p = 0.8 and 1 - p = 0.2, then the general term in the expansion of the binomial formula will be like
(6 r) * 0.2^(6 - r) * 0.8^r
We just sub in integer values of r from 0 to 6 and see the highest one for the most likely integer number of puzzles completed. Of course, adding the probability values for the seven cases must get you 1.
In other words, you must do the 5C5 case in my working for part iv in which I only wrote 5C4.
Wait while I update this around 2 am.