False. It is approximately 1. Theoretically, it is not 1. I used excel, and I know the probability is between 0.999999 and 1. as the probability of Z<6 is 0.999999. I can't calculate the probability exactly because excel only goes to 7 place accuracy.
Yes.
A researcher wants to go from a normal distribution to a standard normal distribution because the latter allows him/her to make the correspondence between the area and the probability. Though events in the real world rarely follow a standard normal distribution, z-scores are convenient calculations of area that can be used with any/all normal distributions. Meaning: once a researcher has translated raw data into a standard normal distribution (z-score), he/she can then find its associated probability.
The normal distribution, also known as the Gaussian distribution, has a familiar "bell curve" shape and approximates many different naturally occurring distributions over real numbers.
If a random variable X has a Normal distribution with mean m and standard deviation s, then z = (X - m)/s has a Standard Normal distribution. That is, Z has a Normal distribution with mean 0 and standard deviation 1. Probabilities for a general Normal distribution are extremely difficult to obtain but values for the Standard Normal have been calculated numerically and are widely tabulated. The z-transformation is, therefore, used to evaluate probabilities for Normally distributed random variables.
The answer will depend on the underlying distribution for the variable. You may not simply assume that the distribution is normal.
It is 0.5
It is 0.1587
0.636 approx.
In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. This means that if x is a random variable that follows a normal distribution, there is about a 68% probability that x will be within one standard deviation of its mean. For distributions that are not normal, the probability may vary and would need to be determined based on the specific characteristics of that distribution.
Yes, to approximately standard normal.If the random variable X is approximately normal with mean m and standard deviation s, then(X - m)/sis approximately standard normal.
P(0 < z < 2.53) refers to the probability that a standard normal random variable (z) falls between 0 and 2.53. To find this probability, you would typically look up the z-scores in a standard normal distribution table or use a calculator. The cumulative probability for z = 2.53 is approximately 0.994, and for z = 0, it is 0.5. Therefore, P(0 < z < 2.53) is approximately 0.994 - 0.5 = 0.494.
True. Due to the symmetry of the normal distribution.
I have included two links. A normal random variable is a random variable whose associated probability distribution is the normal probability distribution. By definition, a random variable has to have an associated distribution. The normal distribution (probability density function) is defined by a mathematical formula with a mean and standard deviation as parameters. The normal distribution is ofter called a bell-shaped curve, because of its symmetrical shape. It is not the only symmetrical distribution. The two links should provide more information beyond this simple definition.
0.97
It is the Standard normal variable.
Approx 0.0027
1