D. Harvey - Modern Analytical Chemistry (794078), страница 24
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This is certainly true for the population of circulatingU.S. pennies. Instead, we select and analyze a limited subset, or sample, of the population. The data in Tables 4.1 and 4.10, for example, give results for two samples drawnat random from the larger population of all U.S.
pennies currently in circulation.sampleThose members of a population that weactually collect and analyze.4D.2 Probability Distributions for PopulationsTo predict the properties of a population on the basis of a sample, it is necessary toknow something about the population’s expected distribution around its centralvalue. The distribution of a population can be represented by plotting the frequencyof occurrence of individual values as a function of the values themselves.
Such plotsare called probability distributions. Unfortunately, we are rarely able to calculatethe exact probability distribution for a chemical system. In fact, the probability distribution can take any shape, depending on the nature of the chemical system beinginvestigated. Fortunately many chemical systems display one of several commonprobability distributions. Two of these distributions, the binomial distribution andthe normal distribution, are discussed next.probability distributionPlot showing frequency of occurrencefor members of a population.711400-CH04 9/8/99 3:54 PM Page 7272Modern Analytical Chemistrybinomial distributionProbability distribution showing chanceof obtaining one of two specificoutcomes in a fixed number of trials.Binomial Distribution The binomial distribution describes a population in whichthe values are the number of times a particular outcome occurs during a fixed number of trials.
Mathematically, the binomial distribution is given asP( X, N ) =N!× p X × (1 − p)N − XX !( N − X )!where P(X,N) is the probability that a given outcome will occur X times during Ntrials, and p is the probability that the outcome will occur in a single trial.* If youflip a coin five times, P(2,5) gives the probability that two of the five trials will turnup “heads.”A binomial distribution has well-defined measures of central tendency andspread. The true mean value, for example, is given asµ = Npand the true spread is given by the varianceσ2 = Np(1 – p)or the standard deviationσ =homogeneousUniform in composition.Np(1 − p)The binomial distribution describes a population whose members have onlycertain, discrete values.
A good example of a population obeying the binomial distribution is the sampling of homogeneous materials. As shown in Example 4.10, thebinomial distribution can be used to calculate the probability of finding a particularisotope in a molecule.EXAMPLE 4.10Carbon has two common isotopes, 12 C and 13 C, with relative isotopicabundances of, respectively, 98.89% and 1.11%. (a) What are the mean andstandard deviation for the number of 13C atoms in a molecule of cholesterol?(b) What is the probability of finding a molecule of cholesterol (C 27H44O)containing no atoms of 13C?SOLUTIONThe probability of finding an atom of 13C in cholesterol follows a binomialdistribution, where X is the sought for frequency of occurrence of 13C atoms, Nis the number of C atoms in a molecule of cholesterol, and p is the probabilityof finding an atom of 13C.(a) The mean number of 13C atoms in a molecule of cholesterol isµ = Np = 27 × 0.0111 = 0.300with a standard deviation ofσ = (27)(0.0111)(1 − 0.0111) = 0.172(b) Since the mean is less than one atom of 13C per molecule, mostmolecules of cholesterol will not have any 13C.
To calculate*N! is read as N-factorial and is the product N × (N – 1) × (N – 2) × . . . × 1. For example, 4! is 4 × 3 × 2 × 1, or 24.Your calculator probably has a key for calculating factorials.1400-CH04 9/8/99 3:54 PM Page 73Chapter 4 Evaluating Analytical Data73the probability, we substitute appropriate values into the binomialequationP(0, 27) =27!× (0.0111)0 × (1 − 0.0111)27 − 0 = 0.7400!(27 − 0)!There is therefore a 74.0% probability that a molecule of cholesterol willnot have an atom of 13C.A portion of the binomial distribution for atoms of 13C in cholesterol isshown in Figure 4.5. Note in particular that there is little probability of findingmore than two atoms of 13C in any molecule of cholesterol.0.80.7Probability0.60.50.40.30.20.1Figure 4.50Portion of the binomial distribution for thenumber of naturally occurring 13C atoms in amolecule of cholesterol.012345Number of atoms of carbon-13 in a molecule of cholesterolNormal Distribution The binomial distribution describes a population whosemembers have only certain, discrete values.
This is the case with the number of 13Catoms in a molecule, which must be an integer number no greater then the numberof carbon atoms in the molecule. A molecule, for example, cannot have 2.5 atoms of13C. Other populations are considered continuous, in that members of the population may take on any value.The most commonly encountered continuous distribution is the Gaussian, ornormal distribution, where the frequency of occurrence for a value, X, is given byf (X) = −( X − µ)2 exp 2 2σ12 πσ 2The shape of a normal distribution is determined by two parameters, the first ofwhich is the population’s central, or true mean value, µ, given as∑i =1 XiNµ =nwhere n is the number of members in the population. The second parameter is thepopulation’s variance, σ2, which is calculated using the following equation*∑i =1 ( XiNσ2 =n− µ)24.8*Note the difference between the equation for a population’s variance, which includes the term n in the denominator,and the similar equation for the variance of a sample (the square of equation 4.3), which includes the term n – 1 in thedenominator.
The reason for this difference is discussed later in the chapter.normal distribution“Bell-shaped” probability distributioncurve for measurements and resultsshowing the effect of random error.1400-CH04 9/8/99 3:54 PM Page 7474Modern Analytical Chemistry(a)f (x )(b)(c)Figure 4.6Normal distributions for (a) µ = 0 andσ2 = 25; (b) µ = 0 and σ2 = 100; and(c) µ = 0 and σ2 = 400.–50–40–30–20–10010Value of x20304050Examples of normal distributions with µ = 0 and σ2 = 25, 100 or 400, areshown in Figure 4.6. Several features of these normal distributions deserve attention. First, note that each normal distribution contains a single maximum corresponding to µ and that the distribution is symmetrical about this value.
Second,increasing the population’s variance increases the distribution’s spread while decreasing its height. Finally, because the normal distribution depends solely on µand σ2, the area, or probability of occurrence between any two limits defined interms of these parameters is the same for all normal distribution curves. For example, 68.26% of the members in a normally distributed population have valueswithin the range µ ±1σ, regardless of the actual values of µ and σ. As shown inExample 4.11, probability tables (Appendix 1A) can be used to determine theprobability of occurrence between any defined limits.EXAMPLE 4.11The amount of aspirin in the analgesic tablets from a particular manufacturer isknown to follow a normal distribution, with µ = 250 mg and σ2 = 25.
In arandom sampling of tablets from the production line, what percentage areexpected to contain between 243 and 262 mg of aspirin?SOLUTIONThe normal distribution for this example is shown in Figure 4.7, with theshaded area representing the percentage of tablets containing between 243 and262 mg of aspirin. To determine the percentage of tablets between these limits,we first determine the percentage of tablets with less than 243 mg of aspirin,and the percentage of tablets having more than 262 mg of aspirin.
This isaccomplished by calculating the deviation, z, of each limit from µ, using thefollowing equationX−µz =σwhere X is the limit in question, and σ, the population standard deviation, is 5.Thus, the deviation for the lower limit is1400-CH04 9/8/99 3:54 PM Page 75f (mg aspirin)Chapter 4 Evaluating Analytical Data0.080.070.060.050.040.030.020.01075Figure 4.7210220240230z low =250260Aspirin (mg)270280290Normal distribution for population ofaspirin tablets with µ = 250 mg aspirinand σ2 = 25.
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