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!!

  • $\begingroup$ You got it. All it means is that it's a normal distribution which it is if you use the formula you learned in class or the calculator's binomcdf. $\endgroup$ – Shahar Mar 27 '14 at 23:25
  • $\begingroup$ Will the calculator's binomcdf account for the continuity correction as well though? $\endgroup$ – Shell Mar 27 '14 at 23:28
  • $\begingroup$ If you want $x>42$, then do binomcdf up to $41$ and do $1-\text{that}$. Is that what you meant by continuity correction? $\endgroup$ – Shahar Mar 27 '14 at 23:29
  • $\begingroup$ In class our professor talked about adding 0.5, the continuity correlation, to get a better estimation. But whether I use the binomcdf on my calculator or code in python, I get a very small number. Much smaller than I was expecting... $\endgroup$ – Shell Mar 27 '14 at 23:32
  • $\begingroup$ Oh, I don't know what that means. But, yeah, I just got a low number too. We're probably not doing something right. Expected value is $np=30$ so it can't be this low. $\endgroup$ – Shahar Mar 27 '14 at 23:35

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 $41$.

  • $\begingroup$ Yeah that's what I got but that seems awfully low. $\endgroup$ – Shahar Mar 29 '14 at 20:22

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