How do you find the probability of a normal distribution

7.2 Normal Probability Distributions

The normal or Gaussian Probability Distribution is most popular and important because of its unique mathematical properties which facilitate its application to practically any physical problem in the real world; if not for the data’s distribution directly, then in terms of the sampling distribution, this will be the discussion in Section 7.3. It constitutes the basis for the development of many of the statistical methods that we will learn in the following chapters. The study of the mathematical properties of the normal probability distribution is beyond the scope of this book; however, we shall concentrate on its usefulness in characterizing the behavior of continuous random variables that frequently occur in daily experience.

The normal probability distribution was discovered by Abraham De Moivre in 1733 as a way of approximating the binomial probability distribution when the number of trials in a given experiment is very large. In 1774, Laplace studied the mathematical properties of the normal probability distribution. Through a historical error, the discovery of the normal distribution was attributed to Gauss who first referred to it in a paper in 1809. In the nineteenth century, many scientists noted that measurement errors in a given experiment followed a pattern (the normal curve of errors) that was closely approximated by this probability distribution. The normal probability distribution is formally defined as follows:

Definition 7.2.1

Normal Probability Distribution

A continuous random variable X is normally distributed or follows a normal probability distribution if its probability distribution is given by the following function:

fx=1σ2πe−x−μ22σ2,

−∞<x<∞,−∞<μ<∞,0<σ2<∞.

The universally accepted notation X~Nμσ 2 is read as “the continuous random variable X is normally distributed with a population mean μ and population variance σ2. Of course in real world problems we do not know the true population parameters, but we estimate them from the sample mean and sample variance. However, first, we must fully understand the normal probability distribution.

The graph of the normal probability distribution is a “bell-shaped” curve, as shown in Figure 7.3. The constants μ and σ2 are the parameters; namely, “μ” is the population true mean (or expected value) of the subject phenomenon characterized by the continuous random variable, X, and “σ2” is the population true variance characterized by the continuous random variable, X. Hence, “σ” the population standard deviation characterized by the continuous random variable X; and the points located at μ−σ and μ+σ are the points of inflection; that is, where the graph changes from cupping up to cupping down.

The area under the bell-shaped curve is so disposed that it represents probability; that is, the total area under the curve is equal to one. The random variable X can assume values anywhere from minus infinity to plus infinity, but in practice we very seldom encounter problems in which random variables have such a wide range. The normal curve graph of the normal probability distribution) is symmetric with respect to the mean μ as the central position. That is, the area between μ and κ units to the left of μ is equal to the area between μ and κ units to the right of μ. This fact is illustrated in Figure 7.4.

Pμ≤X≤μ+κ

There is not a unique normal probability distribution, since the mathematical formula of the graph depends on the two variables, the mean μ and the variance σ2. Figure 7.5 is a graphical representation of the normal distribution for a fixed value of σ2 with μ varying.

You recall that the variance or standard deviation is a measure of the spread or “dispersion” of the random variable X around its expected value or central tendency, μ. Thus, σ2 of the normal distribution determines the shape of the bell-shaped curve. Figure 7.6 is a graphical representation of the normal distribution for a fixed value of μ with varying σ2. Thus, the expected value μ, locates the central tendency of the random variable, X, and the variance σ2 determines the shape of the bell-shaped curve. That is, for small values of σ2, the distribution is clustered close to the mean; as σ2 increases, the distribution deviates away from the mean. Despite the fact that the shapes are different, the total area under each curve which represents probability is equal to one.

Symbolic Computation

Here is code to compute the Gauss normal probability distribution and the corresponding cumulative distribution, given in Eq. (18.71). Because this code is needed for the Exercises, the reader may want to enter it into a workspace and preserve it for future use.

Here are some check values of the probability distribution:

The next two statements give the cumulative probability for x to be within 1.5σ of the mean at x=0:

3.2.4 Normal probability distribution

The single most important distribution in probability and statistics is the normal probability distribution. The density function of a normal probability distribution is bell shaped and symmetric about the mean. The normal probability distribution was introduced by the French mathematician Abraham de Moivre in 1733. He used it to approximate probabilities associated with binomial random variables when n is large. This was later extended by Laplace to the so-called CLT, which is one of the most important results in probability. Carl Friedrich Gauss in 1809 used the normal distribution to solve the important statistical problem of combining observations. Because Gauss played such a prominent role in determining the usefulness of the normal probability distribution, the normal probability distribution is often called the Gaussian distribution. Gauss and Laplace noticed that measurement errors tend to follow a bell-shaped curve, a normal probability distribution. Today, the normal probability distribution arises repeatedly in diverse areas of applications. For example, in biology, it has been observed that the normal probability distribution fits data on the heights and weights of human and animal populations, among others.

We should also mention here that almost all basic statistical inference is based on the normal probability distribution. The question that often arises is, when do we know that our data follow the normal distribution? To answer this question, we have specific statistical procedures that we study in later chapters, but at this point we can obtain some constructive indications of whether the data follow the normal distribution by using descriptive statistics. That is, if the histogram of our data can be capped with a bell-shaped curve (Fig. 3.2), if the stem-and-leaf diagram is fairly symmetrical with respect to its center, and/or by invoking the empirical rule “backward,” we can obtain a good indication of whether our data follow the normal probability distribution.

Mean, Variance, and Moment-Generating Function of a Normal Random Variable

If X ∼ N(μ, σ2), then the z-transform (or z-score) of X,Z=X−μσ∼N(0,1 ). This fact will be used in calculating probabilities for normal random variables. The normal table given in Appendix AV.2 is based on standard normal distribution. Note that for the continuous random variables, the probabilities of strict and nonstrict inequalities are the same, that is, P(X> a)=P(X≥a).

Example 3.2.8

For X ∼ N(0, 1), calculate P(Z ≥ 1.13).

For X ∼ N(5, 4), calculate P(−2.5 < X < 10).

Solution

Using the normal table,

P(Z≥1.13)=0.5−0.3708=0.1292.

The shaded part in the graph represents the P(Z ≥ 1.13).

R-code: pnorm(1.13, mean=0, sd=1, lower.tail=FALSE)

Using the z-transform, we have:

P (−2.5<X<10)=P(−2.5−5 2<Z<10−52)=P(−3.75<Z<2.5)=P(−3.75<Z<0)+P(0<Z<2.5) =0.9938.

That is, we are 99.38% certain the Z will assume a value between −2.5 and 10.

R-code: pnorm(2.5, mean=0, sd=1, lower.tail=TRUE)-pnorm(−3.75, mean=0, sd=1, lower.tail=TRUE) or pnorm(10, mean=5, sd=2, lower.tail=TRUE)-pnorm(−2.5, mean=5, sd=2, lower.tail=TRUE)

In the following example, we will show how to find the z values when the probabilities are given.

Example 3.2.9

For a standard normal random variable Z, find the value of z0 such that:

P(Z > z0) = 0.25

P(Z < z0) = 0.95

P(Z < z0) = 0.12

P(Z > z0) = 0.68

Solution

From the normal table, and using the fact that the shaded area in the figure is 0.25, we have P(Z>z0) =0.5−P(0≤Z≤z0) =0.25. Thus, P(0≤Z≤z0 )=0.25 and hence, we obtain z0 ≈ 0.675.

Because P(Z < z0) = 1 − P(Z ≥ z0) = 0.95 = 0.5 + 0.45. From the normal table, z0=1.645 .

From the normal table, z0 = −1.175.

Using the normal table, we have P(Z > z0) = 0.5 + P(0 < Z < z0) = 0.68.

This implies P(z0<Z<0)=0.18. From the normal table, z0 = −0.465.

Example 3.2.10

The scores of an examination are assumed to be normally distributed with μ = 75 and σ2 = 64. What is the probability that a student score chosen at random will be greater than 85?

Solution

Let X be a randomly chosen score from the exam scores. Then, X ∼ N(75, 64):

P(X>85)=P(X−758>85−758=1.25)=P(Z>1.25)=0.1056.

Thus, there is about a 10.56% chance that the score will be greater than 85.

In practice, whenever a large number of small effects is present and acting additively, it is reasonable to assume that observations will be normal. When the number of data is small, it is risky to assume a normal distribution without a proper testing. Apart from histogram, box-plot, and stem-and-leaf displays, one of the most useful tools for assessing normality is a quantile–quantile or QQ plot. This is a scatterplot with the quantiles of the scores on the horizontal axis and the expected normal scores on the vertical axis. The expected normal scores are calculated by taking the z-scores of (ri − 0.5)/n, where ri is the rank of the ith observation in increasing order. The steps in constructing a QQ plot are as follows: first, we sort the data in ascending order. If the plot of these scores against the expected normal scores is a straight line, then the data can be considered normal. Any curvature of the points indicates departure from normality. This procedure to obtain a normal plot (a QQ plot is similar to a normal plot for a normal distribution) is described in Project 4C. Fig. 3.3 shows a normal probability plot.

If plotted points do not fit the line well, but bend away from it in places, the distribution may be nonnormal. The shapes in Fig. 3.4 will give some indication of the distribution of the data.

Almost all of the statistical software packages include a procedure for obtaining the graph of a normal probability plot that can be used to test the normality of data. Errors in the measurements can also act in a multiplicative (rather than additive) manner. In that case, the assumption of normality is not justified.

A distribution closely related to normal distribution is the log-normal distribution. A variable might be modeled as log-normal if it can be thought of as the multiplicative effect of many small independent factors. This distribution arises in physical problems when the domain of the variate, X, is greater than zero and its histogram is markedly skewed. If a random variable Y is normally distributed, then exp(Y) has a log-normal distribution. Thus, the natural logarithm of a log-normally distributed variable is normally distributed. That is, if X is a random variable with log-normal distribution, then ln(X) is normally distributed. Most biological evidence suggests that the growth processes of living tissue proceed by multiplicative, not additive, increments. Thus, the measures of body size should at most follow a log-normal rather than a normal distribution. Also, the sizes of plants and animals are approximately log-normal. The log-normal distribution is also useful in modeling of claim sizes in the insurance industry.

The pdf of a log-normal random variable, X, is given as:

f(x)={1xσy2πe−(lnx−μy)2/2σy2,x>0,σ y>0,−∞<μy<∞0,otherwise.

where μy and σy are the mean and standard deviation of Y = ln(X). These parameters are related to the parameters of the random variable X as follows:

μy=ln (μx4μx2+σx2) ,σy=ln(μx2+σx2μx2).

We can verify that the expected value X is:

E(X)= eμy+(σy2/2)

and the variance is:

Va r(X)=(eσy2−1)e2 μy+σy2.

The question of when the log-normal distribution is applicable in a given physical problem after a certain amount of data has been obtained can be answered by creating a normal probability plot of ln(X) and testing for normality. Thus, if the natural logarithms of the data show normality, log-normal distribution may be more appropriate.

If X is log-normally distributed with parameters μy and σy, and 0 < a < b, then with Y = ln(X):

P (a≤X≤b)=P(lna≤Y≤ln b)=P( lna−μyσy≤Y−μyσy≤lnb−μyσy )=P(a′≤Z≤b′),

where Z ∼ N(0, 1). This probability can be obtained from the standard normal table.

Example 3.2.11

In an effort to establish a suitable height for the controls of a moving vehicle, information was gathered about X, the amounts by which the heights of the operators vary from 60 in., which is the minimum height. It was verified that the data that were collected followed the log-normal distribution by normal probability plot of Y = ln(X). Assume that μx = 6 in. and σx = 2 in.

What percentage of operators would have a height less than 65.5 in.?

If an operator is chosen at random, what is the probability that his or her height will be between 64 and 66 in.?

Solution

Here, X = 65.5 − 60 = 5.5. Also,

μy=ln(μx4μx2 +σx2)=ln6462+22=1.74,

σy=ln( μx2+σx2μx2)=ln62+2262=0.053.

Thus,

P(X≤5.5)=P(Y≤ln5.5)=P( Z≤(ln5.5)−1.740.053)=P(Z≤−0.67)=0.2514.

Hence, about 25.14% of the heights of the operators vary from 60 in.

Similar to (a), we get:

P(4≤X≤6)=P(ln4≤Y≤ln6)=P((ln4)−1.740.053≤Z≤(ln6) −1.740.053)=P(−6.67≤Z≤0.98)=0.8365.

Thus, 83.65% of the heights of the operators will be between 64 and 66 in.

14.8 Chapter Summary

Solution to Example 14.1 Quality Control

The distribution of the residuals from the ANOVA model for Example 14.1 did not have the assumed normal probability distribution. This leads us to suspect the results of the F test, particularly the p value. This problem, however, does fit the criteria for the use of a Kruskal–Wallis test. The data, the ranks, and the result of using PROC NONPAR1WAY in SAS are given in Table 14.9. Note that the printout gives the Kruskal–Wallis test statistic along with the p value calculated from the χ2 approximation. In this example, the p value is quite small so we reject the null hypothesis of equal treatment distributions.

Table 14.9. Windshield wipers.

Note that the output also gives the sums and means of the ranks (called scores). The sums are the ∑Ri in the formula for the test statistic. Also provided are the expected sum and the standard deviations if the null hypothesis is true. These are identical because the sample sizes are equal (each is 15), and the null hypothesis is that of equality. That is, we expect all four of the treatments to have equal sums of ranks if the populations are identical.

The mean scores given in Table 14.9 can be used to make pairwise comparisons (Section 14.4). The least significant difference between average ranks for α= 0.05 is 6.69. From Table 14.9 we can see that treatment 1 is significantly smaller than the other three, and that treatment 4 is significantly larger than the other three. Treatments 2 and 3 are not significantly different. Since we wanted to minimize the amount of wear, chemical treatment number 1 seems to be the best.

It is interesting to note that these results are quite similar to those obtained by the analysis of variance. This is because, unlike highly skewed or fat-tailed distributions, the uniform distribution of the random error does not pose a very serious violation of asumptions.

Nonparametric methods provide alternative statistical methodology when assumptions necessary for the use of linear model-based methods fail as well as provide procedures for making inferences when the scale of mesurement is ordinal or nominal. Generally, nonparametric methods use functions of observations, such as ranks, and therefore make no assumptions about underlying probability distributions. Previous chapters have presented various nonparametric procedures (for example, Chapter 12) usually used for handling nominal scale data. This chapter discusses rank-based nonparametric methods for one, two, and more than two independent samples, paired samples, randomized block designs, and correlation.

Many nonparametric techniques give answers similar to classical analyses on the ranks of the data rather than the raw data. For example, applying a standard one-way ANOVA to the ranks rather than the raw data will yield something close to the Kruskal-Wallis test. However, the precise details of the test differ somewhat, because we know the population variance for the ranks. For the raw data, this population variance is unknown and has to be estimated using the MSE.

One of the most powerful ideas presented here is the use of a randomization test to assign a p value. We have applied this technique to calculate p values for the standard nonparametric test statistics. However, this is a very general technique, and gives you the option for creating test statistics that you feel are particularly appropriate. For example, rather than using the Wilcoxon rank sum statistic to compare the locations of two groups, you could use the difference in the sample medians as your test statistic. A randomization test would allow you to calculate a p value for the null hypothesis that the distributions are the same, with special sensitivity to the possibility that they differ with respect to their medians.

The bootstrap is also a powerful tool that allows you some creativity in your estimation. Although the original intent of the bootstrap was to develop confidence intervals, it can also be used to calculate p values. Note that bootstraps are very much oriented toward identifying parameters, whereas randomization tests are meant to test for a relationship of an unspecified (nonparametric) nature.

The bootstrap and randomization tests are not all-purpose techniques that supplant the classical inferences. Bear in mind that if the assumptions of a classical analysis (such as least squares regression) are appropriate, then the traditional techniques will not only be more powerful but substantially simpler to employ. Further, the bootstrap has a number of special tricks that have to be understood before it can be applied in any but the simplest situation.

For quantitative dependent variables, we recommend that the first choice for any analysis be one of the standard parametric techniques. If the structure of the variables or an analysis of the residuals reveals problems with assumptions such as normality, then we should try to find a set of transformations that make the assumptions more reasonable. Nonparametric techniques are a useful fallback if no transformation can be found.

3 STATISTICAL MODEL

Each specimen in a population of explosive specimens will have an unobservable voltage threshold, denoted by V. The thresholds may be modeled as obeying a normal probability distribution. The probability distribution V is characterized by

(1)P(V≤t)=Φ(t−μσ),

where

Φ(z)=∫∞Z ϕ(u)du,ϕ(u)=(1/2π )exp(−12u2)

and μ, σ are the population mean and standard deviation of the voltage thresholds, respectively. We wish to estimate the parameters μ and σ as well as the p-th quartile of the distribution,Vp = μ + ϕ−1 (P)σ. The results of tests on a random sample of n specimens tested at voltages t1…tn are summarized by the sequence (x1, t1), (x2, t2),…where

(2)xi={ 1if specimen i detonatesat ti (go) ,0if specimen i fails todetonateatti(nogo).

The Graphical Relationship of Relative Frequency and Probability

Looking at Figure 3.2, relative frequency distributions of a sample of tumor sizes, we see that it is not far from symmetric and bell shaped. Does it arise from a normal probability distribution? Figure 3.4 repeats the frequency distribution with a normal probability distribution superposed. While clearly not exactly in agreement with the normal shape, the frequency distribution is not far off. Later chapters will address ways to test whether the frequency distribution either is normal and slightly off due to the variation of randomness in the sampling or probably did not arise from a normal distribution.

Chapter 4 will examine properties of probability distributions and present the most commonly seen types that are used extensively in statistical methods.

7.2.1 Histograms

We have already seen the use of histograms to assess sample distributions several times in this book, so they are one obvious way to assess if data fit a normal distribution. Figs. 7.1A and 7.1B show histograms of sample data together with normal probability distribution functions with mean and standard deviation equal to the sample mean and standard deviation. Fig. 7.1A shows a sample with 10,000 data values. The sample distribution seems to fit well to a normal distribution. In this case, we might feel confident in applying parametric hypothesis tests to answer questions about the data. However, Fig. 7.1B shows the student height data that we first saw in Chapter 1. Would we be justified in assuming that this sample comes from a normal distribution? We will now consider further ways to help us to answer such questions.

7.3 The Normal Distribution

You already know from Chapter 3 how to tell if a data set is approximately normally distributed. Now it's time to learn how to compute probabilities for this familiar bell-shaped distribution. One reason the normal distribution is particularly useful is the fact that, given only a mean and a standard deviation, you can compute any probability of interest (provided that the distribution really is normal).

The normal distribution, a continuous distribution, is represented by the familiar bell-shaped curve shown in Figure 7.3.1a. Note that there is a normal distribution for each combination of a mean value and a positive standard deviation value.8 Just slide the curve to the right or left until the peak is centered above the mean value; then stretch it wider or narrower until the scale matches the standard deviation. Two different normal distributions are shown in Figure 7.3.1b.

Visualize Probabilities as the Area under the Curve

The bell-shaped curve gives you a guide for visualizing the probabilities for a normal distribution. You are more likely to see values occurring near the middle, where the curve is high. At the edges, where the curve is lower, values are not as likely to occur. Formally, it is the area under the curve that gives you the probability of being within a region, as illustrated in Figure 7.3.2.

Figure 7.3.2. The probability that a normally distributed random variable is between any two values is equal to the area under the normal curve between these two values. You are more likely to see values in regions close to the mean.

Note that a shaded strip near the middle of the curve will have a larger area than a strip of the same width located nearer to the edge. Compare Figure 7.3.2 to Figure 7.3.3 to see this.

Figure 7.3.3. The probability of falling within a region that is farther from the middle of the curve. Since the normal curve is lower here, the probability is smaller than that shown in Figure 7.3.2.

The Standard Normal Distribution Z and Its Probabilities

The standard normal distribution is a normal distribution with mean μ = 0 and standard deviation σ = 1. The letter Z is often used to denote a random variable that follows this standard normal distribution. One way to compute probabilities for a normal distribution is to use tables that give probabilities for the standard one, since it would be impossible to keep different tables for each combination of mean and standard deviation. The standard normal distribution can represent any normal distribution, provided you think in terms of the number of standard deviations above or below the mean instead of the actual units (e.g., dollars) of the situation. The standard normal distribution is shown in Figure 7.3.4.

Figure 7.3.4. The standard normal distribution Z with mean value μ = 0 and standard deviation σ = 1. The standard normal distribution may be used to represent any normal distribution, provided you think in terms of the number of standard deviations above or below the mean.

The standard normal probability table, shown in Table 7.3.1, gives the probability that a standard normal random variable Z is less than any given number z. For example, the probability of being less than 1.38 is 0.9162, illustrated as an area in Figure 7.3.5. Doesn't it look like about 90% of the area? To find this number (0.9162), look up the value z = 1.38 in the standard normal probability table. While you're at it, look up 2.35 (to find 0.9906), 0 (to find 0.5000), and −0.82 (to find 0.2061). What is the probability corresponding to the value z = 0.36?

Table 7.3.1. Standard Normal Probability Table (See Figure 7.3.5)

z Value	Probability	z Value	Probability	z Value	Probability	z Value	Probability	z Value	Probability	z Value	Probability
−2.00	0.0228	−1.00	0.1587	0.00	0.5000	0.00	0.5000	1.00	0.8413	2.00	0.9772
−2.01	0.0222	−1.01	0.1562	−0.01	0.4960	0.01	0.5040	1.01	0.8438	2.01	0.9778
−2.02	0.0217	−1.02	0.1539	−0.02	0.4920	0.02	0.5080	1.02	0.8461	2.02	0.9783
−2.03	0.0212	−1.03	0.1515	−0.03	0.4880	0.03	0.5120	1.03	0.8485	2.03	0.9788
−2.04	0.0207	−1.04	0.1492	−0.04	0.4840	0.04	0.5160	1.04	0.8508	2.04	0.9793
−2.05	0.0202	−1.05	0.1469	−0.05	0.4801	0.05	0.5199	1.05	0.8531	2.05	0.9798
−2.06	0.0197	−1.06	0.1446	−0.06	0.4761	0.06	0.5239	1.06	0.8554	2.06	0.9803
−2.07	0.0192	−1.07	0.1423	−0.07	0.4721	0.07	0.5279	1.07	0.8577	2.07	0.9808
−2.08	0.0188	−1.08	0.1401	−0.08	0.4681	0.08	0.5319	1.08	0.8599	2.08	0.9812
−2.09	0.0183	−1.09	0.1379	−0.09	0.4641	0.09	0.5359	1.09	0.8621	2.09	0.9817
−2.10	0.0179	−1.10	0.1357	−0.10	0.4602	0.10	0.5398	1.10	0.8643	2.10	0.9821
−2.11	0.0174	−1.11	0.1335	−0.11	0.4562	0.11	0.5438	1.11	0.8665	2.11	0.9826
−2.12	0.0170	−1.12	0.1314	−0.12	0.4522	0.12	0.5478	1.12	0.8686	2.12	0.9830
−2.13	0.0166	−1.13	0.1292	−0.13	0.4483	0.13	0.5517	1.13	0.8708	2.13	0.9834
−2.14	0.0162	−1.14	0.1271	−0.14	0.4443	0.14	0.5557	1.14	0.8729	2.14	0.9838
−2.15	0.0158	−1.15	0.1251	−0.15	0.4404	0.15	0.5596	1.15	0.8749	2.15	0.9842
−2.16	0.0154	−1.16	0.1230	−0.16	0.4364	0.16	0.5636	1.16	0.8770	2.16	0.9846
−2.17	0.0150	−1.17	0.1210	−0.17	0.4325	0.17	0.5675	1.17	0.8790	2.17	0.9850
−2.18	0.0146	−1.18	0.1190	−0.18	0.4286	0.18	0.5714	1.18	0.8810	2.18	0.9854
−2.19	0.0143	−1.19	0.1170	−0.19	0.4247	0.19	0.5753	1.19	0.8830	2.19	0.9857
−2.20	0.0139	−1.20	0.1151	−0.20	0.4207	0.20	0.5793	1.20	0.8849	2.20	0.9861
−2.21	0.0136	−1.21	0.1131	−0.21	0.4168	0.21	0.5832	1.21	0.8869	2.21	0.9864
−2.22	0.0132	−1.22	0.1112	−0.22	0.4129	0.22	0.5871	1.22	0.8888	2.22	0.9868
−2.23	0.0129	−1.23	0.1093	−0.23	0.4090	0.23	0.5910	1.23	0.8907	2.23	0.9871
−2.24	0.0125	−1.24	0.1075	−0.24	0.4052	0.24	0.5948	1.24	0.8925	2.24	0.9875
−2.25	0.0122	−1.25	0.1056	−0.25	0.4013	0.25	0.5987	1.25	0.8944	2.25	0.9878
−2.26	0.0119	−1.26	0.1038	−0.26	0.3974	0.26	0.6026	1.26	0.8962	2.26	0.9881
−2.27	0.0116	−1.27	0.1020	−0.27	0.3936	0.27	0.6064	1.27	0.8980	2.27	0.9884
−2.28	0.0113	−1.28	0.1003	−0.28	0.3897	0.28	0.6103	1.28	0.8997	2.28	0.9887
−2.29	0.0110	−1.29	0.0985	−0.29	0.3859	0.29	0.6141	1.29	0.9015	2.29	0.9890
−2.30	0.0107	−1.30	0.0968	−0.30	0.3821	0.30	0.6179	1.30	0.9032	2.30	0.9893
−2.31	0.0104	−1.31	0.0951	−0.31	0.3783	0.31	0.6217	1.31	0.9049	2.31	0.9896
−2.32	0.0102	−1.32	0.0934	−0.32	0.3745	0.32	0.6255	1.32	0.9066	2.32	0.9898
−2.33	0.0099	−1.33	0.0918	−0.33	0.3707	0.33	0.6293	1.33	0.9082	2.33	0.9901
−2.34	0.0096	−1.34	0.0901	−0.34	0.3669	0.34	0.6331	1.34	0.9099	2.34	0.9904
−2.35	0.0094	−1.35	0.0885	−0.35	0.3632	0.35	0.6368	1.35	0.9115	2.35	0.9906
−2.36	0.0091	−1.36	0.0869	−0.36	0.3594	0.36	0.6406	1.36	0.9131	2.36	0.9909
−2.37	0.0089	−1.37	0.0853	−0.37	0.3557	0.37	0.6443	1.37	0.9147	2.37	0.9911
−2.38	0.0087	−1.38	0.0838	−0.38	0.3520	0.38	0.6480	1.38	0.9162	2.38	0.9913
−2.39	0.0084	−1.39	0.0823	−0.39	0.3483	0.39	0.6517	1.39	0.9177	2.39	0.9916
−2.40	0.0082	−1.40	0.0808	−0.40	0.3446	0.40	0.6554	1.40	0.9192	2.40	0.9918
−2.41	0.0080	−1.41	0.0793	−0.41	0.3409	0.41	0.6591	1.41	0.9207	2.41	0.9920
−2.42	0.0078	−1.42	0.0778	−0.42	0.3372	0.42	0.6628	1.42	0.9222	2.42	0.9922
−2.43	0.0075	−1.43	0.0764	−0.43	0.3336	0.43	0.6664	1.43	0.9236	2.43	0.9925
−2.44	0.0073	−1.44	0.0749	−0.44	0.3300	0.44	0.6700	1.44	0.9251	2.44	0.9927
−2.45	0.0071	−1.45	0.0735	−0.45	0.3264	0.45	0.6736	1.45	0.9265	2.45	0.9929
−2.46	0.0069	−1.46	0.0721	−0.46	0.3228	0.46	0.6772	1.46	0.9279	2.46	0.9931
−2.47	0.0068	−1.47	0.0708	−0.47	0.3192	0.47	0.6808	1.47	0.9292	2.47	0.9932
−2.48	0.0066	−1.48	0.0694	−0.48	0.3156	0.48	0.6844	1.48	0.9306	2.48	0.9934
−2.49	0.0064	−1.49	0.0681	−0.49	0.3121	0.49	0.6879	1.49	0.9319	2.49	0.9936
−2.50	0.0062	−1.50	0.0668	−0.50	0.3085	0.50	0.6915	1.50	0.9332	2.50	0.9938
−2.51	0.0060	−1.51	0.0655	−0.51	0.3050	0.51	0.6950	1.51	0.9345	2.51	0.9940
−2.52	0.0059	−1.52	0.0643	−0.52	0.3015	0.52	0.6985	1.52	0.9357	2.52	0.9941
−2.53	0.0057	−1.53	0.0630	−0.53	0.2981	0.53	0.7019	1.53	0.9370	2.53	0.9943
−2.54	0.0055	−1.54	0.0618	−0.54	0.2946	0.54	0.7054	1.54	0.9382	2.54	0.9945
−2.55	0.0054	−1.55	0.0606	−0.55	0.2912	0.55	0.7088	1.55	0.9394	2.55	0.9946
−2.56	0.0052	−1.56	0.0594	−0.56	0.2877	0.56	0.7123	1.56	0.9406	2.56	0.9948
−2.57	0.0051	−1.57	0.0582	−0.57	0.2843	0.57	0.7157	1.57	0.9418	2.57	0.9949
−2.58	0.0049	−1.58	0.0571	−0.58	0.2810	0.58	0.7190	1.58	0.9429	2.58	0.9951
−2.59	0.0048	−1.59	0.0559	−0.59	0.2776	0.59	0.7224	1.59	0.9441	2.59	0.9952
−2.60	0.0047	−1.60	0.0548	−0.60	0.2743	0.60	0.7257	1.60	0.9452	2.60	0.9953
−2.61	0.0045	−1.61	0.0537	−0.61	0.2709	0.61	0.7291	1.61	0.9463	2.61	0.9955
−2.62	0.0044	−1.62	0.0526	−0.62	0.2676	0.62	0.7324	1.62	0.9474	2.62	0.9956
−2.63	0.0043	−1.63	0.0516	−0.63	0.2643	0.63	0.7357	1.63	0.9484	2.63	0.9957
−2.64	0.0041	−1.64	0.0505	−0.64	0.2611	0.64	0.7389	1.64	0.9495	2.64	0.9959
−2.65	0.0040	−1.65	0.0495	−0.65	0.2578	0.65	0.7422	1.65	0.9505	2.65	0.9960
−2.66	0.0039	−1.66	0.0485	−0.66	0.2546	0.66	0.7454	1.66	0.9515	2.66	0.9961
−2.67	0.0038	−1.67	0.0475	−0.67	0.2514	0.67	0.7486	1.67	0.9525	2.67	0.9962
−2.68	0.0037	−1.68	0.0465	−0.68	0.2483	0.68	0.7517	1.68	0.9535	2.68	0.9963
−2.69	0.0036	−1.69	0.0455	−0.69	0.2451	0.69	0.7549	1.69	0.9545	2.69	0.9964
−2.70	0.0035	−1.70	0.0446	−0.70	0.2420	0.70	0.7580	1.70	0.9554	2.70	0.9965
−2.71	0.0034	−1.71	0.0436	−0.71	0.2389	0.71	0.7611	1.71	0.9564	2.71	0.9966
−2.72	0.0033	−1.72	0.0427	−0.72	0.2358	0.72	0.7642	1.72	0.9573	2.72	0.9967
−2.73	0.0032	−1.73	0.0418	−0.73	0.2327	0.73	0.7673	1.73	0.9582	2.73	0.9968
−2.74	0.0031	−1.74	0.0409	−0.74	0.2296	0.74	0.7704	1.74	0.9591	2.74	0.9969
−2.75	0.0030	−1.75	0.0401	−0.75	0.2266	0.75	0.7734	1.75	0.9599	2.75	0.9970
−2.76	0.0029	−1.76	0.0392	−0.76	0.2236	0.76	0.7764	1.76	0.9608	2.76	0.9971
−2.77	0.0028	−1.77	0.0384	−0.77	0.2206	0.77	0.7794	1.77	0.9616	2.77	0.9972
−2.78	0.0027	−1.78	0.0375	−0.78	0.2177	0.78	0.7823	1.78	0.9625	2.78	0.9973
−2.79	0.0026	−1.79	0.0367	−0.79	0.2148	0.79	0.7852	1.79	0.9633	2.79	0.9974
−2.80	0.0026	−1.80	0.0359	−0.80	0.2119	0.80	0.7881	1.80	0.9641	2.80	0.9974
−2.81	0.0025	−1.81	0.0351	−0.81	0.2090	0.81	0.7910	1.81	0.9649	2.81	0.9975
−2.82	0.0024	−1.82	0.0344	−0.82	0.2061	0.82	0.7939	1.82	0.9656	2.82	0.9976
−2.83	0.0023	−1.83	0.0336	−0.83	0.2033	0.83	0.7967	1.83	0.9664	2.83	0.9977
−2.84	0.0023	−1.84	0.0329	−0.84	0.2005	0.84	0.7995	1.84	0.9671	2.84	0.9977
−2.85	0.0022	−1.85	0.0322	−0.85	0.1977	0.85	0.8023	1.85	0.9678	2.85	0.9978
−2.86	0.0021	−1.86	0.0314	−0.86	0.1949	0.86	0.8051	1.86	0.9686	2.86	0.9979
−2.87	0.0021	−1.87	0.0307	−0.87	0.1922	0.87	0.8078	1.87	0.9693	2.87	0.9979
−2.88	0.0020	−1.88	0.0301	−0.88	0.1894	0.88	0.8106	1.88	0.9699	2.88	0.9980
−2.89	0.0019	−1.89	0.0294	−0.89	0.1867	0.89	0.8133	1.89	0.9706	2.89	0.9981
−2.90	0.0019	−1.90	0.0287	−0.90	0.1841	0.90	0.8159	1.90	0.9713	2.90	0.9981
−2.91	0.0018	−1.91	0.0281	−0.91	0.1814	0.91	0.8186	1.91	0.9719	2.91	0.9982
−2.92	0.0018	−1.92	0.0274	−0.92	0.1788	0.92	0.8212	1.92	0.9726	2.92	0.9982
−2.93	0.0017	−1.93	0.0268	−0.93	0.1762	0.93	0.8238	1.93	0.9732	2.93	0.9983
−2.94	0.0016	−1.94	0.0262	−0.94	0.1736	0.94	0.8264	1.94	0.9738	2.94	0.9984
−2.95	0.0016	−1.95	0.0256	−0.95	0.1711	0.95	0.8289	1.95	0.9744	2.95	0.9984
−2.96	0.0015	−1.96	0.0250	−0.96	0.1685	0.96	0.8315	1.96	0.9750	2.96	0.9985
−2.97	0.0015	−1.97	0.0244	−0.97	0.1660	0.97	0.8340	1.97	0.9756	2.97	0.9985
−2.98	0.0014	−1.98	0.0239	−0.98	0.1635	0.98	0.8365	1.98	0.9761	2.98	0.9986
−2.99	0.0014	−1.99	0.0233	−0.99	0.1611	0.99	0.8389	1.99	0.9767	2.99	0.9986
−3.00	0.0013	−2.00	0.0228	−1.00	0.1587	1.00	0.8413	2.00	0.9772	3.00	0.9987

Figure 7.3.5. The probability that a standard normal random variable is less than z = 1.38 is 0.9162, as found in the standard normal probability table. This corresponds to the shaded region to the left of 1.38, which is 91.62% of the total area under the curve.

Solving Word Problems for Normal Probabilities

A typical word problem involving a normal distribution is a story involving some application to business that gives you a value for the mean and one for the standard deviation. Then you are asked to find one or more probabilities of interest. Here is an example of such a word problem:

The upper-management people at Simplified Technologies, Inc., have finally noticed that sales forecasts are usually wrong. Last quarter's sales were forecast as $18 million but came in at $21.3 million. Sales for the next quarter are forecast as $20 million, with a standard deviation (based on previous experience) of $3 million. Assuming a normal distribution centered at the forecast value, find the probability of a “really bad quarter,” which is defined as sales lower than $15 million.

The beginning part sets the stage. The first numbers (18 and 21.3) describe past events but play no further role here. Instead, you should focus on the following facts:

•

There is a normal distribution involved here.

•

Its mean is μ = $20 million.

•

Its standard deviation is σ = $3 million.

•

You are asked to find the probability that sales will be lower than $15 million.

The next step is to convert all of these numbers (except for the mean and standard deviation) into standardized numbers; this has to be done before you can look up the answer in the standard normal probability table. A standardized number (often written as z) is the number of standard deviations above the mean (or below the mean, if the standardized number is negative). This conversion is done as follows:

z=Standardized number =Number−MeanStandard deviation=Number−μσ

In this example, the number $15 million is standardized as follows:

z=15−μσ=15−203=−1.67

This (z = −1.67) tells you that $15 million is 1.67 standard deviations below the mean (the forecast value).9 Your problem has now been reduced to finding a standard normal probability:

Find the probability that a standard normal variable is less than z = −1.67.

From the table, you find the answer:

The probability of a really bad quarter is 0.0475, or about 5%.

Whew! It seems that a really bad quarter is not very likely. However, a 5% chance is an outside possibility that should not be disregarded altogether.

Figures 7.3.6 and 7.3.7 show this probability calculation, both in terms of sales dollars and in terms of standardized sales numbers (standard deviations above or below the mean).

Figure 7.3.6. The probability of a really bad quarter (sales less than $15 million) is represented by the shaded area under the curve. This is based on the forecast of $20 million and the standard deviation of $3 million. The answer is found by standardizing and then using the standard normal probability table.

Figure 7.3.7. The probability of a really bad quarter, in terms of standardized sales numbers. This is the probability that sales will be more than z = −1.67 standard deviations below the mean. The answer is 0.0475.

This was an easy problem, since the answer was found directly from the standard normal probability table. Here is a question that requires a little more care:

Continuing with the sales-forecasting problem, find the probability of a “really good quarter,” which is defined as sales in excess of $24 million.

The first step is to standardize the sales number: $24 million is z = (24 − 20)/3 = 1.33 standard deviations above the mean. Thus, you are asked to solve the following problem:

Find the probability that a standard normal variable exceeds z = 1.33.

Using the complement rule, we know that this probability is 1 minus the probability of being less than z = 1.33. Looking up 1.33 in the table, you find the answer:

Probability of a really good quarter=1−0.9082 =0.0918, or about 9%

This probability is illustrated, in standardized numbers, in Figure 7.3.8.

Figure 7.3.8. The probability of a really good quarter, in terms of standardized sales numbers. The shaded area is 1 minus the unshaded area under the curve, which may be looked up in the table. The answer is 0.0918.

Here's another kind of problem:

Continuing with the sales-forecasting problem, find the probability of a “typical quarter,” which is defined as sales between $16 million and $23 million.

Begin by standardizing both of these numbers, to see that your task is to solve the following problem:

Find the probability that a standard normal is between z1 = −1.33 and z2 = 1.00.

To solve this kind of problem, look up each standardized number in the table and find the difference between the probabilities for the answer. Be sure to subtract the smaller from the larger so that your answer is a positive number and therefore a “legal” probability!

Probability of a typical quarter=0.8413−0.0918= 0.7495, or about 75%

This probability is illustrated, in standardized numbers, in Figure 7.3.9.

Figure 7.3.9. The probability of a typical quarter in terms of standardized sales numbers. The shaded area is found by looking up each standardized number in the table and then subtracting. Subtracting eliminates the unshaded area at the far left. The answer is 0.7495.

Finally, here's yet another kind of problem:

Continuing with the sales-forecasting problem, find the probability of a “surprising quarter,” which is defined as sales either less than $16 million or more than $23 million.

This asks for the probability of not being between two numbers. Using the complement rule, you may simply take 1 minus the probability found in the preceding example, which was the probability of being between these two values. The answer is therefore as follows:

Probability of a surprising quarter=1−0.7495=0.2505, or about 25%

This probability is illustrated, in standardized numbers, in Figure 7.3.10.

Figure 7.3.10. The probability of a surprising quarter, in terms of standardized sales numbers. The shaded area is found by looking up each standardized number in the table, finding the difference between the probabilities, and subtracting the result from 1. The answer is 0.2505.

To use Excel® to compute these first three probabilities, we use the function “NORMDIST(value,mean,standardDeviation,TRUE)” to find the probability that a normal distribution with specified mean and standard deviation is less than some value. There is no need to standardize because Excel will do this for you as part of the calculation. The first calculation is straightforward because it is a probability of being less. The second calculation is one minus the NORMDIST function because it is a probability of being greater. The third calculation is the difference of two NORMDIST calculations because it is the probability of being between two values. Here are the results:

Be Careful: Things Need Not Be Normal!

If you have a normal distribution and you know the mean and standard deviation, you can find correct probabilities by standardizing and then using the standard normal probability table. Fortunately, if the distribution is only approximately normal, your probabilities will still be approximately correct.

However, if the distribution is very far from normal, then any probabilities you might compute based on the mean, the standard deviation, and the normal table could be very wrong indeed.

Example

A Lottery (or Risky Project)

Consider a lottery (or a risky project, if you prefer) that pays back nothing 90% of the time, but pays $500 the remaining 10% of the time. The expected (mean) payoff is $50, and the standard deviation is $150 for this discrete random variable. Note that this does not represent a normal distribution; it's not even close because it's so discrete, with only two possible values.

What is the probability of winning at least $50? The correct answer is 10% because the only way to win anything at all is to win the full amount, $500, which is at least $50.

What if you assumed a normal distribution with this same mean ($50) and standard deviation ($150)? How far from the correct answer (10%) would you be? Very far away, because the probability that a normally distributed random variable exceeds its mean is 0.5 or 50%.

This is a big difference: 10% (the correct answer) versus 50% (computed by wrongly assuming a normal distribution). Figure 7.3.11 shows the large difference between the actual discrete distribution and the normal distribution with the same mean and standard deviation. Always be careful about assuming a normal distribution!

Figure 7.3.11. The discrete distribution of the payoff and the normal distribution having the same mean ($50) and standard deviation ($150). These distributions and their probabilities are very different. The discrete distribution gives the correct answers; the assumption of normality is wrong in this case.

1.7 The Bell-Shaped Curve

The Bell-shaped Curve is commonly called the normal curve and is mathematically referred to as the Gaussian probability distribution. Unlike Bernoulli trials which are based on discrete counts, the normal distribution is used to determine the probability of a continuous random variable.

Continuous

A type of measure such that the outcomes are dense, that is, between any two outcomes, other possible outcomes exist.

The graph of the normal probability distribution is a “bell-shaped” curve, as shown in the figure above. The constants μ and σ are the parameters.

The normal or Gaussian probability distribution is the most popular and important distribution because of its unique mathematical properties, which facilitate its application to practically any physical problem in the real world; if not for the data’s distribution directly, then in terms of the distribution associated with sampling. It constitutes the basis for the development of many of the statistical methods that we will learn in the following chapters.

The area under the curve represents the underlying probability of the situation.

Thus, our goal in studying the Bell-Shaped Curve is to put ourselves in a position to compute and interpret probabilities associated with continuous random variables and address such questions as:

Cancer: What is the probability that in a given group of lung cancer patients, an individual selected at random is Asian?

Education: What is the probability that a student will have a final grade in finite mathematics between 85 and 95?

Sports: What is the probability that a given lineman’s weight on the USF football team will be between 275 and 325 pounds?

Rainfall: What is the probability that the average rainfall in the State of Rhode Island in the year 2012 will be between 16 and 24 inches?

Chemistry: What is the probability that an acid solution made by a specific method will satisfactorily etch a tray?

The objective in learning the mathematical properties of the normal probability distribution is to realize its usefulness in characterizing the behavior of continuous random variables that frequently occur in daily experience.

IV.F Statistical Process Control

Statistical process control (SPC), also called statistical quality control (SQC), involves the application of statistical concepts to determine whether a process is operating satisfactorily. SPC has found widespread application in recent years due to the growing focus on increased productivity. Furthermore, it has become increasingly popular for processes, where feedback control is not possible due to the lack of on-line measurements as is the case in many microelectronics fabrication processes.

If a process is operating satisfactorily, then the variation of product quality falls within acceptable bounds, usually the minimum and maximum values of a specified product property. Normal operating data can be used to compute the mean y¯ and the root mean square (RMS) deviation σ of a given process variable from a series of n observations y1, y2, … ,yn as follows:

(13)y¯=1n∑i=1nyi

(14)σ=[1n∑i=1n(yi−y― )2]12.

The root mean square deviation is a measure of the spread of values for y around their mean. A large value of σ indicates that wide variations in y occur. Assuming the process variable follows a normal probability distribution, then 99.7% of all observations should lie within a deviation of ±3σ of the mean. The upper limit is given by y¯+3σ and the lower limit is equal to y¯−3σ. These limits are used to determine the quality of the control. It can be concluded that nothing unusual has happened during the recorded time period, if all data from a process lie within the ±3σ limits. The process environment is relatively unchanged and the product quality lies within specifications. If, on the other hand, repeated violations of the limits occur, then it can be concluded that the process environment has changed and the process is out of control. There are several specific rules like the Western-Electric rules that determine when a process is out of control. For example the Western-Electric rules state that a process is out of control if any of the following is true:

One measurement outside the ±3σ control limit

Any seven consecutive measurements lying on the same side of the mean

A decreasing or increasing trend for any seven consecutive measurements

Any nonrandom pattern in the process measurements

The above rules are applied to data in a control chart, such as Fig. 6, where pressure measurements are plotted over a time horizon. More details on SPC can be found in Grant and Leavenworth (1996).

A process that is “out of control” has important economic consequences, such as product waste and customer dissatisfaction. Therefore, statistical process control provides a way to continuously monitor process performance and product quality. However, it differs from automatic process control (such as feedback control) in that it does not provide a controller setting that will bring the process back to its desired operating point.