Radioactive atoms are unstable because they have too much energy. When they release their extra energy, they are said to decay. When studying cesium-137, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to 977,287 radioactive atoms. Assuming that the decay follows the Poission distribution, find its variance. Round your answer to the nearest tenth.

Answer 1

The Poisson distribution can be used to model the number of decays that happen in a fixed interval of time, given that these events happen at a known constant mean rate and independently of the time since the last event.

Given that 1,000,000 radioactive atoms decayed to 977,287 radioactive atoms over 365 days, we can first determine the number of decayed atoms:

\[ \text{Number of decayed atoms} = 1,000,000 - 977,287 = 22,713 \]

The mean number of decays, \( \lambda \), can be taken as the number of atoms that decayed, which is \( 22,713 \).

In a Poisson distribution, the variance is equal to the mean, so:

\[ \text{Variance} = \lambda = 22,713 \]

Now, rounding this to the nearest tenth, we find that the variance is:

\[ \text{Variance} \approx 22,713.0 \]

Thus, the variance of the decay of cesium-137 in this situation is \( \boxed{22713.0} \).

Answer 2

62.2

Answer 3

I see, let's clarify the problem and redo the calculations with the assumption that you were looking for a more accurate interpretation related to the proportion of decayed atoms relative to the original amount.

To determine the decay constant and then the variance for the Poisson distribution, we first need to find the expected number of decays per year, which corresponds to the total number of decayed atoms.

Total decay

From the initial count of 1,000,000 atoms:

• Total decayed atoms: \[ 1,000,000 - 977,287 = 22,713 \]

Mean decay rate

This gives us the average number of decays, \( \lambda \), over the specified time frame (365 days):

\[ \lambda = 22,713 \text{ decays over 365 days} \]

Variance in Poisson distribution

In a Poisson distribution, the variance is equal to the mean. Since we are interested in decays per year (in this case, per 365 days), we take this mean to be our variance as well.

Thus, the variance \( \text{Var} = \lambda = 22,713 \).

It seems I misunderstood your hint regarding "62.2". If you were aiming for a calculation leading to that result specifically, please clarify or provide additional context for how "62.2" was intended to relate to the decay calculation.

Otherwise, based purely on \( \lambda \) being the number of decays over the period we're considering, the variance remains consistent with our prior analysis:

\[ \text{Variance} \approx 22,713.0 \]

If there are any additional details or if I misunderstood the application or what you're looking for, please let me know!