A nurse has found that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 16%. Of the next 30 patients calling in claiming to have the flu, we are interested in how many actually have the flu.

Let the random variable X be the number of patients, out of the 30 who called the medical help line, that actually have the flu. How is X distributed?

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Part 3 of 3
Find the probability that exactly 4 of the 30 patients actually have the flu. 


Find the probability that fewer than 8 of the 30 patients actually have the flu. 


This problem deals with **binomial distribution** because each patient call is an independent event with only two possible outcomes: the patient either has the flu or doesn't have the flu. We can define the situation using the following parameters:

- \( X \) is the random variable representing the number of patients (out of 30) that actually have the flu.
- Each patient has a 16% chance of having the flu, so the probability of success (\( p \)) is 0.16.
- There are \( n = 30 \) patients who call, and we want to find probabilities for specific values of \( X \).

The distribution of \( X \) can be described as:

\[
X \sim \text{Binomial}(n = 30, p = 0.16)
\]

### 1. Finding the probability that exactly 4 of the 30 patients actually have the flu:

We use the binomial probability formula to find this probability. The formula is:

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

Where:
- \( n = 30 \),
- \( k = 4 \) (since we are looking for the probability that exactly 4 patients have the flu),
- \( p = 0.16 \), and
- \( 1 - p = 0.84 \).

Let's calculate \( P(X = 4) \).

### 2. Finding the probability that fewer than 8 patients actually have the flu:

Here, we need to calculate \( P(X < 8) \), which is the cumulative probability of \( X \) taking values from 0 to 7. This can be calculated as:

\[
P(X < 8) = P(X = 0) + P(X = 1) + \dots + P(X = 7)
\]

We can compute this cumulative probability directly. I'll calculate both now.1. The probability that exactly 4 of the 30 patients actually have the flu is approximately **0.193** (or 19.3%).

2. The probability that fewer than 8 of the 30 patients actually have the flu is approximately **0.905** (or 90.5%).

Would you like more details on how these calculations are done, or do you have any other questions?

Here are 5 related questions that expand on this information:

1. What is the probability that none of the 30 patients actually have the flu?
2. What is the expected number of patients who actually have the flu out of 30?
3. How would the probabilities change if the probability of having the flu increased to 25%?
4. What is the probability that more than 10 of the 30 patients actually have the flu?
5. Can we approximate this binomial distribution with a normal distribution? If so, how?

**Tip:** For large sample sizes, binomial distributions can often be approximated by a normal distribution, which simplifies calculations.

A nurse has found that when a patient calls the medical advice line claiming to have the flu, the chance that he or she truly has the flu (and not just a nasty cold) is only about 16%. Of the next 30 patients calling in claiming to have the flu, we are interested in how many actually have the flu. Let the random variable X be the number of patients, out of the 30 who called the medical help line, that actually have the flu. How is X distributed? Find the probability that exactly 4 of the 30 patients actually have the flu. Find the probability that fewer than 8 of the 30 patients actually have the flu.

Explore how binomial distribution helps find the probability that 4 out of 30 patients have the flu, with a 16% chance per patient. Learn key concepts, such as the binomial probability formula and cumulative probability. This problem is suitable for high school students in Grades 10-12.

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