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Normality Notes page 138 The heights of the female students at RSH are normally distributed with a What is the zmean of 65 inches. What is the for the standard deviation of this score distribution 63? if 18.5% of the female students are shorter than 63 inches? P(X < 63) = .185 63 65 .9 2 2.22 .9 -0.9 63 The heights of female teachers at RSH 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 & = 3.3634 • What is the probability that a randomly selected male teacher is shorter than a randomly selected female teacher? P(X<0) = .0901 0 4.5 z 1.34 3.3634 4.5 Ways to Assess Normality • Use graphs (dotplots, boxplots, or histograms) • Normal probability (quantile) plot Normal Scores Suppose we have the following To construct a normal probability plot, Sketch a scatterplot by pairing the Think of selecting sample after sample of observations of widths of contact you cansmallest use quantities called normal normal score with the size 10 from a standard normal windows in integrated circuit chips: What should score. The values of the normal scores smallest observation from the the 1distribution. Then -1.539 is data happen if sample size n. The normal depend on the set smallest & so on observation average of the our when data n = 10 are below: scores from each sample & so on . . . is 3.21 set2.49 2.94 4.38 4.02 2 33.34 4 3.81 5 3.62normally 3.301 2.85 distributed? -1 -1.539 -1.001 -0.656 -0.376 -0.123 Contact 1.001 Windows1.539 0.123 Widths 0.376of 0.656 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 called granularity Are these approximately normally distributed? 50 48 54 47 51 52 46 53 What 52 51 48 48 54 55 57is this 45 53 50 47 49 50 56 called? 53 52 Both the histogram & boxplot are approximately symmetrical, so these data are approximately normal. The normal probability plot is approximately linear, so these data are approximately normal. Normal Approximation to the Binomial Before widespread use of technology, binomial probability calculations were very tedious. Let’s see how statisticians estimated these calculations in the past! Premature babies are those born more than 3 weeks early. Newsweek (May 16, 1988) reported that 10% of the live births in the U.S. are premature. Suppose that 250 live births are randomly selected and that the number X of the “preemies” is determined. What is the probability that there are between 15 and 30 preemies, inclusive? (POD, p. 422) 1) Find this probability using the binomial distribution.P(15<X<30) = binomialcdf(250,.1,30) – binomialcdf(250,.1,14) =.866 2) What is the mean and standard deviation of the above distribution? m = 25 & = 4.743 3) If we were to graph a histogram for the above binomial distribution, whatdistribution shape do you Let’s graph this – think it will have? •Put the numbers 1-45 in L1 Since the probability is only 10%, we expect the histogram be •Inwould L2, use binomialpdf to to find strongly skewed right. the probabilities. 4) What do you notice about the shape? Overlay a normal curve on your histogram: •In Y1 = normalpdf(X,m,) Normal distributions can be used to estimate probabilities for binomial distributions when: 1) the probability of success is close to .5 or 2) n is sufficiently large Rule: if n is large enough, then np > 10 & n(1 –p) > 10 Why 10? Normal distributions extend infinitely in both directions; however, binomial distributions are between 0 and n. If we use a normal distribution to estimate a binomial distribution, we must cut off the tails of the normal distribution. This is OK if the mean of the normal distribution (which we use the mean of the binomial) is at least three standard deviations (3) from 0 and from n. (BVD, p. 334) We require: m 3 0 Or m 3 As binomial: np 3 np 1 p Square: n 2 p 2 9np 1 p Simplify: np 91 p Since (1 - p) < 1: np 9 n 1 p 9 And p < 1: Therefore, we say the np should be at least 10 and n (1 – p) should be at least 10. Normal can be used to Thinkdistributions about how discrete histograms estimate probabilities for binomial are made. Each bar is centered distributions when: over the discrete values. The bar 1) the probability of success is close to .5 for “1” actually goes from 0.5 to or 1.5 & the bar for “2” goes from 2) n is sufficiently large 1.5 to 2.5. Therefore, by adding Rule: if n is large enough, or subtracting .5 from the discrete then np > 10 & n(1 –p) > 10 Why? values, you find the actually width of the bars that you need to Since a continuous distribution is used to estimate with the normal curve. estimate the probabilities of a discrete distribution, a continuity correction is used to make the discrete values similar to continuous values.(+.5 to discrete values) (Back to our example) Since P(preemie) = .1 which is not close to .5, is n large enough? np = 250(.1) = 25 & n(1-p) = 250(.9) = 225 Yes, Ok to use normal to approximate binomial 5) Use a normal distribution with the binomial mean and standard deviation above to estimate the probability that between 15 & 30 preemies, inclusive, are born in the 250 randomly selected babies. Binomial written as Normal (w/cont. correction) P(15 < X < 30) P(14.5 < X < 30.5) = Normalcdf(14.5,30.5,25,4.743) = .8635 6) How does the answer in question 6 compare to the answer in question 1 (Binomial answer =0.866)? Homework: •Page 142 (notebook) •Handout “Graded Assignment 2-2” (all)