Question 906865: A die is rolled 720 independent times. Compute, approximately. the
probability that the number of fives that appear will be between 110 and
125 inclusive
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! probability of getting a 5 on each roll is 1/6.
probability of not getting a 5 on each roll is 5/6.
i calculated this using the binomial probability and then approximated it using the normal approximation to the binomial probability.
both answers came very close to each other which is what was expected if the calculations were done correctly.
the excel printout where i did the calculations is shown below.
look below the printout for further comments.
to do the binomial probability, you had to use the binomial probability formula of:
p(x) = c(n,x) * p^x * q^(n-x)
this had to be done for x = 115 to 125
n was equal to 720
p was equal to 1/6
q was equal to 5/6
the actual calculations are shown in the spreadsheet printout.
one of the calculations was:
p(115) = c(720,115) * (1/6)^115 * (5/6)^(720-115)
c(720,115) is the number of possible combinations of 720 things taken 115 at a time.
the formula for c(720,115) = 720! / (115!*(720-115)!)
fyi, my calculator couldn't handle numbers so large, but excel was able to.
to do the normal approximation of the binomial, you had to do the following:
m = mean = n * p = 720 * 1/6 = 720/6 = 120.
se = standard error = sqrt(n*p*q) = sqrt(720*(1/6)*(5/6)) = sqrt((720*5)/36) = sqrt(20*5) = sqrt(100) = 10.
you then needed to find the z factors for 115 and the z factor for 125.
since the binomial formula is discrete and the normal distribution formula is continuous, you also needed to make an adjustment by subtracting .5 from 115 and adding .5 to 125.
this allowed the normal approximation of the binomial to have a result that is closer to the binomial.
so you had to find the z factor for 114.5 and the z factor for 125.5.
once you found those, you needed to find the area under the normal distribution curve that was equal to the area to the left of the z factor for 125.5 and the area to the left of the z factor for 114.5.
that area is the probability that you will get a value between 114.5 and 125.5 using the normal distribution formula.
the details of the calculations are shown below as best i can approximate them.
z factor for 114.5 = (114.5-120)/10 = -.55
z factor for 125.5 = (125.5-120)/10 = .55
area under the normal distribution curve to the left of a z factor of -.55 = .291159...
area under the normal distribution curve to the left of a z factor of .55 = .708840...
subtract .291159... from .708840... and you get .4176806907
there is a slight discrepancy between my calculator numbers and the excel number that is more then likely due to rounding.
the numbers are:
.4176806907 for my calculator
.4176806264 for excel
i assume excel is more accurate since it can handle larger numbers and therefore must have more capacity for each internally stored variable, the details of why not being very important right now.
the bottom line is you can do it either way, but more then likely needed to use the normal approximation as it is very accurate when the number of trials is very large and the probability is close to .5.
the rule for whether to use the normal approximation to the binomial is:"
np > 5 and nq > 5
some may be a little stricter and require np > 10 and nq > 10
in this problem np was equal to 720 * 1/6 = 120 and nq was equal to 720 * 5/6 = 600.
both numbers were well above 10 so there was no problem in using the normal approximation to the binomial.
another more complete reference on using the normal approximation to the binomial can also be found here:
http://onlinestatbook.com/2/normal_distribution/normal_approx.html
a decent reference on normal approximation of the binomial can be found here.
http://www.regentsprep.org/Regents/math/algtrig/ATS7/BLesson3.htm
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