The Poisson distribution is a discrete distribution, with random variable k, related to the number events. The discrete probability function (probability mass function) is given as: f(k; L) where L (lambda) is the mean and square root of lambda is the standard deviation, as given in the link below: http://en.wikipedia.org/wiki/Poisson_distribution
The mean, median, and mode of a normal distribution are equal; in this case, 22. The standard deviation has no bearing on this question.
No. The standard deviation is not exactly a value but rather how far a score deviates from the mean.
Yes.
The mean and standard deviation do not, by themselves, provide enough information to calculate probability. You also need to know the distribution of the variable in question.
There are 16 trials.
The mean and variance are equal in the Poisson distribution. The mean and std deviation would be equal only for the case of mean = 1. See related link.
The distribution of x can be almost anything - apart from the Poisson distribution.
Mean 0, standard deviation 1.
The triangular, uniform, binomial, Poisson, geometric, exponential and Gaussian distributions are some that can be so defined. In fact, the Poisson and exponential need only the mean.
The mean of a distribution gives no information about the standard deviation.
Standard deviation describes the spread of a distribution around its mean.
It depends on what the distribution is. In a Normal or Gaussian distribution, the standard deviation is the square root of the mean, so it could be 3.1 but, again, it depends on the distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1.
The statement is probably: The mean and standard deviation of a distribution are 55 and 4.33 respectively.
The mean and standard deviation often go together because they both describe different but complementary things about a distribution of data. The mean can tell you where the center of the distribution is and the standard deviation can tell you how much the data is spread around the mean.
No.
It is any standardised distribution.