Question

4 the number of radioactive particles emitted per 150-minute period by some material has a poisson distribution with mean 0.7.

(i) find the probability that at most 2 particles will be emitted during a randomly chosen 10-hour period. 3

Explanation:

Define the initial rate parameter

The number of particles emitted per 150-minute period follows a Poisson distribution with mean \(\lambda_1 = 0.7\).

Scale the rate parameter

We need to find the mean for a 10-hour period.
Convert 10 hours to minutes:

$$ 10 \text{ hours} = 10 \times 60 = 600 \text{ minutes} $$

Calculate the scaling factor \(k\):

$$ k = \frac{600}{150} = 4 $$

Calculate the new mean \(\lambda_2\):

$$ \lambda_2 = 4 \times 0.7 = 2.8 $$

State the probability formula

Let \(X\) be the number of particles emitted in a 10-hour period, where \(X \sim \text{Poisson}(2.8)\).
The probability mass function is:

$$ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} = \frac{e^{-2.8} (2.8)^x}{x!} $$

Calculate the target probability

We need to find the probability of at most 2 particles, \(P(X \le 2)\):

$$ P(X \le 2) = P(X = 0) + P(X = 1) + P(X = 2) $$

Compute each term:

$$ P(X = 0) = e^{-2.8} \approx 0.060810 $$

$$ P(X = 1) = e^{-2.8} \times 2.8 \approx 0.170268 $$

$$ P(X = 2) = \frac{e^{-2.8} \times 2.8^2}{2} = e^{-2.8} \times 3.92 \approx 0.238375 $$

Sum the probabilities

Add the individual probabilities:

$$ P(X \le 2) \approx 0.060810 + 0.170268 + 0.238375 = 0.469453 $$

Rounding to three significant figures gives \(0.469\).

Answer:

\(0.469\)