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An auditor samples $100$ of a firm’s travel vouchers to ascertain what percentage of the whole set of vouchers are improperly documented. What is the approximate probability that more than $30$% of the sampled vouchers are improperly documented if, in fact, only $20$% of all the vouchers are improperly documented? If you were the auditor and observed more than $30$% with improper documentation, what would you conclude about the firm’s claim that only $20$% suffered from improper documentation? Why?

I feel like I'm missing something simple in this problem. We want $P(X>.3|X=.2)$ right? Can I say that the distribution is normal?

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Actually, this is not a problem of conditional probability.

Let $X$ be a random variable which can assume two values: $X = I$ if the sample represented by $X$ is improperly documented and $X = C$ if it is correctly documented. We know that $p(X=I) = 0.2$.
Your problem can be easily resolved by using the binomial distribution ( A binomial distribution is used to calculate the probability to have exactly $k$ success in $n$ attempts knowing the probability of the success of a single case $p$.

In our case we have that $n=100$, $p = 0.2$. We are saying that we are inspecting $n=100$ voucher and we say that we are "successful" if a voucher is improperly documented ($X=I$). Since we want to know the probability to obtain "success" in the 30% of the cases, then the number of cases is $k=30$ (which is the 30% of $n=100$).

Using the binomial distribution we have that $$ f(k) = \left( \begin{array}{c} n \\ k \end{array}\right) p^k (1-p)^{(n-k)}$$

$f(k)$ is the probability to have $k$ of success knowing that we examine $n$ voucher and a single voucher is successful with probability $p$.

Using number, we have that $f(30) = \left( \begin{array}{c} 100 \\ 30 \end{array}\right) (0.2)^{30} (1-0.2)^{(100-30)} = 0.0052...$

Clearly if you want to know what is the probability to have more than 30%, you have to sum up all $f(k)$ for $k$ greater to $30$. More precisely

$$ \sum_{k=31}^{100} f(k) = \sum_{k=31}^{100}\left( \begin{array}{c} 100 \\ k \end{array}\right) (0.2)^k (1-0.2)^{(100-k)}$$

Using number (and Matlab) I obtain that $$ \sum_{k=31}^{100} f(k) = 0.0061...$$

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Assume that the probability of improper documentation is indeed $0.2$. Let $X$ be the number of improperly documented vouchers in a random sample of $100$. On the assumption that the number of vouchers we are sampling from is very large, we can assume that $X$ has binomial distribution, with "$n$" equal to $100$, and "$p$" equal to $0.2$.

Then the probability that exactly $k$ vouchers are improperly documented is $\dbinom{100}{k}(0.2)^k(0.8)^{100-k}$.

The probability that $31$ or more vouchers are improperly documented is $$\sum_{k=31}^{100}\binom{100}{k}(0.2)^k(0.8)^{100-k}.$$ Nowadays, one can calculate this without much trouble, with software free or expensive.

In the old days, direct calculation was a lengthy process. So the fact that $X$ has distribution well approximated by the normal would be used. The mean of $X$ is $(100)(0.2)=20$. The variance of $X$ is $(100)(0.2)(0.8)=16$, so $X$ has standard deviation $4$. The standard deviation is very simple, indication perhaps that you are expected to use a normal approximation.

The probability that a normal with mean $20$ and standard deviation $4$ is $\ge 31$ is easy to calculate using tables of the standard normal.

I do not know whether you are expected to use the continuity correction. If we don't, then the probability that $X\le 30$ is approximately the probability that $$Z\le \frac{30-20}{4},\tag{$1$}$$ where $Z$ is standard normal. Then our estimate is $1$ minus the number obtained in $(1)$.

If we use the continuity correction, the $30$ in $(1)$ is replaced by $30.5$.

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I don't think you want $P(X>.3|X=.2)$. Note that $X$ is the proportion of the sample of $100$ vouchers that are defective. So the probability does not really make sense.

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