This problem can be solved using the binomial distribution, since we are dealing with a fixed number of trials (340 users), two possible outcomes (side effect or no side effect), and a constant probability of a side effect occurring (5% or 0.05).

1. Probability of finding between 8 and 12 cases with side effects

Let's define:

• $n = 340$ (the number of users in the sample)
• $p = 0.05$ (the probability of a serious side effect for each user)
• $X$ is a binomial random variable representing the number of users who experience side effects.

We are asked to find $P(8 \leq X \leq 12)$, which can be calculated as:

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

Each term can be computed using the binomial probability formula:

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

Where:

• $\binom{n}{k}$ is the binomial coefficient $\frac{n!}{k!(n-k)!}$
• $p^k$ is the probability of $k$ users having side effects
• $(1 - p)^{n - k}$ is the probability of the remaining users not having side effects.

However, given the large number of trials, it is more efficient to use the normal approximation to the binomial distribution. The normal approximation is valid because $np$ and $n(1 - p)$ are both sufficiently large (which is true in this case, since $np = 340 \times 0.05 = 17$).

Step 1: Apply the normal approximation to the binomial distribution

For a binomial distribution $X \sim B(n, p)$, the normal approximation gives us:

$X \sim N(\mu, \sigma^2)$

Where:

• $\mu = np = 340 \times 0.05 = 17$ (mean)
• $\sigma = \sqrt{np(1 - p)} = \sqrt{340 \times 0.05 \times 0.95} \approx 3.99$ (standard deviation)

Step 2: Use continuity correction

To approximate the probability $P(8 \leq X \leq 12)$, we use continuity correction by adjusting the values by $\pm 0.5$. So we calculate:

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

This becomes:

$P\left( \frac{7.5 - 17}{3.99} \leq Z \leq \frac{12.5 - 17}{3.99} \right)$

$P\left( \frac{-9.5}{3.99} \leq Z \leq \frac{-4.5}{3.99} \right) = P(-2.38 \leq Z \leq -1.13)$

Using a standard normal distribution table or a calculator:

$P(Z \leq -1.13) \approx 0.1292$ $P(Z \leq -2.38) \approx 0.0087$

Thus:

$P(8 \leq X \leq 12) = 0.1292 - 0.0087 = 0.1205$

So, the probability of finding between 8 and 12 cases with side effects is approximately 0.1205 (or 12.05%).

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2. Probability of finding more than 16 cases with side effects

We need to calculate $P(X > 16)$.

Using the normal approximation with continuity correction, we calculate:

$P(X > 16) \approx P(X \geq 16.5)$

This becomes:

$P\left( Z \geq \frac{16.5 - 17}{3.99} \right) = P(Z \geq -0.13)$

From the standard normal distribution table:

$P(Z \geq -0.13) = 1 - P(Z \leq -0.13) \approx 1 - 0.4483 = 0.5517$

Thus, the probability of finding more than 16 cases with side effects is approximately 0.5517 (or 55.17%).

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Would you like more details on any part of the calculations, or do you have any further questions?

Follow-up questions:

1. What is the binomial coefficient, and how is it calculated?
2. How does the normal approximation to the binomial distribution work?
3. Why is a continuity correction necessary in normal approximation?
4. What are the conditions for the normal approximation to be valid for binomial problems?
5. How does the shape of a binomial distribution change as $p$ approaches 0 or 1?

Tip:

When using the normal approximation for binomial distributions, always ensure that $np$ and $n(1 - p)$ are both greater than 5 for an accurate approximation.