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junior college 2 | H2 Maths
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Xiang Ning
Xiang Ning

junior college 2 chevron_right H2 Maths chevron_right Singapore

Pls help me with (ii) onwards. Thanks

Date Posted: 4 years ago
Views: 542
Eric Nicholas K
Eric Nicholas K
4 years ago
Part (i) I forgot what it means. But one look and I can tell that the expected value is 6 x 0.8 = 4.8, since each week has six puzzles and the success rate is 0.8.

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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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
Not 100% sure. My answers are likely wrong.

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.
Xiang Ning
Xiang Ning
4 years ago
Actually, part (iv) answer is 0.737
Eric Nicholas K
Eric Nicholas K
4 years ago
I have this checked when I return
Eric Nicholas K
Eric Nicholas K
4 years ago
Oh, the given answer is correct. I lost the translation along the way and misinterpreted “at least 4 correct for the rest of the week” as “exactly 4 correct for the rest of the week”.

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.
Xiang Ning
Xiang Ning
4 years ago
Okay thanks
Eric Nicholas K
Eric Nicholas K
4 years ago
Updated
Xiang Ning
Xiang Ning
4 years ago
Thanks
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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
Something like this for the fourth part is good.