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Quick S2 question

http://www.ocr.org.uk/Images/136159-question-paper-unit-4733-probability-and-statistics-2.pdf

Question 4 the last bit.

So I've approximated using the normal distribution, but then I'm not sure on the continuity correction. i thought it would become 36.5, but according to the mark-scheme it's 35.5, why?
Reply 1
Original post by Music99
http://www.ocr.org.uk/Images/136159-question-paper-unit-4733-probability-and-statistics-2.pdf

Question 4 the last bit.

So I've approximated using the normal distribution, but then I'm not sure on the continuity correction. i thought it would become 36.5, but according to the mark-scheme it's 35.5, why?


P(x36)P(x\geq 36) use the law of continuity correction to realise that if it were 36.5, it would not include 36, therefore your solution is not valid, and does not satisfy the inequality
by continuity correction P(Y>35.5) is the appropriate correction Y is normal continuous random variable
(edited 11 years ago)
Reply 2
Original post by Robbie242
P(x36)P(x\geq 36) use the law of continuity correction to realise that if it were 36.5, it would not include 36, therefore your solution is not valid, and does not satisfy the inequality
by continuity correction P(Y>35.5) is the appropriate correction Y is normal continuous random variable


But on another question I just did it was find probability it's >17 and the continuity correction was 17.5 :s
Reply 3
Original post by Music99
But on another question I just did it was find probability it's >17 and the continuity correction was 17.5 :s
Yes because its not P(Xx)P(X\geq x) If its greater or equal, the correction must include the number itself, in this case a random x.

Basically for \geq and >> do not have the same continuity correction, can you see why?
Reply 4
Original post by Robbie242
Yes because its not P(Xx)P(X\geq x) If its greater or equal, the correction must include the number itself, in this case a random x.

Basically for \geq and >> do not have the same continuity correction, can you see why?


I think so, I'll draw a sketch should help.

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