The probability that a shooter strikes a target is 0.4.
By using a suitable approximation,find the probability that he will strike the target 220 times out of 500 shots.
Normal approximation of binomial distribution.
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0You should write a better title and add probability as a tag. You should also add your thoughts to the exercise: what have you tried so far? What approximations do you have in mind? – 2017-02-16
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0Thanks for the suggestion. I'm not sure what cld be the suitable title, and this question actually comes with 2 sub question, but I only know the first sub question. The one I asked, I seriously have no thoughts about it >< – 2017-02-16
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This problem is about normal approximation of binomial distribution. This is known as De Moivre - Laplace theorem. When you have very large no. of binomial trials in an experiment it will be difficult to evaluate factorials and thus making it difficult to evaluate exact values. As per the theorem you can approximate the distribution as normal distribution. We have event X: Shooter hits the target. p=0.4 and n=500 => np=200 and npq= 120. So X ~ B (500,0.4) can be approximated as X ~ N (200, 120). Where B is binomial distribution and N is normal distribution with mean 200 and variance 120. Now you need to find P(X=220) in normal distribution . But you know that normal distribution is continuous distribution whereas binomial was discrete. So you do something called continuity correction which tells you that in a normal approximated binomial P(X=n) --> P(n-0.5 < X < n+0.5). So find P(119.5 < X < 220.5) using the standard distribution table method.