Páginas: 4 (906 palabras) Publicado: 11 de octubre de 2010
I. Normal Distribution
1. Normal distribution refers to continuous variables. Most of the values in a normal distribution fall close to the mean. 68.26% of all values lie within one standarddeviation of the mean, 95.44% of all values lie within two standard deviations from the mean, and 99.72% of all values lie within three standard deviations from the mean. The normal distribution iscompletely determined by its mean and standard deviation. The normal curve is symmetric about the mean. Half (0.5) of the probability lies below the mean and half (0.5) above it.
For an example, see:http://www.iejs.com/Research_Methods/normal_distribution.htm
2. Z-scores: It is the number of standard deviations a data value or an observation lies from the mean of the distribution. We can find theprobabilities of occurrence of an observation or a data value by converting the raw data value into a z-score. The formula for z-score is Z = X−μσ.
Example: (From the text book by M,B, &B)
Vince Yosarianworks for the Bureau of forms. His manager has reprimanded Vince because she feels that she has not performed up to standard for processing forms at the bureau, the crucial part of the job. At thebureau, the number of forms processed by employees is normally distributed, with a mean of 67 forms per employee per day, with a standard deviation of 7. His manager has calculated that Vince’s averagerate is 50 forms processed per day. What percentage of employees at the Bureau process fewer forms than Vince? What percentage of employees at the bureau process more forms than Vince? Does themanager’s complaint seem justified or not?
Answer: Here, μ = 67, σ = 7. Forms processed by Vince = X = 50
Therefore, Z-score = (50-67)/7 = -2.43. Associated probability for a z-score of 2.43 = 0.4925. Hence,the percentage of employees who process less than 50 forms per day = (0.5 – 0.4925) = 0.0075 (0.75%). The percentage of employees who process more forms than Vince = 1-0.0075 = 0.9925 (99.25%)....

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