Introduction:
Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space. The mean and variance of a Poisson distribution are equal, and both are denoted by λ.

Given:
Mean of a Poisson variable x is 1.

To find:
Probability that x takes the value at least 1.

Solution:
To find the probability that x takes the value at least 1, we can use the complement rule of probability, which states that the probability of an event happening is equal to one minus the probability of the event not happening.

P(X≥1) = 1 - P(X=0)

We know that the mean of a Poisson distribution is λ, and in this case, λ = 1.

The probability mass function of a Poisson distribution is given by:
P(X=k) = (e^(-λ) * λ^k) / k!

Therefore,
P(X=0) = e^(-λ) * λ^0 / 0! = e^(-1) * 1^0 / 1 = e^(-1) = 0.3679

Using the complement rule, we get:
P(X≥1) = 1 - P(X=0) = 1 - 0.3679 = 0.6321

Therefore, the probability that x takes the value at least 1 is 0.6321.

Conclusion:
The probability that x takes the value at least 1 when the mean of a Poisson variable x is 1 is 0.6321.

ANS: 0.632120311