SOLUTION: Hello Can someone help me with this problem, I did try to work it but I am not sure. Consider a binomial distribution with 13 identical trials, and a probability of success of

Algebra ->  Probability-and-statistics -> SOLUTION: Hello Can someone help me with this problem, I did try to work it but I am not sure. Consider a binomial distribution with 13 identical trials, and a probability of success of      Log On


   



Question 333236: Hello Can someone help me with this problem, I did try to work it but I am not sure.

Consider a binomial distribution with 13 identical trials, and a probability of success of 0.5

i. Consider a binomial distribution with 13 identical trials, and a probability of success of 0.5

i. Find the probability that x = 2 using the binomial tables
ii. Use the normal approximation to find the probability that x = 2. Show all work.
THIS IS WHAT I HAVE SO FAR I AM NOT SURE IF IT IS RIGHT BUT HERE GOES:
P(x = 2) = jpeg will not post
X-=- 2, n=13
p = 0.5
Mean =E(X) = np =13 x 0.5 = 6.5
Standard Dev. = √np (1-p) = √13 x.0.5. x 0.5 =√3.25 =1.8027
√np (1-p) = √13 x.0.5. x 0.5 =√3.25 =1.8027


Please help me that you again.
Thank you.

Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
difficult to help you with binomial tables through this medium
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some binomial tables are constructed using individual probabilites and other bionamial tables are constructed using cumulative probabilities.
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typically the proportion of interest is labelled across at the top, you would be looking for p=0.5
Then under that column, you would go down looking at for an "n" or sample size value along the right hand side (for 13 in your case). Within that sample size you would see events from 0 to the n or 13 in your case. Find for n=13 and x=2, follow that across and where this meets the column value that would represent either P(X=2) or P(X<=2) depending on how your table is constructed.
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To use the normal approximation to the binomial, keep in mind that there is an assumption that np>10, which is not met in your case, but ignoring this, though you shouldnt, then you would compute the mean and standard deviation, just like you did.
Then you would convert The X variable to a Z variable via Z=%28X-np%29%2Fsqrt%28np%2A%281-p%29%29
So you need P(X=2),
problem is the binomial (count of successes) is a discrete distribution and the normal distribution is a continuous distribution. soo .. what do you ?
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instead of P(X=2) you actually want to find P(1.5<=x<=2.5)
when you convert x to z you then have
P%281.5%3C=x%3C=2.5%29=P%28%281.5-6.5%29%2F1.8027%3C=Z%3C=%282.5-6.5%29%2F1.8027%29
this simplies to P%28-2.77%3C=Z%3C=-2.22%29
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and from there you need to use standard normal tables, excel, statistical calculator, manual integration or some method to find this probability.