Mathematics
Empirical Rule
The Empirical Rule, also known as the 68-95-99.7 rule, is a statistical principle that applies to normal distributions. It states that for a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, about 95% within two standard deviations, and almost all (99.7%) within three standard deviations. This rule provides a quick way to estimate the spread of data in a normal distribution.
Written by Perlego with AI-assistance
6 Key excerpts on "Empirical Rule"
Learn about this page
Index pages curate the most relevant extracts from our library of academic textbooks. They’ve been created using an in-house natural language model (NLM), each adding context and meaning to key research topics.
eBook - ePub- Jim Fowler, Lou Cohen, Philip Jarvis(Authors)
- 2013(Publication Date)
- Wiley(Publisher)
Fig. 9.3 .If the vertical axis of the distribution is re-scaled by dividing by the number of observations, it effectively becomes a probability distribution or, strictly, a probability density. The total probability encompassed by the density is 1. If we say that the total area under the curve is 100% then one of the mathematical properties of the normal curve is that the area bounded by one standard deviation on each side of the central axis (that is, µ ± σ ) is approximately 68.26% of the total area. This means that about 68% of observations drawn randomly from a normally distributed population will fall within ±1 standard deviation from the mean. The other 32% fall outside these limits, 16% above 1 standard deviation and 16% below 1 standard deviation. In other words, there is a probability of P = 0.68 that a single observation drawn at random will fall within µ ± σ . By extending the limits to two standard deviations (µ ± 2σ ) the proportion of observations that will be included within them is increased to 95.44%. Similarly, 99.74% of observations fall within µ ± 3σ . These are equivalent to probabilities of 0.9544 and 0.9974. In practice the values of 0.95 and 0.99 are more convenient to deal with. It can be calculated that these probability values fall at µ ± 1.96σ and µ ± 2.58σ , respectively. It follows that the probability of a random observation falling outside these limits is 0.05 and 0.01. It will be recalled from Section 7.8 that these are the critical probability values for assessing whether the outcome of a stated event is unlikely or very unlikely.
eBook - ePub- Debbie L. Hahs-Vaughn, Richard Lomax(Authors)
- 2020(Publication Date)
- Routledge(Publisher)
confidence intervals, such as the 99% CI, enable greater confidence. For example, with a sample mean of 70 and a standard error of the mean of 3, the following confidence intervals result: 68% CI = (67, 73) [i.e., ranging from 67 to 73]; 90% CI = (65.065, 74.935); 95% CI = (64.12, 75.88); and 99% CI = (62.2726, 77.7274). We can see here that to be assured that 99% of the confidence intervals contain the population mean, then our interval must be wider (i.e., ranging from about 62.27 to 77.73, or a range of about 15) than the confidence intervals that are lesser (e.g., the 95% confidence interval ranges from 64.12 to 75.88, or a range of about 11).In general, a confidence interval for any level of confidence (i.e., #% CI) can be computed by the following general formula:# % C I =X ¯±(z)C V(σ)X ¯where zcv is the critical value taken from the standard unit normal distribution table for that particular level of confidence, and the other values are as before.5.2.2.5 Central Limit Theorem
In our discussion of confidence intervals, we used the normal distribution to help determine the width of the intervals. Many inferential statistics assume the population distribution is normal in shape. Because we are looking at sampling distributions in this chapter, does the shape of the original population distribution have any relationship to the sampling distribution of the mean we obtain? For example, if the population distribution is nonnormal, what form does the sampling distribution of the mean take (i.e., is the sampling distribution of the mean also nonnormal)? There is a nice concept, known as the central limit theorem, to assist us here. The central limit theorem states that as sample size n increases, the sampling distribution of the mean from a random sample of size n more closely approximates a normal distribution. If the population distribution is normal in shape, then the sampling distribution of the mean is also normal in shape. If the population distribution is not normal in shape, then the sampling distribution of the mean becomes more nearly normal as sample size increases. This concept is graphically depicted in Figure 5.2
eBook - ePubInterpreting Statistics for Beginners
A Guide for Behavioural and Social Scientists
- Vladimir Hedrih, Andjelka Hedrih(Authors)
- 2022(Publication Date)
- Routledge(Publisher)
As explained in the previous chapter, differences from the mean expressed in standard deviations are z scores and z scores can always be converted to percentiles. This means that we can define an interval on the normal distribution that contains a desired proportion of entities and the borders of which are equally distant from the mean, by simply defining the z scores (i.e. number of standard deviations below or above the mean) between which certain proportion of entities is located on the normal distribution. In the bootstrapping approach, the bootstrap procedure allows us to create the sampling distribution and then we simply define the interval on the sampling distribution that we want to create confidence interval from. The most commonly used proportions of the sampling distribution to be used for the creation of the confidence intervals are 95% and 99%. Although it is equally possible to use any other percentage, there is a strong custom among researchers to apply one of these two percentages. Under the central limit theorem, to create the 95% interval we should multiply the standard error with 1.96 and for the 99% interval the number is 2.56, because on the normal distribution 95% of entities are located between z scores −1.96 and +1.96 and 99% of scores are between z scores −2.56 and +2.56. An interval of 1 z score equals one standard deviation, or one standard error for a sampling distribution, so we could also say that 95% of entities on a normal sampling distribution are located in the interval that starts 1.96 standard errors below the mean and ends at the point that is 1.96 standard errors above the sample mean
- Craig Adam(Author)
- 2011(Publication Date)
- Wiley(Publisher)
Deriving the area from an expression as complex as (Equation 9.1) involves advanced mathematics, so a much simpler method has been devised, which makes such calculations routinely straightforward. This method is based on the tabulation of area values evaluated for the standard normal distribution and using the transformation formula (Equation 9.3) to convert particular parameters into the standard normal equivalent z -values. Before examining this method in detail, there are some useful benchmark probabilities, which are of great value in practice. If we consider all outcomes with values of x within one standard deviation of the mean, the probability of an outcome in this range may be represented by the expression: Evaluation of this area gives a value of 0.683, meaning that 68.3% of outcomes will lie within one standard deviation of the mean. The equivalent calculations for two and three standard deviations result in probabilities of 0.954 and 0.997, respectively. This latter result, indicating that 99.7% of outcomes will be within three standard deviations of the mean, supports the previous benchmark that virtually all the distribution lies within this range. Such ranges are called the statistical confidence limits of the distribution: e.g., the 95% confidence limit corresponds approximately to μ ± 2σ. This means that the expectation is that 95 out of every 100 measurements made will produce an outcome that is within two standard deviations of the mean value. Since areas are additive, we can use these results to derive others; e.g., the probability that an outcome lies between one and two standard deviations of the mean is given by: These ideas form the basis of probability calculations using the cumulative z-probability table given in Appendix IV. This tabulates probability areas for the standard normal distribution evaluated at prescribed values of z
eBook - ePubStatistics
The Essentials for Research
- Henry E. Klugh(Author)
- 2013(Publication Date)
- Psychology Press(Publisher)
The table is designed to give the percentage of the curve between μ(z =.00) and any other value of z directly. Percentages above or below any value of z, or percentages between values of z, can be obtained by subtraction. If these procedures are not clear, review them and review the appropriate practice problems at the end of Chapter 4. The percentage of cases falling between various z scores in a normal distribution can also be given a probability interpretation. For example, if we have 1000 normally distributed measurements we shall find 340 of them, or 34%, between the mean (z =.00) and z = 1.00; therefore, the probability of randomly selecting a measurement with a value of z between.00 and 1.00 is.34. Similarly, the probability of randomly selecting a value of z below −2.58 or above +2.58 is.01; only 1% of a normal distribution lies beyond these values of z. 9.2 Approximation for Frequencies In this section we show just how accurately a theoretical binomial sampling distribution can be approximated by the normal curve. First we shall calculate, from the binomial expansion, the probability of throwing 8 or more tails in 12 tosses of a true coin. We shall let p = the probability of heads and q = the probability of tails. For the solution to the problem we must find the sum of the probabilities represented by the 9th through the 13th terms of the expansion (p + q) 12. This gives the probabilities of obtaining samples with 8 tails, … through 12 tails. We find these five terms are 495 p 4 q 8 + 220 p 3 q 9 + 66 p 2 q 10 + 12 pq 11 + q 12. The sum of the probabilities for the series is.194. Now we shall solve the same problem by using Table N and the normal curve approximation. The procedure is fairly straightforward: the point in the binomial sampling distribution representing 8 tails is converted into a z score. Then we find the proportion of area to the right of such z scores in a normal distribution
eBook - ePubProbability for Kids
Using Model-Eliciting Activities to Investigate Probability Concepts (Grades 4-6)
- Scott Chamberlin(Author)
- 2021(Publication Date)
- Routledge(Publisher)
This is sophisticated statistics and probability jargon to explain that each number, 1–12, should theoretically have exactly the same number of rolls. In this case, each number should have precisely 83.33 rolls because 1,000 divided by 12 is 83.33. (Note: When 83.33 is used in this chapter, it is assumed that the decimal is a repeating one.) Certainly, one cannot have one-third of a roll, so the 83.33 is an exact number that needs to be interpreted by problem solvers in relation to the actual number of rolls per number. What actually happened is referred to as the empirical probability. In this problem, the objective is for problem solvers to compare the empirical (what actually happened) probability to the theoretical (what is supposed to happen) probability. The objective, specified in the problem statement, is to see if a big (a term used in lieu of statistical significance) difference exists. Solution 1 Solution 1 entails calculating the theoretical probability (or expected values) so that the actual (empirical) data can be compared to it. With entry-level mathematical models, it might be common for a group to not have a comprehensively fleshed out comparison method. Highly advanced statisticians might use a process such as a t -test or an ANOVA (Analysis of Variance). It is not logical to think that fourth- to sixth-grade students could conduct, let alone interpret, such results. In most cases, they do not have access to software with such capabilities. Moreover, the notion behind creating early mathematical models is to get modelers to make sense of mathematical information (in this case, the comparison of two data sets). It is also important to remember that figuring out what the second data set, the theoretical probability, is should be considered an accomplishment
