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What requirements are necessary for a normal probability distribution to be a standard normal probability?
For a normal probability distribution to be considered a standard normal probability distribution, it must have a mean of 0 and a standard deviation of 1. This standardization allows for the use of z-scores, which represent the number of standard deviations a data point is from the mean. Any normal distribution can be transformed into a standard normal distribution through the process of standardization.
For a normal distribution what z-score value separates the lowest 10 percent of the scores from the rest of the distribution?
-1.28
Why does a researcher want to go from a normal distribution to a standard normal distributio?
A normal distribution simply enables you to convert your values, which are in some measurement unit, to normal deviates. Normal deviates (i.e. z-scores) allow you to use the table of normal values to compute probabilities under the normal curve.
What role do z scores play in this transformation of data from multiple distributions to standard normal distribution?
Z-scores standardize data from various distributions by transforming individual data points into a common scale based on their mean and standard deviation. This process involves subtracting the mean from each data point and dividing by the standard deviation, resulting in a distribution with a mean of 0 and a standard deviation of 1. This transformation enables comparisons across different datasets by converting them to the standard normal distribution, facilitating statistical analysis and interpretation.
The distribution of ACT scores in recent years has been roughly normal with mean 20.9 and standard deviation 4.4 The quartiles of any distribution are the values with cumulative proportions 0.25 and?
17.7 and 20.9
What percent of the scores in a normal distribution will fall within one standard deviation?
It is 68.3%
In a standard normal distribution about percent of the scores fall above a z-score of 3.00?
0.13
How do you find normal distribution of z-scores?
z-scores are distributed according to the standard normal distribution. That is, with the parameters: mean 0 and variance 1.
What percentage of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution?
99.7% of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution.
For a normal distribution what z-score value separates the lowest 10 percent of the scores from the rest of the distribution?
-1.28
What role do z-scores play in the transformation of data from multiple distributions to the standard normal distribution?
None.z-scores are linear transformations that are used to convert an "ordinary" Normal variable - with mean, m, and standard deviation, s, to a normal variable with mean = 0 and st dev = 1 : the Standard Normal distribution.
Why in a normal distribution the distribution will be less spread out when the standard diviation of the raw scores is small?
The standard deviation (SD) is a measure of spread so small sd = small spread. So the above is true for any distribution, not just the Normal.
Why does a researcher want to go from a normal distribution to a standard normal distribution?
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.
z scores?
If a random variable X has a normal distribution with mean m and standard error s, then the z-score corresponding to the value X = x is (x - m)/s.
For a normal distribution find the z-scores values that separate the middle 60 percent of the distribution for the 40 percent in the tails?
The standard normal table tells us the area under a normal curve to the left of a number z. The tables usually give only the positive value since one can use symmetry to find the corresponding negative values. The middle 60 percent leaves 20 percent on either side. So we want the z scores that correspond to that 80 percentile which is .804. Therefore the values are are between z scores of -.804 and .804 * * * * * I make it -0.8416 to 0.8416
How many of scores will be within 1 standard deviation of the population mean?
Assuming a normal distribution 68 % of the data samples will be with 1 standard deviation of the mean.
How much is 84 percentile equals mean plus 1 standard deviation or mean plus 1.4 standard deviation. Can you give me reference also please?
The cumulative probability up to the mean plus 1 standard deviation for a Normal distribution - not any distribution - is 84%. The reference is any table (or on-line version) of z-scores for the standard normal distribution.
