you can use a binomial distribution calculator to get the answer, but this is not what they want you to do.
it is, however, a good way to determine if you use the right procedure to get the answer.
one such calculator can be found at http://stattrek.com/online-calculator/binomial.aspx
the results from this calculator indicate that the more detailed answer by not using the binomial approximation would be:
p(x>6) = .059899 which is rounded to .0599.
p(x>6) = .136639 which is rounded to .1367.
p(x=6) = .07674 which is rounded to .0767.
when using the binomial approximation, the recommendation is to do the following:
p(x = n), use p(n - .5 < x < n + .5)
p(x > n), use p(x > n + .5)
p(x >= n), use p(x > n - .5)
p(x < n), use p(x < n - .5)
p(x <= n), use p(x < n + .5)
these recommendations are in the following reference, although not in the same order as listed here.
you may check the reference to see that i copied the instructions correctly.
http://www.statisticshowto.com/probability-and-statistics/binomial-theorem/normal-approximation-to-the-binomial/
i tried using a normal approximation to the binomial calculator online but ran into a problem.
the problem was that n*p and n*q were not both greater than 5.
in order to use the binomial approximation, n must be > 30, and n*p and n*q both must be greater than 5.
66 * .053 = .398 which is less than 5.
the calculators would not make the calculation because the recommendation that n * p > 5 was not passed.
that's too bad, because they could have been a resource to see if the calculations were done correctly.
be that as it may, i did the calculations per the recommendations of the reference and got the following.
i used the following:
mean = n * p = 66 * .053 = 3.498.
standard error = sqrt(n * p * q) = sqrt(66 * .053 * .947) = 1.820056593.
the instructions tell you to round the z-score to 2 decimal places and the final answer to 4 decimal places.
unfortunately, they don't tell you how many decimal places to round the standard error.
i decided not to round the standard error and to use it as shown in my calculator.
using the recommendations in the reference, this is what i did.
for p > 6), i used p(x > 6.5).
the z-score calculation was z = (6.5 - 3.498) / 1.820056593.
this results in a z-score of 1.649399262 which i rounded to 1.65.
i then looked for the area to the left of a z-score of 1.65 and got .9505.
the area to the right of that z-score would be 1 - .9505 = .0495.
since that's what you got, then congratulations on that one.
for at least 6 errors, you are looking for p (x >= 6).
for p(x >= 6), i used p(x > 5.5).
the z-score calculation was z = (5.5 - 3.498) / 1.820056593.
this results in a z-score of 1.099965797 which i rounded to 1.10.
i then looked for the area to the left of a z-score of 1.10 and got .8643.
the area to the right of that z-score would be 1 - .8643 = .1357.
for p(x = 6), i used p(5.5 < x < 6.5).
to do this, you needed to find p(x < 5.5) and p(x < 6.5) and then subtract the smaller area from the larger area.
the z-score for p(x < 5.5) is z = (5.5 - 3.498) / 1.820056593 = 1.099965797.
i rounded that to 1.10.
the area to the left of that is .8643.
the z-score for p(x < 6.5) is z = (6.5 - 3.498) / 1.820056593 = 1.649399262.
i rounded that to 1.65.
the area to the left of that is .9505.
the area in between is .9505 - .8643 = .0862.
a comparison between what the normal approximation got and what the more detailed binomial distribution got is shown below:
binomial distribution normal approximation
p(x > 6) .0599 .0495
p(x >= 6) .1367 .1357
p(x = 6) .0767 .0862
despite the fact that the binomial approximation was not recommended because it failed the happiness test, the results are still not as far off as you might have otherwise thought.
i checked these answers a couple of times and so i think they're good.
try to duplicate the method and see if you get the same.
let me know if you don't get the same, and, if you don't, what you got, so i can see where one of us went wrong.