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Normal Distributions. Normal Distributions. Symmetrical bell-shaped ( unimodal ) density curve Above the horizontal axis N( m , s ) The transition points occur at m + s (Points of inflection) Probability is calculated by finding the area under the curve
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Normal Distributions • Symmetrical bell-shaped (unimodal) density curve • Abovethe horizontal axis • N(m, s) • The transition points occur at m+s (Points of inflection) • Probability is calculated by finding the area under the curve • As sincreases, the curve flattens & spreads out • As sdecreases, the curve gets taller and thinner How is this done mathematically?
Normal distributions occur frequently. • Length of newborns • Height • Weight • ACT or SAT scores • Intelligence • Number of typing errors • Chemical processes
s s A B 6 Do these two normal curves have the same mean? If so, what is it? Which normal curve has a standard deviation of 3? Which normal curve has a standard deviation of 1? YES B A
AP StatisticsFriday, 10 January 2014 • OBJECTIVETSW investigate normal distributions. • You need to have the following out: • Blue chart (Table A) • Calculator (sign up for a new/old number, if needed) • QUIZ: Continuous & Uniform Distributions on Monday, 13 January 2014. • ASSIGNMENTS DUE • WS Normal Distributions Monday, 01/13/2014 • WS Continuous Distributions Review Tuesday, 01/14/2014 • Bookwork: 7.67, 7.69, 7.71 Monday, 01/13/2014 • Bookwork: 7.73, 7.77, 7.80 Monday, 01/13/2014
WS Uniform Distributions 1)a) μ = 2 min, σ = 1.15470 min b)0.375 2)a) 10 min b)0.05 3)a) 0.666 b) 0.333 c) 82.5 degrees 4)a) continuous b) μ = 7 oz., σ = 0.28868 oz. c) 0.75 d) 0.25 e)0.57735 5) 0.00024414
Empirical Rule • Approximately 68% of the observations fall within s of m • Approximately 95% of the observations fall within 2s of m • Approximately 99.7% of the observations fall within 3s of m
68% 71 Suppose that the height of male students at JVHS is normally distributed with a mean of 71 inches and standard deviation of 2.5 inches. What is the probability that the height of a randomly selected male student is more than 73.5 inches? 1 - 0.68 = 0.32 P(X > 73.5) = 0.16
Standard Normal Density Curves Always has m = 0 & s = 1 To standardize: Must have this memorized!
Strategies for finding probabilities or proportions in normal distributions • State the probability statement • Draw a picture • Calculate the z-score • Look up the probability (proportion) in the table
The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last less than 220 hours? Write the probability statement Draw & shade the curve P(X < 220) = 0.9082 Look up z-score in table Calculate z-score
The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last more than 220 hours? P(X>220) = 1 - 0.9082 = 0.0918
The lifetime of a certain type of battery is normally distributed with a mean of 200 hours and a standard deviation of 15 hours. How long must a battery last to be in the top 5%? Look up in table 0.95 to find z- score P(X > ?) = 0.05 .95 .05 1.645 224.675 hours Label units when given
The heights of the female students at JVHS are normally distributed with a mean of 65 inches. What is the standard deviation of this distribution if 18.5% of the female students are shorter than 63 inches? What is the z-score for the 63? P(X < 63) = 0.185 -0.9 63
The heights of female teachers at JVHS are normally distributed with mean of 65.5 inches and standard deviation of 2.25 inches. The heights of male teachers are normally distributed with mean of 70 inches and standard deviation of 2.5 inches. • Describe the distribution of differences of heights (male – female) teachers. Normal distribution with m= 4.5 inches & s= 3.3634 inches
4.5 • What is the probability that a randomly selected male teacher is shorter than a randomly selected female teacher? P(X<0) = 0.0901
Will my calculator do any of this normal stuff? • Normalpdf – use for graphing ONLY • Normalcdf – will find probability of area from lower bound to upper bound • Invnorm (inverse normal) – will find X-value for probability
Ways to Assess Normality Use graphs (dotplots, boxplots, or histograms) Use the Empirical Rule Normal probability (quantile) plot
Normal Probability (Quantile) plots The observation (x) is plotted against known normal z-scores If the points on the quantile plot lie close to a straight line, then the data is normally distributed Deviations on the quantile plot indicate nonnormal data Points far away from the plot indicate outliers Vertical stacks of points (repeated observations of the same number) is calledgranularity
Normal Scores Widths of Contact Windows Suppose we have the following observations of widths of contact windows in integrated circuit chips: 3.21 2.49 2.94 4.38 4.02 3.62 3.30 2.85 3.34 3.81 Think of selecting sample after sample of size 10 from a standard normal distribution. Then -1.539 is the average of the smallest observation from each sample & so on . . . Sketch a scatterplot by pairing the smallest normal score with the smallest observation from the data set & so on What should happen if our data set is normally distributed? To construct a normal probability plot, you can use quantities called normal score. The values of the normal scores depend on the sample size n. The normal scores when n = 10 are below: -1.539 -1.001 -0.656 -0.376 -0.123 0.123 0.376 0.656 1.001 1.539
Are these approximately normally distributed? 50 48 54 47 51 52 46 53 52 51 48 48 54 55 57 45 53 50 47 49 50 56 53 52 What is this called? The normal probability plot is approximately linear, so these data are approximately normal. Both the histogram & boxplot are approximately symmetrical, so these data are approximately normal.
Assignment • WS Normal Distributions • Due on Monday, 13 January 2014 • WS Continuous Distributions Review • Due on Tuesday, 14 January 2014 • Bookwork: 7.67, 7.69, 7.71, 7.73, 7.77, 7.80 • Due on Monday, 13 January 2014