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:
=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:
=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:
=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:
=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:
=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: \( \sigma \approx 2.686 \)
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