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Q: How do you calculate standard deviation with the help of z-score?

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The standard deviation of a set of data is a measure of the random variability present in the data. Given any two sets of data it is extremely unlikely that their means will be exactly the same. The standard deviation is used to determine whether the difference between the means of the two data sets is something that could happen purely by chance (ie is reasonable) or not.Also, if you wish to take samples of a population, then the inherent variability - as measured by the standard deviation - is a useful measure to help determine the optimum sample size.

The purpose of obtaining the standard deviation is to measure the dispersion data has from the mean. Data sets can be widely dispersed, or narrowly dispersed. The standard deviation measures the degree of dispersion. Each standard deviation has a percentage probability that a single datum will fall within that distance from the mean. One standard deviation of a normal distribution contains 66.67% of all data in a particular data set. Therefore, any single datum in the data has a 66.67% chance of falling within one standard deviation from the mean. 95% of all data in the data set will fall within two standard deviations of the mean. So, how does this help us in the real world? Well, I will use the world of finance/investments to illustrate real world application. In finance, we use the standard deviation and variance to measure risk of a particular investment. Assume the mean is 15%. That would indicate that we expect to earn a 15% return on an investment. However, we never earn what we expect, so we use the standard deviation to measure the likelihood the expected return will fall away from that expected return (or mean). If the standard deviation is 2%, we have a 66.67% chance the return will actually be between 13% and 17%. We expect a 95% chance that the return on the investment will yield an 11% to 19% return. The larger the standard deviation, the greater the risk involved with a particular investment. That is a real world example of how we use the standard deviation to measure risk, and expected return on an investment.

Karl Pearson simplified the topic of skewness and gave us some formulas to help. The first is the Pearson mode or first skewness coefficient. It is defined by the (mean-median)/standard deviation. So in this case the Pearson mode is: (8-6)/2 =1 There is also the Pearson Median. This is also called second skewness coefficient. It is defined as 3(mean-median)/standard deviation which in this case is 6/2 =3 hence the distribution is positive skewed

That's on page 126 in your statistics textbook...... DO YOUR OWN HOMEWORK!!! K i obv needed help with how to do it, ass. I didn't want just the answer.

5000 students participated in a certain test yielding a result that follows the normal distribution with means of 65 points and standard deviation of 10 points.(1) Find the probability of a certain student marking more than 75 points and less than 85 points inclusive.(2) A student needs more than what point to be positioned within top 5% of the participants in this test?(3) A student with more than what point can be positioned within top 100 students?i dont understand the question.. could you help me??pls....

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No. It's the number that will help you work out the percentage.

I donβt know if this is correct or not but, I think itβs standard deviation 1.581138830084 I hope this helps but can you help me with each product of the same facor 10/4 =. I hope you can help with my problem an again I hope this helpsππ

Standard deviation helps you identify the relative level of variation from the mean or equation approximating the relationship in the data set. In a normal distribution 1 standard deviation left or right of the mean = 68.2% of the data 2 standard deviations left or right of the mean = 95.4% of the data 3 standard deviations left or right of the mean = 99.6% of the data

Can someone help me find the answer for a sample n=36 with a population mean of of 76 and a mean of 79.4 with a standard deviation of 18?

The standard deviation of a set of data is a measure of the random variability present in the data. Given any two sets of data it is extremely unlikely that their means will be exactly the same. The standard deviation is used to determine whether the difference between the means of the two data sets is something that could happen purely by chance (ie is reasonable) or not.Also, if you wish to take samples of a population, then the inherent variability - as measured by the standard deviation - is a useful measure to help determine the optimum sample size.

The purpose of obtaining the standard deviation is to measure the dispersion data has from the mean. Data sets can be widely dispersed, or narrowly dispersed. The standard deviation measures the degree of dispersion. Each standard deviation has a percentage probability that a single datum will fall within that distance from the mean. One standard deviation of a normal distribution contains 66.67% of all data in a particular data set. Therefore, any single datum in the data has a 66.67% chance of falling within one standard deviation from the mean. 95% of all data in the data set will fall within two standard deviations of the mean. So, how does this help us in the real world? Well, I will use the world of finance/investments to illustrate real world application. In finance, we use the standard deviation and variance to measure risk of a particular investment. Assume the mean is 15%. That would indicate that we expect to earn a 15% return on an investment. However, we never earn what we expect, so we use the standard deviation to measure the likelihood the expected return will fall away from that expected return (or mean). If the standard deviation is 2%, we have a 66.67% chance the return will actually be between 13% and 17%. We expect a 95% chance that the return on the investment will yield an 11% to 19% return. The larger the standard deviation, the greater the risk involved with a particular investment. That is a real world example of how we use the standard deviation to measure risk, and expected return on an investment.

Karl Pearson simplified the topic of skewness and gave us some formulas to help. The first is the Pearson mode or first skewness coefficient. It is defined by the (mean-median)/standard deviation. So in this case the Pearson mode is: (8-6)/2 =1 There is also the Pearson Median. This is also called second skewness coefficient. It is defined as 3(mean-median)/standard deviation which in this case is 6/2 =3 hence the distribution is positive skewed

okay wikianswers messed up the question. H0: M=23; H1: M not 23; n=50; x-bar = 21.25; standard deviation = 5. please help!

This helps to show where things may not follow the norm. Quartiles help you to keep data organized and so a deviation would show how it would vary.

The study material of CMRP Dumps is very trustworthy and accurate. By using this study material, I have passed my CMRP Certification Exam easily. I suggest all of you to take help from qualified and experienced experts of Dumpspass4sure. Question: 1 What relationship should Maintenance and Reliability Teams have with customers and suppliers for optimum effectiveness? Please choose the correct answer. A. Purchasing should be the only communicators with suppliers B. Management should be the only communicators with customers and suppliers C. Team members should be involved in communicating with customers and suppliers D. Sales should be the only department communicating with customers. Answer: C Question: 2 What is the statistical average and the population standard deviation of the following series of numbers? 1.30, 1.25, 1.50, 1.12, 1.35, 1.38 Please choose the correct answer. A. Statistical average = 1.32; standard deviation= 0.128 B. Statistical average = 1.50; standard deviation= 0.25 C. Statistical average = 1.32; standard deviation= 1.50 D. Statistical average = 0.128: standard deviation= 1.32 Answer: A

z- statistics is applied under two conditions: 1. when the population standard deviation is known. 2. when the sample size is large. In the absence of the parameter sigma when we use its estimate s, the distribution of z remains no longer normal but changes to t distribution. this modification depends on the degrees of freedom available for the estimation of sigma or standard deviation. hope this will help u.... mona upreti.. :)

There is a website called bmicalculator.org that can help you. It is an online tool. It is also completely free. You can also use it to calculate the BMI of your child. You can set the setting to standard for adults or teens and to child for your child.

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