CAPÍTULO IV ANÁLISIS DE SELECCIÓN DEL PAÍS META PARA EXPORTAR CAFÉ ORGÁNICO MEXICANO
4.2 MATRIZ DE EVALUACIÓN DE FACTORES INTERNOS (MEFI) 23
Among the special probability densities we study in this chapter, the normal prob-ability density, usually referred to simply as the normal distribution, is by far the most important.1 It was studied first in the eighteenth century when scientists ob-served an astonishing degree of regularity in errors of measurement. They found that the patterns (distributions) they observed were closely approximated by a continu-ous distribution, which they referred to as the “normal curve of errors” and attributed to the laws of chance. The equation of the normal probability density, whose graph (shaped like the cross section of a bell) is shown in Figure 5.3, is
Normal distribution f ( x; μ, σ2)= 1
√2πσ e−( x−μ )2/2σ2 − ∞ < x < ∞
In Exercises 5.42 and 5.43, the reader will be asked to verify that its parametersμ andσ are indeed its mean and its standard deviation.
Figure 5.3
1The words density and distribution are often used interchangeably in the literature of applied statistics.
Sec 5.2 The Normal Distribution 141
Figure 5.4
The standard normal
probabilities F (z)= P( Z ≤ z ) 0 z
F(z)
Since the normal probability density cannot be integrated in closed form be-tween every pair of limits a and b, probabilities relating to normal distributions are usually obtained from special tables, such as Table 3 at the back endpapers of this book. This table pertains to the standard normal distribution, namely, the normal distribution withμ = 0 and σ = 1, and its entries are the values of
a b
F(b) 2 F(a)
Figure 5.5
The standard normal probability F (b)− F(a) = P( a< Z ≤ b )
F (z)= 1
√2π
z
−∞ e−t2/2dt= P( Z ≤ z )
for positive or negative z= 0.00, 0.01, 0.02, . . . , 3.49, and also z = 3.50, z = 4.00, and z = 5.00. The cumulative probabilities F(z) correspond to the area under the standard normal density to the left of z, as shown by the shaded area in Figure 5.4.
To find the probability that a random variable having the standard normal dis-tribution will take on a value between a and b, we use the equation
P( a< Z ≤ b ) = F(b) − F(a)
as shown by the shaded area in Figure 5.5. We also sometimes make use of the identity F (−z) = 1 − F(z), which holds for all symmetric distributions centered around 0. The reader is asked to verify this in Exercise 5.41.
Given access to statistical software or a statistical calculator, that approach is preferable to looking in tables. The solution to Example 5 includes the R commands (see Appendix C on R and Exercise 5.44 for MINITAB).
EXAMPLE 5 Calculating some standard normal probabilities
Find the probabilities that a random variable having the standard normal distribution will take on a value
(a) between 0.87 and 1.28;
(b) between−0.34 and 0.62;
(c) greater than 0.85;
(d) greater than−0.65.
Solution It is helpful to first indicate the area of interest in a graph as in Figure 5.6.
Figure 5.6
P( 0.87 < Z < 1.28 ) 0.87
z 0.0919
1.28 0
Looking up the necessary values in Table 3, for part (a) we get F (1.28) − F(0.87) = 0.8997 − 0.8078
= 0.0919 As indicated in Figure 5.7 for part (b),
F (0.62) − F(−0.34) = 0.7324 − 0.3669
= 0.3655
Figure 5.7
P(−0.34 < Z < 0.62 ) 20.34
z 0.3655
0 0.62
As indicated in Figure 5.8 for part (c),
1− F(0.85) = 1 − 0.8023
= 0.1977
Figure 5.8
P( Z> 0.85 ) 0.85
z 0.1977
0
As indicated in Figure 5.9 for part (d)
1− F(−0.65) = 1 − 0.2578 = 0.7422 or, alternatively,
1− F(−0.65) = 1 − [1 − F(0.65)]
= F(0.65)
= 0.7422
Figure 5.9
P( Z> −0.65 ) 20.65
z 0.7422
0 j
[ Using R: (a) pnorm(1.28) - pnorm(.87) (b) 1 - pnorm(.85) ]
There are also problems in which we are given probabilities relating to standard normal distributions and asked to find the corresponding values of z.
Sec 5.2 The Normal Distribution 143
Figure 5.10
The zαnotation for a standard
normal distribution 0 za
a
Let zαbe such that the probability isα that it will be exceeded by a random vari-able having the standard normal distribution. That is,α = P( Z > zα) as illustrated in Figure 5.10.
The results of the next example are used extensively in subsequent chapters.
EXAMPLE 6 Two important values for zα Find (a) z0.01; (b) z0.05.
Solution (a) Since F (z0.01)= 0.99, we look for the entry in Table 3 which is closest to 0.99 and get 0.9901 corresponding to z= 2.33. Thus z0.01= 2.33.
(b) Since F (z0.05)= 0.95, we look for the entry in Table 3 which is closest to 0.95 and get 0.9495 and 0.9505 corresponding to z= 1.64 and z = 1.65. Thus,
by interpolation, z0.05= 1.645. j
[ Using R: (a) qnorm(.99) (b) qnorm(.95) ]
To use Table 3 in connection with a random variable X which has a normal distribution with the mean μ and the variance σ2, we refer to the corresponding standardized random variable,
Z= X− μ σ
which can be shown to have the standard normal distribution. Thus, to find the prob-ability that the original random variable will take on a value less than or equal to a, in Table 3 we look up
F
a− μ σ
Also, to find the probability that a random variable having the normal distri-bution with the meanμ and the variance σ2will take on a value between a and b, we have only to calculate the probability that a random variable having the standard normal distribution will take on a value between
a− μ
σ and b− μ
σ
That is, to find probabilities concerning X , we convert its values to z scores using Z= X− μ
σ
Normal probabilities
When X has the normal distribution with meanμ and standard deviation σ.
P( a< X ≤ b ) = F
b− μ σ
− F
a− μ σ
According to Figure 2.9 on page 30, the observations on the strength of an alu-minum alloy appear to be normally distributed. The normal distribution is often used to model variation when the distribution is symmetric and has a single mode.
EXAMPLE 7 Calculation of probabilities using a normal distribution
With an eye toward improving performance, industrial engineers study the ability of scanners to read the bar codes of various food and household products. The maxi-mum reduction in power, occurring just before the scanner cannot read the bar code at a fixed distance, is called the maximum attenuation. This quantity, measured in decibels, varies from product to product. After collecting considerable data, the en-gineers decided to model the variation in maximum attenuation as a normal distri-bution with mean 10.1 dB and standard deviation 2.7 dB.
(a) For the next food or product, what is the probability that its maximum attenuation is between 8.5 dB and 13.0 dB?
(b) According to the normal model, what proportion of the products have maximum attenuation between 8.5 dB and 13.0 dB?
(c) What proportion of the products have maximum attenuation greater than 15.1 dB?
Solution (a) We treat the maximum attenuation of the next product, X , as a random selection for the normal distribution withμ = 10.1 and σ = 2.7.
Consequently, Z= (X − 10.1)/2.7 and, from Table 3, we get
F
13.0 − 10.1 2.7
− F
8.5 − 10.1 2.7
= F(1.07) − F(−0.59)
= 0.8577 − 0.2776
= 0.5801
as illustrated in Figure 5.11.
Figure 5.11
P( 8.5 < X < 13.0 ) =
P(−0.59 < Z < 1.07 ) 23 22 21
z 0.5801
0 1 2 3
z 10
5 0
.1 .2
15
Sec 5.2 The Normal Distribution 145
(b) The variation in maximum attenuation for the vast, but finite, collection of all different products is modeled by a normal distribution. The proportion of products having maximum attenuation between 8.5 and 13.0 dB corresponds to the probability in part (a). When we consider the even larger infinite population of all existing products and those that could have been made, we still refer to 0.5801 as the proportion having maximum attenuation between 8.5 and 13.0 dB.
(c) Looking up the necessary value in Table 3, 1− F
15.1 − 10.1 2.7
= 1 − F(1.85)
= 1 − 0.9678
= 0.0322 corresponding to the shaded area in Figure 5.12.
Figure 5.12
P( X ≥ 15.1 ) 0
z 0.0322
1 2 3
21 22 23
[ Using R: (a) pnorm(13.0, 10.1, 2.7)− pnorm(8.5, 10.1, 2.7)
(c) 1− pnorm(15.1, 10.1, 2.7) ] j
EXAMPLE 8 Normal Distribution as a Population Distribution
A major manufacturer of processed meats monitors the amount of each ingredient.
The weight(lb) of cheese per run is measured on n= 80 occasions. (courtesy of David Brauch))
72.2 67.8 78.0 64.4 76.3 72.3 73.1 71.7 66.2 63.3 85.4 67.4 66.3 76.3 57.7 50.3 77.4 63.1 73.9 67.4 74.7 68.2 87.4 86.4 69.4 58.0 63.3 72.7 73.6 68.8 63.3 63.3 73.0 64.8 73.1 70.9 85.9 74.4 75.9 72.3 84.3 61.8 79.2 64.3 65.4 66.7 77.2 50.0 70.3 90.4 63.9 62.1 68.2 55.1 52.6 68.5 55.2 73.5 53.7 61.7 47.9 72.3 61.1 71.8 83.1 71.2 58.8 61.8 86.8 64.5 52.3 58.3 65.9 80.2 75.1 59.9 62.3 48.8 64.3 75.4
Figure 5.13 suggests that the histogram, and therefore the population distribu-tion, is well approximated by a normal distribution with meanμ = 68.4 and stan-dard deviation σ = 9.6 pounds. You are asked to examine the assumption of a normal distribution more closely in Exercise 5.102.
Using the normal population distribution,
(a) Find the probability of using 80 or more pounds of cheese.
(b) Set a limit so that only 10 % of production runs have less than L pounds of cheese.
(c) Determine a new mean for the distribution so that only 5 % of the runs have less than L pounds.
Figure 5.13
A normal distribution models
weight of cheese. Weigh (lb)
Density
40 50 60 70 80 90 100
.05
.025
0
Solution (a) Z = ( X − 68.4 ) / 9.6 and, from Table 3, we get 1 − F
80 − 68.4 9.6
= 1 − F ( 1.208) = 1 − .8865 = .1135 About 1 out of 9 production runs will result in more than 80 pounds of cheese.
(b) From Table 3, the entry with probability closest to .1000 is z0.10 = 1.28. The limit L is given by
L = μ − σ z0.10 = 68.4 − 9.6 × 1.28 = 56.1 pounds (c) The new value of the meanμ must satisfy
− z.05 = L − μ 9.6 where z0.05= 1.645 so
μ = L + 9.6 × z0.05 = 56.1 + 9.6 × 1.645 = 71.9 pounds The mean must be increased by 3.5 pounds to decrease the percentage of units
below the limit L from 10% to 5%. j
[ Using R: (a) 1-pnorm(80,68.4,9.6) (b) L = 68.4+9.6*qnorm(.10) (c) L+9.6*qnorm(.95) ]
EXAMPLE 9 Calculating probabilities when ln X has a normal distribution
After collecting a large number of assays of the gold content in rocks from an open pit mine, a mining engineer postulates that the natural log of the gold content (oz/st gold) follows a normal distribution with mean−4.6 and variance 1.21. Under this distribution, would it be unusual to get 0.0015 oz/st gold or less in an assay?
Solution Because it is ln X that has a normal distribution, the question concerns the standard-ized value
ln ( 0.0015 ) − (−4.6)
√1.21 = −1.729
The standard normal probability of obtaining this value or smaller (see Figure 5.14) is F
ln ( 0.0015 ) − (−4.6)
√1.21
= F(−1.73) = 0.0419
Sec 5.2 The Normal Distribution 147
Figure 5.14 P( Z≤ −1.73 ) = P( X≤ 0.0015 )
0.0419
0 1 2 3
21 22 23
z
This probability is small, so we suspect that the postulated normal distribution with mean−4.6 does not apply. An assay with this small amount of gold content could suggest that the specimen was collected outside of the vein. j Although the normal distribution applies to continuous random variables, it is often used to approximate distributions of discrete random variables. Quite often, this yields satisfactory results, provided that we make the continuity correction illustrated in the following example.
EXAMPLE 10 A continuity correction to improve the normal approximation to a count variable
In a certain city, the number of power outages per month is a random variable, having a distribution withμ = 11.6 and σ = 3.3. If this distribution can be approximated closely with a normal distribution, what is the probability that there will be at least 8 outages in any one month?
Solution The answer is given by the area of the shaded region of Figure 5.15—the area to the right of 7.5, not 8. The reason for this is that the number of outages is a dis-crete random variable, and if we want to approximate its distribution with a normal distribution, we must “spread” its values over a continuous scale. We do this by representing each integer k by the interval from k− 12 to k+ 12. For instance, 3 is represented by the interval from 2.5 to 3.5, 10 is represented by the interval from 9.5 to 10.5, and “at least 8” is represented by the interval to the right of 7.5. Thus the desired probability is approximated by
1− F
7.5 − 11.6 3.3
= 1 − F(−1.24)
= F(1.24)
= 0.8925 j
Figure 5.15
Diagram for example dealing with power outages
0.8925
7.5 11.6
Number outages