Binomial probability (1 Viewer)

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for part (ii) in the ans they take P(X<9) but shouldn't it be P(X<8) cause thats less than 9 and cause its discrete you can have 8.5 ppl?

 

[WeiWeiMan]

View attachment 39888
for part (ii) in the ans they take P(X<9) but shouldn't it be P(X<8) cause thats less than 9 and cause its discrete you can have 8.5 ppl?

James ruse 2023?
The question says to use a normal approximation to solve it as the sample size is sufficiently large enough for it to be reasonable. The normal approximation is continuous btw

 

[=)(=]

James ruse 2023?
The question says to use a normal approximation to solve it as the sample size is sufficiently large enough for it to be reasonable. The normal approximation is continuous btw

Yea lol
Ohh right okay but if it was say binomial probabilty with like np and nq< 10 then you would do P(x≤8) right?

 

[Luukas.2]

I agree. It should be:

 

[howcanibesmarter]

is continuity correction part of hsc or nah

 

[Luukas.2]

Solutions from last year accepted answers both with and without the continuity correction.

In this case, though the result changes by a significant amount by the way the solution is constructed, which to me shows that including the correction is the better answer.

I wonder, in a case like this, whether the results for n following from the normal approximation applied to each of:

• 
• 
• 

would all be acceptable in the HSC. I'm sure views would vary between schools.

 

[WeiWeiMan]

Yea lol
Ohh right okay but if it was say binomial probabilty with like np and nq< 10 then you would do P(x≤8) right?

No, just do what the question tells you to. If it asks for smaller than 9, do <9, not <= 8. You can do continuity correction if you want but it seems it’s not necessary for hsc.

 

[Luukas.2]

No, just do what the question tells you to. If it asks for smaller than 9, do <9, not <= 8. You can do continuity correction if you want but it seems it’s not necessary for hsc.

As a straight binomial probability, and are the same thing because .

In taking a normal approximation, a discrete distribution is approximated by a continuous one, which is why the continuity correction is needed.

Suppose you toss a fair coin 10,000 times. The most likely outcome is 5,000 heads, with probability



You have no hope of finding a value like 10,000! with a calculator, and the approximation I've used here (using Stirling's formula for n!) would need to be given in a question.

It can be estimated easily with a normal distribution, however, but only if a continuity correction is applued. Without it:



but, after applying the appropriate correction,



the probability can be found from a z-score table.

 

[=)(=]

ahh interesting thanks for that insight!