SampleofRandomVariablexiiXiXi1110.981.022121.071.263130.861.0841.16141.025150.960.946161.110.6871.34170.998181.040.7891.21191.06100.86200.960.650.750.850.951.051.151.251.351.45x
Sample of Random Variable x
HistogramThe abscissa is divided between the maximum and minimummeasured values of x into K small intervalsThe number of times, n, that a measured value assumes avalue within an interval defined byx - Sx ≤x≤x + Sx plotted on the ordinateThe resulting plot of nj versus x is called a histogram ofthe variable x.Interval numberK = 1.87(N - 1)0.40 + 1
Histogram The abscissa is divided between the maximum and minimum measured values of x into K small intervals. The number of times, nj , that a measured value assumes a value within an interval defined by plotted on the ordinate The resulting plot of nj versus x is called a histogram of the variable x. Interval number
Case StudyiXiXi1110.98105021.0730.86945j41.16408一50.9610.660.683520.72771.3430.830681.0440.991.2125551.C100.86061.120inn471.2N = 20153102K = 1.87(N - 1)0.40 + 1 = 7518x=0.0500.551.151.251.351.400.650.750.850.951.05x
Case Study N = 20 dx = 0.05
ConclusionsThe total number of measurements, N, equals the sum ofthe number of occurrencesKZN=njj=1The area under the percent frequency distribution curveequals 100%KZf, = 100%100 ×j-1
Conclusions The total number of measurements, N, equals the sum of the number of occurrences The area under the percent frequency distribution curve equals 100%
ProbabilityDensityFunction ()As N → co, we can estimate the probability densityfunction, p(x), of the population of variable x.nj8x → 0 andlimp(x)= ox-0 N(28x)
Probability Density Function (1) • As N ∞, we can estimate the probability density function, p(x), of the population of variable x. • dx 0 and