~If you had a problem where the coin is tossed 80 times & were asked to find the probability of getting 50 or more correct, it would be a nightmare!

~You would have to calculate combinations & use the binomial formula at least 30 times..ugh! It becomes even worse if the problem had different values for p and q.

~Is there a better way? Yes!

~Using Bernoulli's back door along with the A key on the menu 2nd VARS (i.e., binomcdf( ), which gives the cumulative probabilies up to a given number of trials).

~How does it work?. Let's take our problem of getting 50 or more successes in 80 trials. The backdoor approach gives 1 - p(complement of this event), that is, 1 - p(at most 49 successes). Note that "at most" 49 successes is cumulative since it starts at 0. This can be calculated by using the binomcdf key once simply by inserting 80, .5, 49 & pressing enter. However, you must remember to subtract this number from 1. That will give you an answer of .0165.

~Is there an alternative method if you didn't have a calculator available? Yes!, using the standard normal curve to approximate the binomal distribution.

~As stated, A normal distribution can be used to approximate a binomial distribution provided np≥5 and nq≥5 with some minor adjustments for continuity. In any case, the bigger n, the better the approximation. (see Central Limit Theorem link)

~If, for example, we wanted the probability of getting 50 or more correct out of 80 tosses of a coin, we would do it this way. We need z scores for the standard normal curve to find probabilities (from a table), so we need to convert x ≥ 50 to z scores.

~We make a minor adjustment to the x scores before we convert them. This is called the "continuity correction" since we are using a continuous variable (the standard normal curve) to approximate a discrete variable (binomial distribution).

~For a lot more detail on this, see our textbook.

~So, the score 50 converts to 49.5 to 50.5. Also, the score of 80 would be 79.5 to 80.5. So, to get scores from 50 to 80, we subtact .5 from the lowest score & add .5 to the highest one. So, we use x's from 49.5 to 80.5 when considering the normal curve approximation.

~Now, convert them to z scores by z = (x-mean)/s.d., the mean = np = 80(1/2)= 40. The s.d. = √(npq) = √[80 (1/2)(1/2)] = √(20). So, the z scores are (49.5-40)/√(20) to (80.5-40)/√(20). (z=2.12 to z=9.06) Then find the probability using the table for z-scores. You would get .0216 (they would agree, if we round off to two decimal places, but a lot more work).

~Another way: you can avoid finding the z-scores completely by going to 2nd Vars, menu 2 (normalcdf) and entering (49.5, 10000, (80)(.5), sqr[(80)(.5)(.5)] to get .0168 (better result). The 49.5 was used since we are using a normal distribution to approximate the binomial distribution (continuity correction). It would be silly to use this method since we can get the exact answer from menu A under 2nd Vars (see upper part of discussion).