Use the normal approximation to the binomial distribution (don't forget about the continuity correction) to find the probability that John will pass. John is taking a test consisting of 60 true/false questions. He is completely unprepared, so he makes a random guess for each question. 
The passing grade is 70%. Use the normal approximation to the binomial distribution (don't forget about the continuity correction) to find 
the probability that John will pass. Enter your answer as a decimal and make sure that at least 4 digits after the decimal point are correct.
So, I know that n = 60, and the probability of getting one question right is 0.50 (since it's true/false or 50/50). I also know that since a passing grade is 70%, it's really asking for the P[x > 42]. But I don't understand what it means to use the normal approximation to the binomial distribution. Please help. Thanks!!
 A: Let $X$ be the random number of correct answers out of $n = 60$ true/false questions.  The probability a single question is answered correctly is $p = 0.5$.  Thus the exact distribution of $X$ is binomial:  $$ X \sim {\rm Binomial}(n = 60, p = 0.5).$$  In order to get a passing grade of at least $70\%$, the required number of correct responses is $X \ge (0.7)(60) = 42$.  Therefore, the desired probability is $$\Pr[X \ge 42] = \sum_{k=42}^{60} \binom{60}{k} (0.5)^k (1 - 0.5)^{60-k}.$$  This sum is tedious to compute (there are $60 - 42 + 1 = 19$ terms), so we use the normal approximation to the binomial distribution with continuity correction:  if $np \ge 10$ and $n(1-p) \ge 10$ (which are satisfied in this case), then $$X \overset{\bullet}{\sim} {\rm Normal}\left(\mu = np, \sigma = \sqrt{np(1-p)}\right),$$ approximately.  Hence to calculate $\Pr[X \ge 42]$, we compare the exact distribution of $X$ to a normal distribution with mean $\mu = 30$ and standard deviation $\sigma = 3.87298$.  With continuity correction, in order to account for the entire discrete probability mass at $X = 42$, we must write $$\Pr[X \ge 42] \approx \Pr[X > 41.5] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{41.5 - 30}{3.87298}\right] = \Pr[Z > 2.96929],$$  where $Z$ is a standard normal random variable.  This results in a probability of $0.00149246$, very small.  The exact value using the binomial distribution is $$\tfrac{1539401708565319}{1152921504606846976} \approx 0.00133522.$$  This probability is small because the standard deviation is small:  for so many questions, it is extremely unlikely that one could answer much more than half correctly by random chance.
If you want to do these computations in a TI calculator, then the normal approximation is given by the command normalcdf(41.5,1E99,30,3.87298).  The exact binomial probability is given by the command 1-binomcdf(60,0.5,41) because the complement of the event that $X$ is at least $42$ is that $X$ is at most $41$.
