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Add the values of the variable for all elements in the sample and divide by the number of elements on the sample.

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Population has a mean of and micro80 and standard deviation of and sigma10. If a sample is taken from the population with a sample size of and 57345 25 find the percentage that the sample mean is samp?

To find the percentage that the sample mean is within a certain range, we can use the Central Limit Theorem. Given a population mean (μ) of 80 and a standard deviation (σ) of 10, for a sample size (n) of 25, the standard error (SE) is calculated as σ/√n = 10/√25 = 2. To find the percentage of sample means that fall within a specific range, you would use the z-score formula and standard normal distribution tables. However, without specifying the range for the sample mean, we cannot provide a specific percentage.


Why wont a sample statistics change from sample to sample?

The sample mean may differ from the population mean, especially for small samples.


What is the probability that a sample of 120 female graduates will provide a sample mean more than .30 below the population mean?

To determine the probability that a sample mean from 120 female graduates is more than 0.30 below the population mean, you would need information about the population standard deviation or the standard error of the sample mean. Assuming a normal distribution, you can use the Central Limit Theorem to find the standard error by dividing the population standard deviation by the square root of the sample size (120). Then, you can calculate the z-score corresponding to a sample mean that is 0.30 below the population mean and use a standard normal distribution table or calculator to find the probability associated with that z-score.


A population has mean 128 and standard deviation 22. find the mean and the standard deviation of mean for sample of size 36?

The mean of the sample means remains the same as the population mean, which is 128. The standard deviation of the sample means, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size. Therefore, the standard error is ( \frac{22}{\sqrt{36}} = \frac{22}{6} \approx 3.67 ). Thus, the mean is 128 and the standard deviation of the sample means is approximately 3.67.


What is the estimated population mean with a sample of 36 and a standard deviation of 0.03?

What is the sample mean?

Related Questions

How do you find the sample size when given the standard deviation and the mean with a sample value?

You cannot from the information provided.


What happens to the sample mean as the sample size increases?

With a good sample, the sample mean gets closer to the population mean.


Population has a mean of and micro80 and standard deviation of and sigma10. If a sample is taken from the population with a sample size of and 57345 25 find the percentage that the sample mean is samp?

To find the percentage that the sample mean is within a certain range, we can use the Central Limit Theorem. Given a population mean (μ) of 80 and a standard deviation (σ) of 10, for a sample size (n) of 25, the standard error (SE) is calculated as σ/√n = 10/√25 = 2. To find the percentage of sample means that fall within a specific range, you would use the z-score formula and standard normal distribution tables. However, without specifying the range for the sample mean, we cannot provide a specific percentage.


What is the difference between the population mean and sample mean?

The population mean is the mean calculated over every member of the set of subjects being studied. It is usually not available and a survey is used to find an estimate for the population mean. The mean value of the variable in question, calculated from only the subjects included in the sample (or survey) is the sample mean. Provided some basic statistical requirements are met, the sample mean is a "good" estimate of the population mean.


Why wont a sample statistics change from sample to sample?

The sample mean may differ from the population mean, especially for small samples.


How do you determine your sample score on the comparison distribution?

To determine your sample score on the comparison distribution, you first need to calculate the sample mean and standard deviation. Then, you can use these statistics to find the z-score, which indicates how many standard deviations your sample mean is from the population mean. By comparing this z-score to critical values from the standard normal distribution, you can assess the significance of your sample score in relation to the comparison distribution.


When using the distribution of sample mean to estimate the population mean what is the benefit of using larger sample sizes?

The variance decreases with a larger sample so that the sample mean is likely to be closer to the population mean.


A Numerical Measure from a Sample Such as a Sample Mean Is Known?

sample statistic


What is the difference between calculating the sample mean and the population mean?

You calculate the actual sample mean, and from that number, you then estimate the probable mean (or the range) of the population from which that sample was drawn.


A population has mean 128 and standard deviation 22. find the mean and the standard deviation of mean for sample of size 36?

The mean of the sample means remains the same as the population mean, which is 128. The standard deviation of the sample means, also known as the standard error, is calculated by dividing the population standard deviation by the square root of the sample size. Therefore, the standard error is ( \frac{22}{\sqrt{36}} = \frac{22}{6} \approx 3.67 ). Thus, the mean is 128 and the standard deviation of the sample means is approximately 3.67.


What is the estimated population mean with a sample of 36 and a standard deviation of 0.03?

What is the sample mean?


Why is the sample mean an unbiased estimator of the population mean?

The sample mean is an unbiased estimator of the population mean because the average of all the possible sample means of size n is equal to the population mean.