0
When the standard deviation of a population is known, the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution, due to the Central Limit Theorem. The mean of this sampling distribution will be equal to the population mean, while the standard deviation (known as the standard error) will be the population standard deviation divided by the square root of the sample size. This allows for the construction of confidence intervals and hypothesis testing using z-scores.
What else can I help you with?
When the population standard deviation is known the sampling distribution is known as what?
normal distribution
When the population standard deviation is known the sampling distribution is a?
normal distribution
When the population standard deviation is known the sample distribution is a?
When the population standard deviation is known, the sample distribution is a normal distribution if the sample size is sufficiently large, typically due to the Central Limit Theorem. If the sample size is small and the population from which the sample is drawn is normally distributed, the sample distribution will also be normal. In such cases, statistical inference can be performed using z-scores.
How do you standardise a value?
There are many different bases for standardisation - even if you only consider the Gaussian (Normal) distribution. If a variable X has a Gaussian distribution, then the corresponding Standard Normal deviate, Z, is obtained from X by subtracting the mean of X and then dividing the result by the standard deviation of X. The variable, Z, more commonly known as the Z-score, has a Gaussian distribution with mean 0 and standard deviation 1. But, if X is an IQ score, for example, different measures of the X variable are used so that the resulting variable has mean 100 and standard deviation 15.
How standard deviation and Mean deviation differ from each other?
There is 1) standard deviation, 2) mean deviation and 3) mean absolute deviation. The standard deviation is calculated most of the time. If our objective is to estimate the variance of the overall population from a representative random sample, then it has been shown theoretically that the standard deviation is the best estimate (most efficient). The mean deviation is calculated by first calculating the mean of the data and then calculating the deviation (value - mean) for each value. If we then sum these deviations, we calculate the mean deviation which will always be zero. So this statistic has little value. The individual deviations may however be of interest. See related link. To obtain the means absolute deviation (MAD), we sum the absolute value of the individual deviations. We will obtain a value that is similar to the standard deviation, a measure of dispersal of the data values. The MAD may be transformed to a standard deviation, if the distribution is known. The MAD has been shown to be less efficient in estimating the standard deviation, but a more robust estimator (not as influenced by erroneous data) as the standard deviation. See related link. Most of the time we use the standard deviation to provide the best estimate of the variance of the population.
When the population standard deviation is known the sampling distribution is known as what?
normal distribution
When the population standard deviation is known the sampling distribution is a?
normal distribution
When the population standard deviation is not known the sampling distribution is a?
If the samples are drawn frm a normal population, when the population standard deviation is unknown and estimated by the sample standard deviation, the sampling distribution of the sample means follow a t-distribution.
The standard deviation of the distribution of means is also known as the population standard deviation?
Yes.
When to use z or t-distribution?
If the sample size is large (>30) or the population standard deviation is known, we use the z-distribution.If the sample sie is small and the population standard deviation is unknown, we use the t-distribution
Is it possible that in standard normal distribution standard deviation known and mean unknown?
it is possible to distribute standard deviation and mean but you dont have to understand how the mouse runs up the clock hicorky dickory dock.
What are importance of mean and standard deviation in the use of normal distribution?
For data sets having a normal distribution, the following properties depend on the mean and the standard deviation. This is known as the Empirical rule. About 68% of all values fall within 1 standard deviation of the mean About 95% of all values fall within 2 standard deviation of the mean About 99.7% of all values fall within 3 standard deviation of the mean. So given any value and given the mean and standard deviation, one can say right away where that value is compared to 60, 95 and 99 percent of the other values. The mean of the any distribution is a measure of centrality, but in case of the normal distribution, it is equal to the mode and median of the distribtion. The standard deviation is a measure of data dispersion or variability. In the case of the normal distribution, the mean and the standard deviation are the two parameters of the distribution, therefore they completely define the distribution. See: http://en.wikipedia.org/wiki/Normal_distribution
When do you know when to use t-distribution opposed to the z-distribution?
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.. :)
How do you standardise a value?
There are many different bases for standardisation - even if you only consider the Gaussian (Normal) distribution. If a variable X has a Gaussian distribution, then the corresponding Standard Normal deviate, Z, is obtained from X by subtracting the mean of X and then dividing the result by the standard deviation of X. The variable, Z, more commonly known as the Z-score, has a Gaussian distribution with mean 0 and standard deviation 1. But, if X is an IQ score, for example, different measures of the X variable are used so that the resulting variable has mean 100 and standard deviation 15.
How standard deviation and Mean deviation differ from each other?
There is 1) standard deviation, 2) mean deviation and 3) mean absolute deviation. The standard deviation is calculated most of the time. If our objective is to estimate the variance of the overall population from a representative random sample, then it has been shown theoretically that the standard deviation is the best estimate (most efficient). The mean deviation is calculated by first calculating the mean of the data and then calculating the deviation (value - mean) for each value. If we then sum these deviations, we calculate the mean deviation which will always be zero. So this statistic has little value. The individual deviations may however be of interest. See related link. To obtain the means absolute deviation (MAD), we sum the absolute value of the individual deviations. We will obtain a value that is similar to the standard deviation, a measure of dispersal of the data values. The MAD may be transformed to a standard deviation, if the distribution is known. The MAD has been shown to be less efficient in estimating the standard deviation, but a more robust estimator (not as influenced by erroneous data) as the standard deviation. See related link. Most of the time we use the standard deviation to provide the best estimate of the variance of the population.
How do you calculate mean median and standard deviation?
The mean is the sum of each sample divided by the number of samples.The median is the middle sample in a ranked list of samples, or the mean of the middle two samples if the number of samples is even.The standard deviation is the square root of the sum of the squares of the difference between the mean and each of the samples, such sum then divided by either N or by N-1, before the square root is taken. N is used for population standard deviation, where the mean is known independently of the calculation of the standard deviation. N-1 is used for sample standard deviation, where the mean is calculated along with the standard deviation, and the "-1" compensates for the loss of a "degree of freedom" that such a procedure entails.Not asked, but answered for completeness sake, the mode is the most probable value, and does not necessarily represent the mean such as in an asymmetrically skewed distribution, such as a Poisson distribution.
How do you solve if a population has a mean of μ80 and a standard deviation of σ20?
The question gives summary statistics for a population. If the underlying distribution is Gaussian, or some other known distribution, then the probability density function can be calculated. Even so, there is no question and so nothing to "solve".
