According to a study done by Nick Wilson of Otago University Wellington, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267
Suppose you sit on a bench in a mall and observe people's habits as they sneeze.

1. Using the binomial distribution, what is the probability that among 10 randomly observed individuals, exactly 6 do not cover their mouth when sneezing?

2. Using the binomial distribution, what is the probability that among 10 randomly observed individuals, fewer than 5 do not cover their mouth when sneezing?

3. Using the binomial distribution, would you be surprised if, after observing 10 individuals, fewer than half covered their mouth when sneezing? Why?

Let's tackle the problems using the binomial distribution.

### 1. Probability that exactly 6 out of 10 do not cover their mouth when sneezing

The binomial distribution formula is given by:

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

Where:
- \( n = 10 \) (number of trials)
- \( k = 6 \) (number of successes)
- \( p = 0.267 \) (probability of success)

\[
P(X = 6) = \binom{10}{6} (0.267)^6 (1 - 0.267)^{10 - 6}
\]

Calculating the components:
\[
\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = 210
\]

\[
0.267^6 \approx 0.0004
\]

\[
0.733^4 \approx 0.287
\]

So:
\[
P(X = 6) = 210 \cdot 0.0004 \cdot 0.287 \approx 0.0241
\]

Thus, the probability that exactly 6 out of 10 do not cover their mouth when sneezing is approximately 0.0241 or 2.41%.

### 2. Probability that fewer than 5 out of 10 do not cover their mouth when sneezing

To find the probability that fewer than 5 do not cover their mouth, we sum the probabilities of \( k = 0, 1, 2, 3, \) and \( 4 \):

\[
P(X < 5) = \sum_{k=0}^{4} P(X = k)
\]

Using the binomial formula for each \( k \):

\[
P(X = 0) = \binom{10}{0} (0.267)^0 (0.733)^{10} = 1 \cdot 1 \cdot 0.733^{10} \approx 0.0317
\]

\[
P(X = 1) = \binom{10}{1} (0.267)^1 (0.733)^9 = 10 \cdot 0.267 \cdot 0.733^9 \approx 0.1154
\]

\[
P(X = 2) = \binom{10}{2} (0.267)^2 (0.733)^8 = 45 \cdot 0.0713 \cdot 0.733^8 \approx 0.2164
\]

\[
P(X = 3) = \binom{10}{3} (0.267)^3 (0.733)^7 = 120 \cdot 0.019 \cdot 0.733^7 \approx 0.2638
\]

\[
P(X = 4) = \binom{10}{4} (0.267)^4 (0.733)^6 = 210 \cdot 0.0051 \cdot 0.733^6 \approx 0.2141
\]

Summing these probabilities:

\[
P(X < 5) = 0.0317 + 0.1154 + 0.2164 + 0.2638 + 0.2141 \approx 0.8414
\]

So, the probability that fewer than 5 out of 10 do not cover their mouth when sneezing is approximately 0.8414 or 84.14%.

### 3. Would you be surprised if fewer than half covered their mouth?

To determine if it's surprising, we need to calculate the probability of fewer than 5 out of 10 covering their mouth. The probability that fewer than 5 do not cover their mouth is 0.8414. Therefore, the probability that fewer than 5 cover their mouth is \( 1 - 0.8414 = 0.1586 \).

Since the probability is 0.1586 (15.86%), it is relatively unlikely, but not extremely so. Statistically, events with probabilities less than 0.05 (5%) are often considered surprising. Therefore, a 15.86% probability does not typically fall into the "surprising" category.

### Questions and Tip

1. What is the binomial distribution formula?
2. How do you calculate \(\binom{n}{k}\)?
3. What is the probability that exactly 4 out of 10 do not cover their mouth when sneezing?
4. How would you calculate \(P(X = 5)\) for this problem?
5. What is the cumulative distribution function for a binomial distribution?
6. How do you determine if a probability outcome is surprising?
7. What would the probability be if the sample size was 20 instead of 10 for the same number of successes?
8. How would the probabilities change if the probability of not covering the mouth was 0.5 instead of 0.267?

**Tip:** When using the binomial distribution, remember that the probability of success \(p\) and the number of trials \(n\) are crucial parameters. Adjusting these can significantly affect the probabilities of different outcomes.

Explore the binomial distribution in probability theory through examples related to people not covering their mouth when sneezing. Calculate probabilities for exactly 6 out of 10 individuals and fewer than 5 out of 10 individuals based on given conditions.

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