This problem can be solved using the binomial distribution, since we are looking for the probability of a specific number of occurrences (side effects) in a sample of users.

The binomial distribution formula is:

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

where:

• $n = 230$ is the number of users in the sample,
• $p = 0.05$ is the probability of a user experiencing side effects,
• $k$ is the number of users experiencing side effects (in this case, between 8 and 12).

The problem asks for the probability of finding between 8 and 12 cases with side effects, so we need to find:

$P(8 \leq X \leq 12) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

This is quite tedious to calculate manually, so we can use the binomial cumulative distribution to solve this. However, an approximation can also be made using the normal distribution since $n$ is large and $p$ is small, making the binomial distribution approximately normal.

Step 1: Approximating Binomial with Normal Distribution

The mean and standard deviation of the binomial distribution are: $\mu = np = 230 \times 0.05 = 11.5$ $\sigma = \sqrt{np(1 - p)} = \sqrt{230 \times 0.05 \times 0.95} \approx 3.297$

Now, we can approximate the probability by converting the binomial problem to the normal distribution.

Step 2: Applying Normal Approximation

Using the continuity correction, we want to find:

$P(7.5 \leq X \leq 12.5)$

Standardizing this using the z-score formula:

$Z = \frac{X - \mu}{\sigma}$

For $X = 7.5$: $Z = \frac{7.5 - 11.5}{3.297} \approx -1.215$

For $X = 12.5$: $Z = \frac{12.5 - 11.5}{3.297} \approx 0.303$

Step 3: Finding the Probabilities

Using standard normal distribution tables or a calculator, we find:

$P(Z \leq -1.215) \approx 0.1122$ $P(Z \leq 0.303) \approx 0.6180$

Thus, the probability of finding between 8 and 12 cases with side effects is approximately:

$P(7.5 \leq X \leq 12.5) = 0.6180 - 0.1122 = 0.5058$

Final Answer:

The probability of finding between 8 and 12 cases with side effects is approximately 0.5058.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How do you calculate the binomial probability directly without approximation?
2. What is the exact probability for each value of $k$ using the binomial formula?
3. How does the continuity correction improve normal approximation for discrete distributions?
4. What is the z-score and how is it used in normal distribution calculations?
5. What factors determine when a binomial distribution can be approximated by a normal distribution?

Tip: When using normal approximation to a binomial distribution, don't forget to apply the continuity correction for more accurate results!