# Table For Z Value

### Z Score helps us compare results to the normal population or mean Q: 300 college student’s exam scores are tallied at the end of the semester. (Source: www.dummies.com)

Contents

The average score for the batch was 700 (µ) and the standard deviation was 180 (). Let’s find out how well Eric scored compared to his batch mates.

De Moiré came about to create the normal distribution through his scientific and math based approach to the gambling. He was trying to come up with a mathematical expression for finding the probabilities of coin flips and various inquisitive aspects of gambling.

He discovered that although data sets can have a wide range of values, we can ‘standardize’ it using a bell shaped distribution curve which makes it easier to analyze data by setting it to a mean of zero and a standard deviation of one. It was realized that normal distribution applied to many mathematical and real life phenomena.

For example, Belgian astronomer, Lambert Outlet (22nd February 1796 to 17th February 1874) discovered that despite people’s height, weight and strength presents a big range of datasets with people’s height ranging from 3 to 8 feet and with weight’s ranging from few pounds too few hundred pounds, there was a strong link between people’s height, weight and strength following a standard normal distribution curve. For example, the normal curve was used to analyze errors in astronomical observation measurements.

Whereas in probability theory a special case of the central limit theorem known as the DE Moivre-Laplace theorem states that the normal distribution may be used as an approximation to the binomial distribution under certain conditions. This theorem appears in the second edition pf the book published in 1738 by Abraham de Moivre titled ‘Doctrine of Chances’. (Source: ramsey1.chem.uic.edu)

The label for rows contains the integer part and the first decimal place of Z. Because the normal distribution curve is symmetrical, probabilities for only positive values of Z are typically given.

The user has to use a complementary operation on the absolute value of Z, as in the example below. Since this is the portion of the area above Z, the proportion that is greater than Z is found by subtracting Z from 1.

The values correspond to the shaded area for given Z This table gives a probability that a statistic is between 0 (the mean) and Z. F(z)=(z)12{\display style f(z)=\Phi (z)-{\franc {1}{2}}} Note that for z = 1, 2, 3, one obtains (after multiplying by 2 to account for the interval) the results f(z) = 0.6827, 0.9545, 0.9974, characteristic of the 68–95–99.7 rule.

The values are calculated using the cumulative distribution function of a standard normal distribution with mean of zero and standard deviation of one, usually denoted with the capital Greek letter {\display style \Phi} (phi), is the integral F(z)=1(z){\display style f(z)=1-\Phi (z)} This table gives a probability that a statistic is greater than Z, for large integer Z values.

It is a way to compare the results from a test to a “normal” population. If X is a random variable from a normal distribution with mean () and standard deviation (), its Z -score may be calculated by subtracting mean from X and dividing the whole by standard deviation. (Source: www.dataanalysisclassroom.com)

For the average of a sample from a population ‘n’, the mean is and the standard deviation is . A z -score equal to 1 represents an element, which is 1 standard deviation greater than the mean; a z -score equal to 2 signifies 2 standard deviations greater than the mean; etc.

What is the probable percentage of students scored more than 85? The z score table helps to know the percentage of values below (to the left) a z -score in a standard normal distribution.

A z score is simply defined as the number of standard deviation from the mean. You can use the Z -score table to find a full set of “less-than” probabilities for a wide range of z -values using the z -score formula.

To get this area of `0.4265`, we read down the left side of the table for the standard deviation's first 2 digits (the whole number and the first number after the decimal point, in this case `1.4`), then we read across the table for the “`0.05`” part (the top row represents the 2nd decimal place of the standard deviation that we are interested in.) (left column) `1.4\ +` (top row) `0.05 = 1.45` standard deviations The area represented by `1.45` standard deviations to the right of the mean is shaded in green in the standard normal curve above.

Intersect that row and column to find the probability: 0.9834. Observe that this happens to equal p (Z >+2.13). The reason for this is ‘ because the normal distribution is symmetric. (Source: statstutorstl.blogspot.com)

About the Book Author Deborah J. Ramsey, PhD, is Professor of Statistics and Statistics Education Specialist at The Ohio State University.

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