Question

The article, “Should You Report That Fender-Bender?” (Consumer Reports, 2013:15), reported that 7 in 10 automobile accidents involve a single vehicle. A random sample of 25 accident reports is selected.

Answer the following questions. Include your calculator functions or Excel commands for each probability question and round your answers to 4 digits.

a. What is the probability that exactly 8 accidents involve a single vehicle?

b. What is the probability that at most 13 accidents involve a single vehicle?

c. What is the probability that at least 5 accidents involve a single vehicle?

d. What is the probability that fewer than 15 accidents involve a single vehicle?

e. What is the probability that more than 18 accidents involve a single vehicle?

f. Find the mean and standard deviation of the accidents that involve a single vehicle. Show your formula setup and your work. Use proper symbols.

Answer 1

To solve these problems related to the probability of automobile accidents involving a single vehicle, we can utilize the binomial probability formula. The issues presented meet the criteria for a binomial distribution where:

- \( n = 25 \) (the number of trials, i.e., accident reports),
- \( p = 0.7 \) (the probability of success, i.e., accidents involving a single vehicle), and
- \( q = 1 - p = 0.3 \) (the probability of failure, i.e., accidents involving more than one vehicle).

The general formula for the binomial probability \( P(X = k) \) is given by:

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

Where:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient,
- \( k \) is the number of successes.

### a. Probability that exactly 8 accidents involve a single vehicle

Let \( k = 8 \):

\[
P(X = 8) = \binom{25}{8} (0.7)^8 (0.3)^{25 - 8}
\]

**Using Excel**:

To calculate this in Excel, you can use the `BINOM.DIST` function:

```excel
=BINOM.DIST(8, 25, 0.7, FALSE)
```

**Calculating**:

\[
P(X = 8) = \binom{25}{8} (0.7)^8 (0.3)^{17} \approx 0.1078
\]

### b. Probability that at most 13 accidents involve a single vehicle

This is the sum of the probabilities from 0 to 13:

\[
P(X \leq 13) = \sum_{k=0}^{13} P(X = k)
\]

**Using Excel**:

```excel
=BINOM.DIST(13, 25, 0.7, TRUE)
```

**Calculating**:

\[
P(X \leq 13) \approx 0.7949
\]

### c. Probability that at least 5 accidents involve a single vehicle

This is 1 minus the probability of fewer than 5 accidents:

\[
P(X \geq 5) = 1 - P(X < 5) = 1 - P(X \leq 4)
\]

**Using Excel**:

```excel
=1 - BINOM.DIST(4, 25, 0.7, TRUE)
```

**Calculating**:

\[
P(X \geq 5) \approx 0.9795
\]

### d. Probability that fewer than 15 accidents involve a single vehicle

Similar to part b, we want:

\[
P(X < 15) = P(X \leq 14)
\]

**Using Excel**:

```excel
=BINOM.DIST(14, 25, 0.7, TRUE)
```

**Calculating**:

\[
P(X < 15) \approx 0.8890
\]

### e. Probability that more than 18 accidents involve a single vehicle

This is:

\[
P(X > 18) = 1 - P(X \leq 18)
\]

**Using Excel**:

```excel
=1 - BINOM.DIST(18, 25, 0.7, TRUE)
```

**Calculating**:

\[
P(X > 18) \approx 0.1390
\]

### f. Mean and standard deviation of the accidents that involve a single vehicle

The mean \( \mu \) and standard deviation \( \sigma \) for a binomial distribution are given by:

\[
\mu = np = 25 \times 0.7 = 17.5
\]

\[
\sigma = \sqrt{npq} = \sqrt{25 \times 0.7 \times 0.3} \approx 2.686
\]

**Final Results**:

- a. \( P(X = 8) \approx 0.1078 \)
- b. \( P(X \leq 13) \approx 0.7949 \)
- c. \( P(X \geq 5) \approx 0.9795 \)
- d. \( P(X < 15) \approx 0.8890 \)
- e. \( P(X > 18) \approx 0.1390 \)
- **Mean**: \( \mu = 17.5 \)
- **Standard Deviation**: \( \sigma \approx 2.686 \)