Normality Notes page 138

The heights of the female students at RSH 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? P(X < 63) = What is the z- score for the 63? -0.9

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  = 4.5 &  =

What is the probability that a randomly selected male teacher is shorter than a randomly selected female teacher? 4.5 P(X<0) =.0901

Ways to Assess Normality Use graphs (dotplots, boxplots, or histograms) Normal probability (quantile) plot

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: Think of selecting sample after sample of size 10 from a standard normal distribution. Then is the average of the smallest observation from each sample & so on... Suppose we have the following observations of widths of contact windows in integrated circuit chips: Sketch a scatterplot by pairing the smallest normal score with the smallest observation from the data set & so on Normal Scores Widths of Contact Windows What should happen if our data set is normally distributed?

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? 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. What is this called?

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. 2) What is the mean and standard deviation of the above distribution? P(15<X<30) = binomialcdf(250,.1,30) – binomialcdf(250,.1,14) =.866  = 25 &  = 4.743

3) If we were to graph a histogram for the above binomial distribution, what shape do you think it will have? 4) What do you notice about the shape? Since the probability is only 10%, we would expect the histogram to be strongly skewed right. Let ’ s graph this distribution – Put the numbers 1-45 in L1 In L2, use binomialpdf to find the probabilities. Overlay a normal curve on your histogram: In Y1 = normalpdf(X, ,  )

Normal distributions can be used to estimate probabilities for binomial distributions when: 1) the probability of success is close to.5or 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: Or As binomial: Square: Simplify: Since (1 - p) < 1: And p < 1: Therefore, we say the np should be at least 10 and n (1 – p) should be at least 10.

Normal distributions can be used to estimate probabilities for binomial distributions when: 1) the probability of success is close to.5or 2) n is sufficiently large Rule: if n is large enough, then np > 10 & n(1 –p) > 10 Since a continuous distribution is used to 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) Why? Think about how discrete histograms are made. Each bar is centered over the discrete values. The bar for “ 1 ” actually goes from 0.5 to 1.5 & the bar for “ 2 ” goes from 1.5 to 2.5. Therefore, by adding or subtracting.5 from the discrete values, you find the actually width of the bars that you need to estimate with the normal curve.

(Back to our example) Since P(preemie) =.1 which is not close to.5, is n large enough? 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. Binomialwritten as Normal (w/cont. correction) P(15 < X < 30) 6) How does the answer in question 6 compare to the answer in question 1 (Binomial answer =0.866)? Normalcdf(14.5,30.5,25,4.743) =.8635 np = 250(.1) = 25 & n(1-p) = 250(.9) = 225 Yes, Ok to use normal to approximate binomial  P(14.5 < X < 30.5) =

Homework: Page 142 (notebook) Handout “Graded Assignment 2-2” (all)