Refer to the mean and SD which you computed in previous assignments for your assigned variable. Using this information, select an extreme value from the NCS data file for the variable. Indicate the value.For example, if you previously used the variable, depression, an extreme score for a person in the sample would be 90, if the mean depression score you previously computed was 50 and the SD = 20. This score is much greater than one SD above that mean, 50 + 20 = 70.By hand, compute the z score for the value you selected using the mean and SD, and show your work.The z score for the depression score above would be found as follows:Z = (X - M) / SD = 90 - 50 / 20 = 2Use the table in the module to find the percentile rank for this z score assuming a normal distribution.The table in the module for this unit allows you to find the percentage of people in a normal distribution that score below any given z score, a percentile rank. The first column contains z scores, and the other columns indicate proportions below those z scores. The second column has the percentages for z scores taken to one decimal place. For example, a z score of 2.0 has 97.7% of the scores below it (a proportion of .97725 x 100).What is the probability that a score with this value, or higher would be found in the population by chance?The percentages in the normal distribution indicate probabilities of a score occurring in a normal distribution of scores from a random sample. For example, there is a 97.2% probability that someone will have a lower depression score than the subject with a z = 2, and a 100 - 97.2 = 1.8% probability that someone will have a higher depression score than the subject with a z = 2 in a normal distributionWhat is the general range of z values most likely to be found in a random sample?Z scores near the mean (z=0) are most common, and most likely to be selected in a random sample. In contrast, extreme scores are less common, and least likely to be sampled at random (by chance). for the mean of 3.09 and the standard deviation of 6.998