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Section 2.5 The Normal Distribution
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68% of values lie within 1 SD of the mean. Including to the right and left 90% of the values lie with 1.645 SDs of the mean. 95% lie within about 2 SDs (actually 1.96 SDs) of the mean. 99.7% of the data lie within 3 SDs of the mean.
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Think of a case variable that when measured will give results that are normally distributed. Determine how you would measure the variable: Measure a time Measure length or distance Survey people Actually run an experiment or get survey results and collect data sufficient enough to determine if the data is roughly normally distributed. Plot the data with a histogram or dot plot and create a smooth curve over it to show its basic shape. Find a measure a center and spread for the data. If it is roughly normal, check to see if approximately 68% of the data is within 1 SD of the mean, 95% within 2 SDs and 99.7% within 3 SDs. If it is not normal, find the quartiles and create a box plot of the data.
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Very common distribution of data throughout many disciplines. SAT / ACT scores Measure of diameter of tennis balls Heights / weights of people Once we know a distribution is Normal, there is a tremendous amount of information we can determine or predict about it.
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All normal distributions have the same basic shape. The difference: tall and thin vs. short and fat However, we could easily stretch the scale of the tall thin curve to make it identical to the short fat one. The area under the curve can be thought of in terms of proportions or percentage of data. The total area under the curve is 1.0 (100%)
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We can standardize any normal curve to be identical. We do this by treating the mean as Zero and the SD as One. The variable along the x-axis becomes what we call a z score. The z score is the number of SDs away from the mean.
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Practice problems: 1) Normal distribution with: mean = 45 and SD = 5 Find the z score for a data value of 19 Find the z score for a data value of 52 2) Normal distribution with: Mean = 212 and SD = 24 Find the z score for a data value of 236
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We can use the standard normal curve to find proportion of data in a range of values. Normal Curve example: SAT I Math scores Mean = 500 SD = 40 Find the proportion of data in the score range 575 or less. Using z tables: Table A very back of book Find the proportion of data above 575. Find the proportion of data between 490 and 550.
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Read all of 2.5. Be prepared for quiz on Tuesday.
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You have collected data regarding the weights of boys in a local middle school. The distribution is roughly normal. The mean is 113 lbs and the SD is 10 lbs. A) What proportion of boys are below 100 lbs? B) What proportion are above 120 lbs? C) What proportion are in between 90 & 120 lbs?
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You can also use the TI-83 or higher to find these same proportions: 2 nd, Distr, normalcdf(low, high,mean,SD) When using z scores you can leave mean,SD blank. normalcdf(low, high) it will default to mean=0 and SD=1. This will give you the same area under the curve (proportion of data) as the z table.
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If you know the percent of data covered under a normal distribution, you can find the z-score. Simply look up the percent (proportion) in the z table and relate it to the corresponding z score. Find the value that is closest to the percent given Another method is with the calculator. 2 nd,Distr, invNorm(proportion, mean, SD)
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Find the z-score that has the given percent of values below it in a standard normal distribution: a) 32% b) 41% (use the z-table) c) 87% d) 94%(use your calculator)
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If you know how many SDs a value is from the mean, you can use this (z-score) to find the actual data value: x = mean + (z SD) Example: The mean weight of the boys at a middle school is 113 lbs, with a SD of 10 lbs. One boy is determined to be 2.2 SDs above the mean. How much did the boy weigh?
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So now, if you know the percentage of data above or below a data value and you know the mean and SD, you can figure out that data value: Use z-table to find the z-score, then use the z score with mean and SD to find the data value. Or you can use the invNorm function on your calc. ▪ invNorm(proportion, mean, SD)
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The heights of U.S. 18-24 yr old females is roughly normally distributed with a mean of 64.8 in. and a SD of 2.5 in. Estimate the percent of women above 5’8” What height would a US female be if she was 1.5 SDs below the mean? Give your answer in ft & in. What height would a US female be if she was considered to be in the 80 th percentile?
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What percentage of US females is above 5’7”? What percent are between 5’7” and 5’0”?
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The cars in Clunkerville have a mean age of 12 years and a SD of 8 years. What percentage of cars are more than 4 years old? Why is this a trick question?
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