Statistics question - normality?

Question

I am trying to find out some ways in determining if a set of data is normally distributed. I know that one way you can find out is by testing if the mean = median, also by creating an normal probability plot, making a histogram, and constructing a boxplot, but is the following guideline true for determining normality?

Determine if 2/3 of the data lie within 1 standard deviation from the mean, if approximately 4/5 lie within 1.28 standard deviations from the mean, and if approximately 19/20 lie between 2 standard deviations from the mean.

Thank you

Answer 1

The normal probability plot is your best tool. Its linearity, or lack of linearity, tells you instantly about the normality of your data. The histogram is also useful for determining the symmetry of your data. The criteria you mention would be useful, but only after you've established that you have a symmetric distribution.

Answer 2

purely placed, normality potential the information comes from a inhabitants that follows the conventional distribution. once you have a small pattern self belief era and use the pupil t statistic you'll have general records for the era to be valid. on your question here, with purely 20 records factors, if the information did no longer come from a classic inhabitants this is possible the pattern length isn't great sufficient for the recommend to act as a classic random variable. This assumption of normality may be left out to a pair quantity in great pattern self belief durations via fact the powerful shrink theorem tells use that the recommend of any pattern with a sufficiently great pattern length is approximately general. permit X1, X2, ... , Xn be an elementary random pattern from a inhabitants with recommend ? and variance ?². permit Xbar be the pattern recommend = a million/n * ?Xi permit Sn be the sum of pattern observations: Sn = ?Xi then, if n is satisfactorily great: Xbar has the conventional distribution with recommend ? and variance ?² / n Xbar ~ general(? , ?² / n) Sn has the conventional distribution with recommend n? and variance n?² Sn ~ general(n? , n?²) the large element is that it is not important what the under mendacity distribution is, the powerful shrink theorem holds. It became shown by employing Markov employing persevering with fractions. if the pattern comes from a uniform distribution the sufficient pattern length is as small as 12 if the pattern comes from an exponential distribution the sufficient pattern length must be quite a few hundred to quite a few thousand. if the information comes from a classic distribution at the start then any pattern length is sufficient. for n < 30, if the pattern is from a classic distribution we use the pupil t statistic to estimate the distribution. We do this via fact the pupil t takes under consideration the uncertainty interior the estimate for the wide-unfold deviation. if we now the inhabitants wide-unfold deviation then we are able to apply the z statistic from the initiating. the value of 30 became empirically defined via fact at around that pattern length, the quantiles of the pupil t are very close the quantiles of the wide-unfold general.

Answer 3

merely placed, normality capability the records comes from a inhabitants that follows the traditional distribution. once you have a small pattern self belief era and use the pupil t statistic you're able to desire to have familiar records for the era to be valid. on your question right here, with in basic terms 20 records factors, if the records did not come from a classic inhabitants that's attainable the pattern length isn't sufficiently super for the advise to act as a classic random variable. This assumption of normality could be surpassed over to 3 volume in super pattern self belief periods by using fact the needed shrink theorem tells use that the advise of any pattern with a sufficiently super pattern length is approximately familiar. permit X1, X2, ... , Xn be an trouble-free random pattern from a inhabitants with advise ? and variance ?². permit Xbar be the pattern advise = one million/n * ?Xi permit Sn be the sum of pattern observations: Sn = ?Xi then, if n is adequately super: Xbar has the traditional distribution with advise ? and variance ?² / n Xbar ~ familiar(? , ?² / n) Sn has the traditional distribution with advise n? and variance n?² Sn ~ familiar(n? , n?²) the super ingredient is that it does not count what the fewer than mendacity distribution is, the needed shrink theorem holds. It grew to become into shown by using Markov making use of continuing fractions. if the pattern comes from a uniform distribution the sufficient pattern length is as small as 12 if the pattern comes from an exponential distribution the sufficient pattern length ought to be various hundred to various thousand. if the records comes from a classic distribution to initiate with then any pattern length is sufficient. for n < 30, if the pattern is from a classic distribution we use the pupil t statistic to estimate the distribution. We try this by using fact the pupil t takes under consideration the uncertainty interior the estimate for the time-honored deviation. if we now the inhabitants time-honored deviation then we are able to apply the z statistic from the start. the value of 30 grew to become into empirically defined by using fact at around that pattern length, the quantiles of the pupil t are very close the quantiles of the time-honored familiar.