A Z score of 300 is an extremely large number as the z scores very rarely fall above 4 or below -4. About 0 percent of the scores fall above a z score of 300.
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.
-1.28
To find the proportion of a normal distribution corresponding to z-scores greater than +1.04, you can use the standard normal distribution table or a calculator. The area to the left of z = 1.04 is approximately 0.8508. Therefore, the proportion of the distribution that corresponds to z-scores greater than +1.04 is 1 - 0.8508, which is approximately 0.1492, or 14.92%.
When the normal curve is plotted using standard deviation units, each with a value of 1.00, it is referred to as the standard normal distribution. In this distribution, the mean is 0 and the standard deviation is 1, allowing for easy comparison of different data sets by transforming them into z-scores. The standard normal distribution is often represented by the symbol Z.
If the mean score is 100 and the standard deviation is 15, the distribution of scores is likely to follow a normal distribution, also known as a bell curve. In this distribution, approximately 68% of scores fall within one standard deviation of the mean (between 85 and 115), about 95% fall within two standard deviations (between 70 and 130), and about 99.7% fall within three standard deviations (between 55 and 145). This pattern indicates that most scores cluster around the mean, with fewer scores appearing as you move away from the center.
It is 68.3%
0.13
z-scores are distributed according to the standard normal distribution. That is, with the parameters: mean 0 and variance 1.
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.
99.7% of scores fall within -3 and plus 3 standard deviations around the mean in a normal distribution.
-1.28
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.
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.
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.
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.
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
To find the proportion of a normal distribution corresponding to z-scores greater than +1.04, you can use the standard normal distribution table or a calculator. The area to the left of z = 1.04 is approximately 0.8508. Therefore, the proportion of the distribution that corresponds to z-scores greater than +1.04 is 1 - 0.8508, which is approximately 0.1492, or 14.92%.