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S2 question Sampling Watch

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    Could I have help with 6a please?Name:  IMG_0472.jpg
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    (Original post by swagmister)
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Size:  486.8 KBAttachment 583034583036Could I have help with 6a please?
    Unless you've misread the value from the table I see nothing wrong with your working
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    (Original post by SeanFM)
    Unless you've misread the value from the table I see nothing wrong with your working
    The answers say it's 0.155 I tried 6b which was a similar question but also got the wrong answer
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    (Original post by swagmister)
    The answers say it's 0.155 I tried 6b which was a similar question but also got the wrong answer
    Beats me, sorry. I agree with your answer
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    (Original post by SeanFM)
    Beats me, sorry. I agree with your answer
    Maybe the textbook is wrong? I'll ask my teacher tomorrow, thanks for taking a look anyway 👍
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    Shouldn't there be a continuity correction, since it's a binomial distribution?

    (Original post by swagmister)
    Could I have help with 6a please?Name:  IMG_0472.jpg
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Size:  486.8 KBAttachment 583034583036
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    (Original post by NotNotBatman)
    Shouldn't there be a continuity correction, since it's a binomial distribution?
    Idk that would make sense but I didn't use it in any of the other questions that similar and got the right answer
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    (Original post by swagmister)
    Idk that would make sense but I didn't use it in any of the other questions and got the right answer
    Try it again with a continuity correction and see if you get the answer in the textbook.
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    (Original post by NotNotBatman)
    Try it again with a continuity correction and see if you get the answer in the textbook.
    I got 0.9292 Name:  image.jpg
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    (Original post by NotNotBatman)
    Try it again with a continuity correction and see if you get the answer in the textbook.
    Do you not need a continuity correction because the means and sampling mean aren't discrete?
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    (Original post by swagmister)
    I got 0.9292 Name:  image.jpg
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    You'd need to apply the continuity correction to the total score (mean multiplied by number of throws) and then divide by 70.
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    (Original post by NotNotBatman)
    You'd need to apply the continuity correction to the total score (mean multiplied by number of throws) and then divide by 70.
    So ((35/12)+0.5)/70??
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    (Original post by swagmister)
    So ((32/15)+0.5)/70??
    Where's 32 and 15 coming from?

    You do 3.3 x 70 = 231, then apply the continuity correction and divide by 70, then you find your standardized z value.
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    (Original post by NotNotBatman)
    Where's 32 and 15 coming from?

    You do 3.3 x 70 = 231, then apply the continuity correction and divide by 70, then you find your standardized z value.
    Oh I was taught to add or minus 0.5 onto x where eg. P(X<x), I got 0.1723 this timeName:  image.jpg
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    (Original post by swagmister)
    Oh I was taught to add or minus 0.5 onto x where eg. P(X<x), I got 0.1723 this timeName:  image.jpg
Views: 29
Size:  419.6 KB
    In this case it's different because it's a mean.

    You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

    230.5/70 = 3.29...
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    (Original post by NotNotBatman)
    In this case it's different because it's a mean.

    You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

    230.5/70 = 3.29...
    Oh right I haven't learnt that yet, oh yeah, I got 0.1585
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    Nevermind I read the tables wrong
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    Got 0.155 now!
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    How come you use the continuity correction in Q6 but not Q5?
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    (Original post by NotNotBatman)
    In this case it's different because it's a mean.

    You've done the continuity correction the wrong way around; it's less than x, not less than or equal to, so subtract 0.5.

    230.5/70 = 3.29...
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    (Original post by swagmister)
    How come you use the continuity correction in Q6 but not Q5?
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    I'm not sure, I think you should use a continuity correction.
 
 
 
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