The square of a standard normal variate is a
Chi-square distribution
A standard normal variate, often denoted by \(Z\), is a random variable that follows the standard normal distribution. This distribution has a mean of 0 and a variance of 1. In mathematical notation, we write \(Z \sim N(0, 1)\).
The question asks about the distribution of the square of such a variable, which is \(Z^2\).
In probability and statistics, there is a well-known result concerning the distribution of the square of a standard normal variate. This result is fundamental in the construction of several important statistical tests and distributions.
Thus, the distribution of the square of a standard normal variate is a chi-square distribution with 1 degree of freedom, denoted as \(\chi^2(1)\).
Let's briefly look at the other options provided to understand why they are not the correct distribution for the square of a standard normal variate:
Based on the definition and properties, the square of a standard normal variate is indeed a chi-square distribution with 1 degree of freedom.
The memory-less property is followed by which of the following continuous distribution:
A Poisson distribution has a double mode at x = 1 and x = 2. The probability for x = 1 or for x = 2 of these two value is:
If a discrete random variable X follows uniform distribution and assume only the values 8, 9, 11, 15, 18, 20, the value of P(|X - 14| < 5) will be:
The probability density function of a random variable X is f(x) = \(\frac{\pi}{10} sin \frac{\pi x}{5}\) ; 0 ≤ x ≤ 5. The first quartile of X is:
The mode of a geometric distribution with parameter p is: