A sample of a radioactive material is studied in a lab. There are 1053 gamma ray emissions over 390 seconds.Use the Poisson distribution to find the probability that exactly 2 gamma rays are emitted in a single second.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)
Question
A sample of a radioactive material is studied in a lab. There are 1053 gamma ray emissions over 390 seconds.Use the Poisson distribution to find the probability that exactly 2 gamma rays are emitted in a single second.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)
Solution
To solve this problem, we first need to find the average rate of gamma ray emissions per second. This is given by the total number of emissions divided by the total time.
Step 1: Calculate the average rate (λ) λ = Total number of emissions / Total time λ = 1053 / 390 λ = 2.7 (approximately)
This means that on average, there are 2.7 gamma ray emissions per second.
Step 2: Use the Poisson distribution formula to find the probability of exactly 2 emissions in a second. The formula is:
P(X=k) = (λ^k * e^-λ) / k!
where:
- P(X=k) is the probability we want to find,
- λ is the average rate,
- k is the number of occurrences we are interested in (in this case, 2),
- e is the base of the natural logarithm (approximately 2.71828),
- and "!" denotes factorial.
Step 3: Substitute the values into the formula:
P(X=2) = (2.7^2 * e^-2.7) / 2! P(X=2) = (7.29 * 0.067) / 2 P(X=2) = 0.488 / 2 P(X=2) = 0.244
So, the probability that exactly 2 gamma rays are emitted in a single second is approximately 0.244, or 24.4% when rounded to three decimal places.
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