If you light the fuse of a firecracker the time it will take to explode is N(3.2, 1.22) seconds. What is the probability that a firecracker will take longer than 4.2 seconds to explode? Use 4 decimal places What is the probability that two firecrackers will take longer than 4.2 seconds to explode on average? Use 4 decimal places What is the probability that three firecrackers will take longer than 4.2 seconds to explode on average? Use 4 decimal places
The Correct Answer and Explanation is:
To solve this, we will apply concepts from probability, specifically the normal distribution. The time it takes for a firecracker to explode follows a normal distribution with the following parameters:
- Mean (μ) = 3.2 seconds
- Standard deviation (σ) = 1.22 seconds
We will use the z-score formula to find the probabilities:z=x−μσz = \frac{x – \mu}{\sigma}z=σx−μ
Where:
- xxx is the value we are comparing to (in this case, 4.2 seconds),
- μ\muμ is the mean (3.2 seconds),
- σ\sigmaσ is the standard deviation (1.22 seconds).
1. Probability for one firecracker
We are interested in finding the probability that a firecracker will take longer than 4.2 seconds to explode.
First, we calculate the z-score:z=4.2−3.21.22=1.01.22≈0.8197z = \frac{4.2 – 3.2}{1.22} = \frac{1.0}{1.22} \approx 0.8197z=1.224.2−3.2=1.221.0≈0.8197
Next, we use a z-table or a standard normal distribution calculator to find the cumulative probability for a z-score of 0.8197. This gives:P(Z<0.8197)≈0.7939P(Z < 0.8197) \approx 0.7939P(Z<0.8197)≈0.7939
Now, since we want the probability that the firecracker will take longer than 4.2 seconds, we subtract this cumulative probability from 1:P(X>4.2)=1−0.7939=0.2061P(X > 4.2) = 1 – 0.7939 = 0.2061P(X>4.2)=1−0.7939=0.2061
So, the probability that a firecracker will take longer than 4.2 seconds to explode is approximately 0.2061.
2. Probability for two firecrackers
For two firecrackers, we are looking for the probability that both will take longer than 4.2 seconds to explode on average. The key assumption here is that the times are independent, so the probability for each firecracker is independent.
The probability for each firecracker to take longer than 4.2 seconds is 0.2061, and for two firecrackers, we multiply this probability by itself:P(both)=0.2061×0.2061=0.0425P(\text{both}) = 0.2061 \times 0.2061 = 0.0425P(both)=0.2061×0.2061=0.0425
So, the probability that both firecrackers will take longer than 4.2 seconds to explode on average is approximately 0.0425.
3. Probability for three firecrackers
For three firecrackers, the same logic applies, and we simply multiply the individual probability three times:P(all three)=0.2061×0.2061×0.2061=0.0088P(\text{all three}) = 0.2061 \times 0.2061 \times 0.2061 = 0.0088P(all three)=0.2061×0.2061×0.2061=0.0088
So, the probability that all three firecrackers will take longer than 4.2 seconds to explode on average is approximately 0.0088.
Summary:
- Probability that one firecracker will take longer than 4.2 seconds: 0.2061
- Probability that two firecrackers will take longer than 4.2 seconds to explode on average: 0.0425
- Probability that three firecrackers will take longer than 4.2 seconds to explode on average: 0.0088
These probabilities make sense because as the number of firecrackers increases, the likelihood that all will take longer than 4.2 seconds decreases exponentially.
