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The percent of Arts students (other than students in Theater and English) who are doing a minor in english is 20%. If 300 students are selected, what is the approximate probability that:

i. more than 150 are taking a minor in english?

ii. between 20 and 60 are taking a minor in english?

iii. exactly 50 are taking a minor in english?

I've asked this question before, but the person who answered gave me i) 0, ii) 0.50, and iii) 0.0203. They didn't provide any steps, and no matter what I tried, I couldn't figure out how they got those values. I would really appreciate it if someone could explain step by step how to get these values?

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  • $\begingroup$ You seem to be missing some information. How many Arts students are there in total? "An incredibly large number, several orders of magnitude larger than 300" perhaps? Or a small number on the same order of magnitude as $300$ such as $600$? $\endgroup$ – JMoravitz Mar 20 at 0:15
  • $\begingroup$ Depending on your answer to that, for part (3) you will either use the binomial distribution (as a good approximation in the case of a massive number of students) or the hypergeometric distribution (in the case of a relatively small number of students). For the first two, you could find exact answers using these as well, but what is intended is probably to use a normal approximation. $\endgroup$ – JMoravitz Mar 20 at 0:19
  • $\begingroup$ That's all the question seems to provide in the textbook. They don't mention the number of Arts students. $\endgroup$ – GilmoreGirling Mar 20 at 0:19
  • $\begingroup$ Those answers seem to be based on the assumption that each student independently has a $\frac 15$ probability of taking an English minor. Of course, that assumption only makes sense if you assume that you have a very large number of Arts students. $\endgroup$ – lulu Mar 20 at 0:20
  • $\begingroup$ In the extreme case, if there were in fact only $300$ students to begin with, of course every time you select $300$ students from them you will have exactly $60$ of them who minor in english, so the answer to (iii) would be exactly zero as it is an impossibility. The textbook should have been more careful with how the question was phrased, but that isn't really your fault. Just keep it in mind for when you decide to write your own questions. $\endgroup$ – JMoravitz Mar 20 at 0:24
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The binomial distribution is a distribution commonly used to model repeated experiments with a clear success/fail where each trial is independent of one another and each trial has the same chance of success. For example, when flipping an unfair coin $n$ times and counting the number of heads you see, or selecting balls from a bag with replacement.

Here for our problem, we need to make the assumption that the number of students we are selecting from is a "large" number compared to $300$. In doing so, by modelling the process by a binomial distribution, we can get a good estimates to our probability calculations which are accurate to several decimal points of precision. The idea being, with so many people the difference between whether we were to have selected with replacement or without is so minuscule that it doesn't affect our probability by a significant amount.

Letting $X$ be the random variable which counts the number of successes, the probability of getting exactly $k$ successes out of $n$ independent trials where chance of success is $p$ for each trial is:

$$Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$

By plugging in $p=0.2, n=300$ and $k=50$ to the above, we get an immediate answer for (iii) of

$$Pr(X=50)=\binom{300}{50}\cdot 0.2^{50}\cdot (1-0.2)^{300-50}\approx 0.0207557$$

Now, for the earlier parts, if we have a strong enough calculator, we could just loop through a sum and get an exact answer such as this, but I expect that the intended solution is instead to approach using the normal approximation to the binomial.

It is well known that a binomially distributed random variable has average $np$ and standard deviation $\sqrt{np(1-p)}$, so in our case we have an average of $60$ and a standard deviation of $\sqrt{48}\approx 6.93$.

You may approximate the discrete binomial distribution for given values of $n,p$ by using the continuous normal distribution with mean $np$ and standard deviation $\sqrt{np(1-p)}$, the areas under the graph for each over a region will be very similar so long as $n$ and $np(1-p)$ are large enough. The larger they are, the more accurate the approximation will be.

To continue, to find the approximation for $Pr(a\leq X\leq b)$, we first apply a slight continuity correction by padding the distance away from the region we are interested in for each bound by adding or subtracting $\frac{1}{2}$ so we look instead at $Pr(a-\frac{1}{2}\leq X\leq b+\frac{1}{2})$. This, again, because the binomial distribution was discrete while the normal distribution is continuous. (With large enough standard deviation, this step might even be skipped)

Next, we find "how many standard deviations above the mean $a-\frac{1}{2}$ and $b+\frac{1}{2}$ are respectively (a negative value corresponding to actually being that amount below the mean). Through a bit of algebra, you will find that they are $\dfrac{a-\frac{1}{2}-np}{\sqrt{np(1-p)}}$ and $\dfrac{b+\frac{1}{2}-np}{\sqrt{np(1-p)}}$ respectively. (You may see this instead written as $\frac{a-\mu}{\sigma}$.) You will commonly hear these referred to as "$Z$-Scores" or "Standard Scores".

Now that we have found those values, we can ask the question for a standard normal distribution, asking what the probability is that it lies within that range of standard deviations away from those values. This can be done by looking at a table or using a calculator.

When using a table, remember that $Pr(m\leq Z\leq n) = Pr(Z\leq n) - Pr(Z< m)$


For the example of finding the probability between $20$ and $60$ people are English majors, we go through the effort of calculating the appropriate z-scores, in this case being $\approx\dfrac{19.5-60}{6.93}\approx -5.844$ and $\approx\dfrac{60.5-60}{6.93}\approx 0.072$ respectively.

Finding the probabilities of these, you can look on a table to find $Pr(Z\leq -5.844)\approx 0.0000$ and $Pr(Z\leq 0.072)\approx 0.5287$ respectively, so the probability of being between $20$ and $60$ people is approximately $0.5287$ (compare to the exact answer of $\approx 0.534476$ we found earlier with exact calculation). (compare also to the answer of $0.5$ that we would have gotten had we not used the continuity correction).

The other part can be calculated similarly.

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  • $\begingroup$ Hi! for finding the probability between 20 and 60 people are English minors, I was wondering where the 19.5 - 60 came from? I thought to calculate the z score you would subtract the mean from x and divide that by the standard dev? $\endgroup$ – GilmoreGirling Mar 20 at 8:12
  • $\begingroup$ Also when calculating C(300,50) I keep getting an error on my calculator? $\endgroup$ – GilmoreGirling Mar 20 at 8:21
  • $\begingroup$ @Gilmore 300 choose 50 is a big number, some 58 digits long. If you are using a cheap calculator, I'm not surprised if you get an overflow error. In that case, just don't give the answer in decimal form. As for the z-score what are you not understanding? I did subtract the mean from 20. I also then subtracted an extra 0.5 for good measure. I then divided by the standard deviation. Are you confused by the extra 0.5? As mentioned already above it is a continuity correction and just so happens to be helpful. Search elsewhere on the site for discussion as to why that is. $\endgroup$ – JMoravitz Mar 20 at 9:59
  • $\begingroup$ See math.stackexchange.com/q/416150/179297 $\endgroup$ – JMoravitz Mar 20 at 10:01
  • $\begingroup$ Ah thank you for replying! To confirm, is continuity correlation also needed for the example when calculating more than 150 are taking a minor in stats? $\endgroup$ – GilmoreGirling Mar 20 at 11:39

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