# Not sure what i did wrong binomial probability

In a 22-item true–false examination, a student guesses on each question.

If 14 correct answers constitute a passing grade, what is the probability the student will pass?

i did c(22,14)* (1/2)^14 * (1/2)^8

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mathman
You have to add in the probabilities for more than 14 correct guesses. Your answer is for exactly 14 correct.

so i would have to add every single probability up to 22?
is there any other way i could do this?

so i would have to add every single probability up to 22?
is there any other way i could do this?
You can use an online calculator for p=0.5, n=22, x=14 and solve for $P(X\geq x)$

http://stattrek.com/tables/binomial.aspx

so i would have to add every single probability up to 22?
is there any other way i could do this?
You might also use a Normal approximation to the Binomial distribution.

You might also use a Normal approximation to the Binomial distribution.
The normal approximation gives p=0.1004 whereas the presumably exact binomial gives $(P(X\geq x)=0.1431$ for x=14.

For the normal approximation I'm using mean 11 and $SD = \sqrt {11(1-0.5)} = 2.345$

The normal approximation gives p=0.1004 whereas the presumably exact binomial gives $(P(X\geq x)=0.1431$ for x=14.

For the normal approximation I'm using mean 11 and $SD = \sqrt {11(1-0.5)} = 2.345$
I get $P(X \geq 13.5) = 0.1432$ using the Normal distribution adjusted for continuity.

I get $P(X \geq 13.5) = 0.1432$ using the Normal distribution adjusted for continuity.
I did too, but when Ted Williams was told his 0.3995 batting average would go into the record books as 0.400, he said that wasn't really 0.400 and played through two final season games ending up with a 0.406 batting average. Is 13.5 a passing grade or is 14 a threshold value? I understand the continuity correction and it's fine for some applications but for n=22 and a "threshold" value, why not use an exact calculation? In either case, you will likely use tables or a calculator.

Having said that, it's closer than I would have thought, but I wouldn't have been comfortable without doing the exact approach.

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I agree, in this case the approximation works better than we have any right to expect. Still, it's a useful tool to have around.