Why is the normal distribution important?
Why is the normal distribution consider as the most important probability distribution?Suppose you have a large sample from a non-normal distribution. Now suppose you estimate a parameter of the distribution such as the mean or the median. This statistic itself has a distribution. That is, if we were to keep repeating the sampling and estimation process we would get a set of different estimates and these have a distribution. (We don’t repeat the sampling process, we can deduce the distribution from theory.)If the sample is large enough this sampling distribution will often be approximately normal.The conditions for this to occur are that the estimate is some sort of sum, or approximately a sum, or a measure near the centre of the original distribution.This could require a sample much larger than we are willing to use, so we usually require that the original data have a distribution that doesn’t produce many outliers (very different values from the rest) and is not too skewed.
Why does the sampling distribution of the mean follow a normal distribution for a large enough sample size even though it may not be normally distributed?
Essentially yes. The Central Limit Theorem states that for a large sample size, the distribution of the sample mean approaches a normal distribution for any underlying population distribution with a finite standard deviation. In practice, if the sample size is at least 30, then the sampling distribution of the sample means is well approximated by a normal distribution.
When can we assume that a sampling distribution is normally distributed (differs by the type of a sample statistic)?
Short answers:In theory, only when you know the underlying population is normal itself and you are looking at averages; or when you have a very large sample average from any population (that’s the Central Limit Theorem).In practice, if your population isn’t obviously multi-modal and you have a reasonably large sample you are trying to estimate the average of; 50 or more is pretty safe.
How is "The sampling distribution of x has mean equal to the population mean mu even if the population is not normally distributed" a true statement?
According to the CLT, as you increase sample size, the distribution of its average approaches a normal distribution. The mean must, obviously, be the same as in the original distribution. However, as the sample size increases, the variance of its average decreases (roughly proportionally to [math]1/\sqrt{n}[/math]). With large enough sample size, this makes the bell curve narrow enough. So even if the original random variable is nonnegative, things will just work out: the entire statistically significant peak of the bell curve will be on the positive side, it will not get "cut off at zero".
What is the sampling distribution of the mean percent x_bar that the sample spends on housing if null is true?
For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - μ ) / σ Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed with mean μ and standard deviation σ /√(n) An applet for finding the values http://www-stat.stanford.edu/~naras/jsm/... calculator http://stattrek.com/Tables/normal.aspx how to read the tables http://rlbroderson.tripod.com/statistics... In this question we have Xbar ~ Normal( μ = 0.31 , σ² = 0.009216 / 40 ) Xbar ~ Normal( μ = 0.31 , σ² = 0.0002304 ) Xbar ~ Normal( μ = 0.31 , σ = 0.096 / sqrt( 40 ) ) Xbar ~ Normal( μ = 0.31 , σ = 0.01517893 )
If selecting samples of size n ≤30 from a population with known mean and standard deviation, what requirement?
If selecting samples of size n ≤30 from a population with known mean and standard deviation, what requirement, if any must be satisfied in order to assume that the distribution of the sample means is a normal distribution.
Time spent using the phone (per session) is normally distributed with u = 8 minutes and o = 2 minutes.?
For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - μ ) / σ Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed with mean μ and standard deviation σ /√(n) An applet for finding the values http://www-stat.stanford.edu/~naras/jsm/... calculator http://stattrek.com/Tables/normal.aspx how to read the tables http://rlbroderson.tripod.com/statistics... In this question we have Xbar ~ Normal( μ = 8 , σ² = 4 / 25 ) Xbar ~ Normal( μ = 8 , σ² = 0.16 ) Xbar ~ Normal( μ = 8 , σ = 2 / sqrt( 25 ) ) Xbar ~ Normal( μ = 8 , σ = 0.4 ) Find P( 7.8 < Xbar < 8.2 ) = P( ( 7.8 - 8 ) / 0.4 < ( Xbar - μ ) / σ < ( 8.2 - 8 ) / 0.4 ) = P( -0.5 < Z < 0.5 ) = P( Z < 0.5 ) - P( Z < -0.5 ) = 0.6914625 - 0.3085375 = 0.3829249 Find P( 7.5 < Xbar < 8.2 ) = P( ( 7.5 - 8 ) / 0.4 < ( Xbar - μ ) / σ < ( 8.2 - 8 ) / 0.4 ) = P( -1.25 < Z < 0.5 ) = P( Z < 0.5 ) - P( Z < -1.25 ) = 0.6914625 - 0.1056498 = 0.5858127 For a sample size of 100 we have Xbar ~ Normal( μ = 8 , σ² = 4 / 100 ) Xbar ~ Normal( μ = 8 , σ² = 0.04 ) Xbar ~ Normal( μ = 8 , σ = 2 / sqrt( 100 ) ) Xbar ~ Normal( μ = 8 , σ = 0.2 ) Find P( 7.8 < Xbar < 8.2 ) = P( ( 7.8 - 8 ) / 0.2 < ( Xbar - μ ) / σ < ( 8.2 - 8 ) / 0.2 ) = P( -1 < Z < 1 ) = P( Z < 1 ) - P( Z < -1 ) = 0.8413447 - 0.1586553 = 0.6826895
A population has a mean of 84, standard deviation of 12. Sample of 36 observation will be taken. PROBABLITY?
We are given that the population mean-U and population standard deviation-δ of the distribution which are normally distributed are 84 and 2.8 respectively. We are also given that n=36 and we are asked to find the probability the sample mean-x-bar will be between 80.54 and 88.9. So, it follows that P(80.54
What is the difference between a sampling distribution and the distribution of a sample?
When we refer to the distribution of a sample (assumed numerical data in this answer) we're simply discussing descriptive statistics: is there evidence the data are skewed? Are there outliers? What are some basic descriptive statistics (sample mean, standard deviation, five-number summary, etc.)A sampling distribution is a set of probability rules that tells us how some sample statistic behaves from one sample to the next. Suppose this is my "population": 10, 20, 30If I list all possible samples of size two from this population (randomly selected, duplicates allowed) and calculate the average in each case:10, 10 1010,20 1510,30 2020,10 1520,20 2020,30 2530,10 2030,20 2530,30 30I could, from that set, create the probability distribution for the sample mean based on samples of size 2 from that population: that distribution is the sampling distribution.More generally: the sampling distribution for a statistic is often too difficult to work out by "brute force" and some theory is required. The Central Limit Theorem is an example: it tells us that in certain circumstances that the sampling distribution of the sample mean is approximately normal, and provides methods for determining the appropriate mean and standard deviation for that normal distribution.
Given a sample mean, why do we need to calculate the probability of a population mean being within a certain confidence interval? Doesn't it follow that if we know the population's standard deviation then we already know the mean?
In practice, yes, but it is RARE that you will know either. You may have an upper bound for population s.d. and you can use that as a w.c.s. In practice you know neither, so you use the sample s.dev to estimate the sample and use that for your CIs.