FM REQUENCY EASURES OF D C ISTRIBUTION ENTRAL
RETPAH2CFREQUENCY DISTRIBUTIONMEASURES OF CENTRAL VALUESAND DISPERSIONINTRODUCTIONVariability is the most common characteristic of the data with reference toagriculture, biological and physical science research. The need for statisticalmethod arises from the variability.Examples where variation is observed1. Number of seedlings2. Blood group in man and their frequency3. Plaque morphology of Bacteriophages infecting different strains of E. coli.4. Plant height5. Number of insects with specific eye colour6. Eye colour in Drosophila.Characteristics which show variation of this sort are called random variables or variates.Continuous variable: A continuous variable is one that can take any valuein a given range which may be finite or infinite. For example, the yield of acrop, height of a plant, animal height, human height are continuous variables.Discrete variables: A discrete variable is one for which the possible valuesare not observed on a continuous scale.For example, number of children in a family, number of fingers, number ofplants bearing reel flowers, number of Drosophila with red eyes, number of wildtype and mutant colonies of a microorganism, number of lethals, (dominant orrecessive lethals), number of fragments in a cell observed during cell division.POPULATION AND SAMPLEA population is defined as the total set of actual or possible values of thevariable. A population may be finite or infinite, and the variable, continuousor discrete.
78Fundamentals of BiostatisticsThe idea of infinite populations distributed in a frequency distribution inrespect of one or more characters is fundamental to all statistical work (Fisher-1948). One of the principle objectives of statistics is to draw inferences withrespect to populations by the study of groups of individuals forming part ofthe populations.A sample is therefore any finite set of items drawn from a population. Thepurpose of drawing samples is to obtain information about the populationsfrom which they are drawn. A random sample from a given population is asample, chosen in such a manner that each possible sample has an equal chanceof being drawn.Quantities which characterise populations are known as parameters. Characters of sample are called statistics. A parameter is a fixed quantity, statisticis a variate. Generally, a statistic is sought which ‘best’ estimate the corresponding population parameter.FREQUENCY DISTRIBUTIONSThe assemblage of xi with their associated frequencies f is called a frequencydistribution. A typical frequency distribution is presented in table.Frequency distribution of variable XValue of variableFrequencyX1X2XiXnf1f2fifnTotal frequencyNExample 2.1 Prepare a frequency distribution for the following data of heightof 20 7144133133143128129Frequency y 15743Total20
Frequency Distribution — Measures of Central Values and Dispersion79HISTOGRAMGrouped data can be displayed in a histogram.In a histogram rectangles are drawn so that the area of each rectangle isproportional to the frequency in the range covered by it.Example 2.2 The lengths of 30 plant leaves of species A were measured andthe information grouped as shown. Measurements were taken correct to thenearest cm. Draw a histogram to illustrate the data.Length of leaf 27Frequency12Frequency10864209.5 14.5 19.5 24.5 29.5Length of leafFREQUENCY POLYGONSA frequency distribution may be displayed as a frequency polygon.A frequency polygon may be superimposed on a histogram by joining themidpoints of the tops of the rectangles. This is in grouped data.(a) Ungrouped data: The fin length in (cm) of particular type of fish is given.Draw a frequency polygon to illustrate this information.Fin length (cm)246789Frequency3854218Frequency6420246 7 8 9Fin length (cm)
80Fundamentals of BiostatisticsNo. of cases (Frequency)(b) Grouped data: The following table shows the age distribution of cases ofa certain disease reported during a year in a particular state. Prepare afrequency polygon.AgeNumber of 22213525201510504.5 14.5 24.5 34.5 44.5 54.5AgeExample 2.3 Pollen grain length in microns is given below. Prepare frequencytable. Represent cumulative frequency, draw histogram, frequency s ��5960–69 3 11 9 9 7 933 11 1414 9 2323 9 3232 7 3939 9 48Total 48
Frequency Distribution — Measures of Central Values and Dispersion811210864209.5 19.5 29.5 39.5 49.5 59.5Frequency polygon1210864209.5 19.5 29.5 39.5 49.5 59.5 69.5HistogramMEASURES OF CENTRAL TENDENCYOne of the most important objectives of statistical analysis is to get one singlevalue that describes the characteristic of the entire mass of data.There are three main statistical measures which attempt to locate a ‘typical’value. These are1. Arithmetic mean (A.M.)2. Median3. Mode
82Fundamentals of BiostatisticsOther measures of central tendency are1. Geometric Mean (G.M.)2. Harmonic Mean (H.M.)ARITHMETIC MEANA numerical value which indicates the centre of the distribution is called theArithmetic Mean.For ungrouped dataX̄ n xii 1nGrouped data fxX̄ n where n f ; (Sigma) Summation; X̄(X-Bar) MeanExample 2.4 Find the mean of set of numbers 63, 65, 67, 68, 69, 70, 71, 72,74, 75.n 10 x 63 65 67 68 69 70 71 72 74 75 694 694X 69.4X̄ n10Frequency Distribution No. of flowers (X)12345No. of plants (f )1110531XffX12345111053111201512563fX 2.1n30 where n fX̄ f 30 f X 63
Frequency Distribution — Measures of Central Values and Dispersion83Grouped Frequency DistributionThe lengths of 32 leaves were measured correct to the nearest mm. Find themean length.Length (mm)FrequencyLength int 6–2829–3132–34 867fX 27.1 mn32 where n fX̄ f 32 f X 867Merits and demerits of arithmetic mean1. It is the simplest to understand and the easiest to compute.2. It is affected by the value of every item in series.Demerit1. Arithmetic mean is not always a good measure of central tendency, as forinstance in extremely symmetrical distribution.MEDIANThe median refers to the middle value in a distribution. The median is called apositional average. The median is the middle value of a set of numbers arranged1in order of magnitude. The median is the (n 1)th value. The median is2that value of the variate for which 50% of the observations, when arranged inorder of magnitude, lie on each side. Median is the value of the variate whichdivides the total frequency in the whole range into two equal parts. Median isnot affected by extreme values.If the total frequency is even, the median is the arithmetic mean of the twomiddle values. Compared with the arithmetic mean, the median places lessemphasis on the minimum or maximum value in the sample or population.Example 2.5Find the median of each of the sets.(a) 7 , 7 , 2 , 3 , 4 , 2 , 7 , 9 , 31(b) 36 , 41 , 27 , 32 , 29 , 38 , 39 , 43
84Fundamentals of BiostatisticsSolution:(a) 2, 2, 3, 4, 7, 7, 7, 9, 311n 9, and the median is the (9 1) the value, i.e., the 5th value.2Median 7(b) Arranging in order of magnitude27, 29, 32, 36, 38, 39,41,4311n 8 and the median is average of the (8 1) the value, i.e., the 422value. This does not exist, so we consider the 4th and 5th values.Median 1(36 38) 372In general, if n is odd then there is a middle value, and this is the median.If n is even and the two middle value are c and d, then the median is1(c d).2Calculation of medianGrouped dataMedian L (n/2) m Cfwhere L Lower limit of median class; m Cumulative frequency abovemedian class; f Frequency of median class; C Class interval.Example 2.6Find median of the following distributionClass 6565–7575–85361132153136516Class interval �85Frequency (F)Cumulative frequency (CF)361132153 F136516364196 m349485536542 NN542 27122(This is nearer to 349 under CF) Median class 45-55.Median L (n/2) m271 196 C 45 10 49.9f153
Frequency Distribution — Measures of Central Values and Dispersion85MODEMode or modal value of a given data is the one which has maximum frequencyor any value of the data which occurs repeatedly.Example 2.72, 4Find the mode of the following 2, 1, 0, 3, 5, 2, 3, 1, 0, 2, 2,Value012345Frequency2252112 is the mode (unimodal) as this is having high frequency.Example 2.8Find the mode of the following 1, 1, 2, 3, 0, 2, 4.Value01234Frequency12211These are two modal values in the data they are 1 and 2 (bimodal) havinghigh frequency.Mode 3 Median 2 MeanExample 2.9Find the Mode of the followingClass interval �85Mode L Frequency (F)361132 f1153 f0136 f2516 1 C 1 2By inspection the highest frequency is (153) and the modal class is 45–55where l lower limit of the modal class 1 f0 f1 (153 132 21) 2 f0 f2 (153 136 17)C length of class interval 1021021 10 45 50.5Mode 45 21 1738
86Fundamentals of BiostatisticsHARMONIC MEAN FOR GROUPED DATA AND UNGROUPED DATAExample 2.10Calculate Harmonic mean for the following grouped data,Class interval (CI)0–1010–2020–3030–404557Frequency (f ) 7(1/5)(1/15)(1/25)(1/35)(1/5) 4 (4/5)(1/15) 5 (1/3)(1/25) 5 (1/5)(1/35) 7 (1/5)231 1.53x15N21H.M. 13.72 11.53f x where N ff.Harmonic mean for ungrouped dataHarmonic mean is calculated by the following formula:H.M. (N)111 .x1x2xnx1 , x2 , . . . , xn are variables.Example 2.11 Calculate the harmonic mean of the following1 , 0 .5 , 10 , 45 .0 , 175 , 0 .01 , 4 .0 , 11 .2(B.Com., Mysore, 00.02220.0057100.000.25000.0893 1/x 103.4672N8 0.0777H.M. 1103.467x
Frequency Distribution — Measures of Central Values and DispersionHence,89 X1 X̄ X2 X̄ . . . Xn X̄ Mean deviation Xi X̄ M.D. Modulusnn(i.e., absolute values are taken e.g.: 3-8 or (8-3) is written as 5)Example 2.14 Find the mean deviation of the scores 3, 5, 7, 9, 11 and 13from the arithmetic mean.3 5 7 9 11 1348 866 X X̄ d orMean deviation nN 3 8 5 8 7 8 9 8 11 8 13 8 65 3 1 1 3 5 3 6Arithmetic mean X̄ Example 2.15distributionFind the mean deviation from the A.M. for the followingClass 8152593Midvalue XFrequencyffX d x 32.33 x x̄ f. d 22.67138.64109.9566.75114.0368.01601940497.38 1940fX Mean 60f 497.38f d 8.3Mean Deviation N60QUARTILE DEVIATIONCalculate the quartile deviation for the following frequency distribution.Class 405660968468
90Fundamentals of BiostatisticsAgeFrequencyCumulative frequency5–66–77–88–99–1010–11405660 (f1 )9684 (f3 )684096(m1 )156 Q1 , Class252(m3 )336 Q3 , Class404 NHere N 404; 25% of N N/4 101 and 75% of N 3N/4 303 (whereN 404) Q1 class 7 8l1 7, m1 96, f1 60 and C 1][n/4 m1 CQ 1 l1 f1][101 96Q1 7 1 7 0.08 7.0860][3n/4 m3Q 3 l3 C3f3Q3 Class 9 109 (303 252) 1 9 0.60784Q3 9.6079.607 7.08Q3 Q1 Quartile deviation Q 222.527 1.263Q.D. 2Q3 VARIANCE AND STANDARD DEVIATIONOne of the important measures of variation is that of variance, which indicateshow the different values of the data are scattered away from the centre of thedistribution. It is usually denoted by the symbol σ 2 . The positive square rootof variance is called the standard deviation and is denoted by σ. 22d(x x̄)orn 1n 1 d2Standard deviation Variance n 1Variance σ 2 Example 2.1611 and 13. Find the standard deviation of the following scores 3, 5, 7, 9,3 5 7 9 11 13Mean 86 222222(x x̄) (3 8) (5 8) (9 8) (11 8) (13 8)
Frequency Distribution — Measures of Central Values and Dispersion93different units. For example, we may wish to know, for a certain population,whether serum cholesterol levels, measured in mg per 100 ml, are more variablethan body weight, measured in pounds.What is needed in situations like these is a measure of relative variationrather than absolute variation. Such a measure is found in the coefficient ofvariation which expresses the standard deviation as a percentage of the mean.The formula is given byS.DC.V. 100x̄We see that, since the mean and standard deviation are expressed in thesame unit of measurement, the unit of measurement cancels out in computingthe coefficient of variation. What we have, then, is a measure that is independent of the unit of measurement.AgeMean WeightStandard DeviationSample 125 years145 pounds10 poundsSample 211 years80 pounds10 poundsA comparison of the standard deviations might lead one to conclude thatthe two samples possess equal variability. If we compute the coefficients ofvariation, however, we have for the twenty-five years old.C.V. 10 100 6.9145and for the eleven-year old10 100 12.580If we compare these results, we get quite a different impression.C.V. Example 2.19Calculate Mean, Median, Mode of the following 60MidpointxFrequency(f )CFfx51525354555121416 f128 f010 f28122642m708088n60210400980450440882540
94Fundamentals of BiostatisticsMean l 30,m 42, 2540fx 28.86 (where N f)N88(n/2) mMedian L c (n/2 88/2 44)fCheck under ‘cf ’ which is same as n/2 or just above, i.e., 70; againstthat is the median class 30–40.C 10,n88 22f 28(88/2) 42 102844 42 10 30 282 10 30 28 30 30 0.07142 10 30 0.7142 30.7142 1 CMode L 1 2L Lower limit of modal class 1 f0 f1 28 16 12 2 f0 f2 28 10 18C (Class interval) 10. The highest among frequency 28.Against that modal class 30 40 30 12 1012 18 30 12 1030 30 4 34Construct Frequency Distribution TableExample 2.2058477094846188876875Draw 07092749672747066907290617880
96Fundamentals of Biostatistics8Frequency64201Example 2.2223 4 5 6 7Height in (cm)8910Prepare relative frequency histogramClass intervalFrequencyRelative frequency468512154/50 0.086/50 0.128/50 0.165/50 0.1012/50 0.2415/50 0.300-1010-2020-3030-4040-5050-600.300.30Relative frequency0.240.200.160.120.10.100.080.0101020 30 40 50Class interval60Example 2.23 The following table gives the number of yeast cells in 100squares of a haemocytometer. Calculate standard deviation, variance, coefficient of variation (C.V.) and standard error.Class intervalFrequency0-23-56-89-1112-1415-17661240306
Frequency Distribution — Measures of Central Values and Dispersion97ClassFrequency Midpoint f X Deviation from (x x̄)2 f (x X̄)2interval(f )XMeanf (x 027021610001296 1000fX 10 (where n f)n100 2 2fd1296f (x x̄)Variance S 2 or 13.1n 1n 199 S.D. 13.1 3.63.63.6S.D. 0.36 S.E. 10n100X̄ S.D. 100X̄3.6 100 36 10Coefficient of variation Example 2.244, 8, 4 Calculate mean for white blood counts: (X 100) 5, 6, 4, 5, 4,Xi 5 6 4 5 4 4 8 4 40n 8;X̄ 40/8 5Example 2.25Weight of 11 tables removed from box for quality controlpurpose is given below. Calculate median251, 231, 245, 250, 250, 251, 255, 260, 265, 275, 300(Hint: Arrange data in ascending order)Median n 1/2 11 1/2 12/2 251 (this is the 6th position value)Example 2.26 Concentration of a therapeutic agent in vials of commerciallyavailable products given below is mg/ml. Calculate mode.Vial no.Cone. (mg/ml)12345678910200205205201199199202205205206
98Fundamentals of BiostatisticsThe most common value in the above data is 205 mg/ml. Therefore, themode is 205.Example 2.2731, 33.Find mode 20, 22, 23, 25, 25, 25, 28, 29, 30, 31, 31, 31, 31,Mode is 25, 31. Two modes exist, 31 is major mode and 25 is minor mode.Example 2.28Calculate mode for the following data:Class interval 0–l0 10–20 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100Frequency48101525201816181 C Mode class 40 501 2 40 10/10 5 10Mode L 40 20/3 46.66Example 2.29Calculate mean, median and mode.XffXCumulative 5337.5340332.5315575312.5270410183044 m6070798794100110115119 an f X 3927.5f X/n 3927.5/119 33Median L (n/2 m)/f c 25 59.5 44/16 5 29.84Mode L 1/ 1 2 C 25 2/8 5 25 1.25 26.2510
Frequency Distribution — Measures of Central Values and DispersionExample 2.30Calculate harmonic meanClass interval 0–4 4–8 8–12 12–16 16–20 70.050.041.50.640.60.350.250.32 31H.M. N/f (1/x)8f (l/x) 3.66 31/3.66 8.47Example 2.31Calculate mean and mode for the following dataClass 20.0115.0 Mean f 69 f X 2177.5f x/n 2177.5/69 31.56Mode L 1/ 1 2 C 25 6/6 8 5 25 6/14 5 25 15/7 25 2.14 27.14Mode class 25–3099
102Fundamentals of BiostatisticsExample 2.36Calculate the mean deviation from the following data.CIXffX d f d 780450372717070313233318521620410560182276198 f 92 f X 3870f d 1426Mean 3870/92 42Mean deviation Σf d /(n 1) 1426/(92 1) 1426/91 15.67Example 2.37data.Calculate the median, mode and variance from the followingCIXffXCFdd2fd 4991695291089684558323468735180236663486534 f 92 f X 3870Mean 3870/92 42Median L n/2 m/f c 40 46 40/20 10 40 6/20 10 43 Variance f d2 /n 1 32308/92 1 32308/91 355.03Mode 40 5/(5 6) 10 40 0.45 10 40 4.5 44.5 f d2 32308
104Fundamentals of BiostatisticsMedian 140 140/2 140Mean deviation Σ d /N 300/10 30 Standard deviation 26600/10 1 26600/9 2955.5 54.36Mode is 140.Example 2.40 Following is the information on the distribution obtained forthe character plant height (cm) in castor in the control data. Calculate 10.0 f 150 f X 15750Mean 15750/150 105Mode 75 5/5 9 15 75 5/14 15 75 0.35 15 75 5.25 80.25Example 2.41 Following is the information on the distribution obtained forthe character no. of capsules per main raceme in castor in the control data.Calculate C.V.CIXffXdd2fd 49922210112343110437422027 9.5 1.50992.2523.00665.501102.50240.25 f 180 f X 2080 f d2 3565
Frequency Distribution — Measures of Central Values and Dispersion105Mean 2080/180 11.5 S.D. 3565/179 19.91 4.4627C.V. 4.4627/11.5 100 0.3880 100 38.80Example 2.42Calculate mean, variance, S.D., S.E. for the following data.No. of daysXffXdd2fd 912465 10 7 4 588 f 150 f X 21168S.D. Mean 21168/150 141 (Approx.) S.E. S.D/ n 5.5/ 150 0.45 4512/149 30.28No. of 35–4040–4545–50 30.28 5.5Variance Σf d2 /n 1Example 2.43f d2 4512Calculate mean, variance, S.D., 482693100025382.51600455202.582.532.5000 f 148 f X 1780dd2fd 29.5290.254.5220.430.480.235.4830.0310.48 109.8315.48 239.6320.48 419.4325.48 647.2330.48 929.0335.48 1258.83902.51041.911.0780.7988.4718.8419.4000 f d2 4862.7
Frequency Distribution — Measures of Central Values and 071.53Mean I/b 16/10 1.6Example 2.46Find the harmonic mean of 4,8,10,12,14,16, and .055 H.M. N/ I/X 0.7469I/X 7/0.7469 9.37Example 2.47 AIDS is due to Retroviridae. The size in (nm) is given below.Calculate G.M.xflog xf log 67.92CI80–8585–9090–9595–100G.M. Antilog 46.8/24 1.95 f log x 46.8 Antilog 1.95 90.15Example 2.48S.E.Root length (cm) in safflower is given below. Calculate mean,CIXffXdd2fd 90.390.160.040.00010.030.151.340.520.00080.720.91 f 54 f X 330.2 f d2 3.4908
108Fundamentals of BiostatisticsMean 330.2/54 6.11 S.D. 3.49/53 0.06 0.244 S.E. 0.244/ 54 0.244/7.34 0.034Example 2.49following.Calculate the mean temperature and mean absorbance for theTemperature (C)10 20 30405060Absorbance (A 420) 0.2 0.3 0.3 0.35 0.37 0.39Mean Temperature 210/6 35Mean Absorbance 1.91/6 0.32Example 2.50 The no. of E. coli (X 104) observed every 40 min is givenbelow 60, 80, 100,125,145. Calculate G.M.Xlog X60801001251451.7781.9032.0002.0962.161log X 9.938G.M Antilog 9.938 97.185EXERCISE1. Calculate the Mean deviation from the following data.(B.Com., Andhra Univ.; �5050–6060–7070–805812152014126Also calculate (a) Median (b) Mode (c) Variance.
Frequency Distribution — Measures of Central Values and Dispersion1092. Compute Quartile Deviation from the following data.(B.Com., Kakatiya Univ.; 219510963. From the results given below calculate Mean, Standard Deviation, Variance and Coefficient of Variation (C.V.)No. of seeds per podFrequency234567842211841014. The following measurements of weight (in grams) have been recorded fora common strain of rats.(a) Choose appropriate class intervals and group the data into a frequency distribution;(b) Calculate the relative frequency of each class interval;(c) Plot the relative frequency 5. Calculate the Mean, Median, Variance, Standard Deviation and Rangefor each of the following sets.(a) 5, 10, 15, 20, 25(b) 2, 4, 2, 2, 6(c) 4, 6, 8, 10(d) 2, 1, 1, 0, 4, 2, 36. Blood cholesterol levels mg/ml were recorded in a 65175135155125120(a) Group the data into a frequency distribution.(b) Compute mean and standard deviation from the ungrouped data.(c) Compute the mean and standard deviation from the grouped data.
Frequency Distribution — Measures of Central Values and Dispersion111(c) Median(d) Standard Deviation(e) Mean Deviation12. The table shows a distribution of bristle number in Drosophoila,Bristle numberNo. of individuals12345672383050166Calculate Mean, Variance, S.D. of the distribution.13. Following is the information on the distribution obtained for the characterplant height(cm) in castor in the control data.(a) Calculate Mean, Mode(b) Calculate S.D., C.V., Variance, S.E.Height in 6312217FrequencyHeight in cm 120–135 135–150 150–165 165–180 180–195 195–210Frequency1410755414. Following is the information on the distribution obtained for the character number of capsules per main raceme in castor in the control data.Calculate C.V.Number of capsules per 0–4 5–9 10–14 15–19 20–24 25–29main racemeFrequency649622210115. Following is the information on the distribution obtained for the characternumber of days to maturity in castor in the control data.Calculate Mean, Variance, S.D. and S.E.Number of days to 130–132 133–135 136–138 139–141 142–144maturityFrequency615352725
112Fundamentals of BiostatisticsNumber of days to 145–147 148–150 151–153 154–156maturityFrequency2496316. Following is the information on the distribution obtained for the character number of capsules/main raceme in castor for different treatments.Calculate Mean, Variance, S.D. and S.E.(a)Class interval0–5 5–10 10–15 15–20 20–25 25–30 30–35Frequency 0.1% Hz105148Class interval2693135–40 40–45 45–50Frequency 0.1% Hz000(b)Class interval0–5 5–10 10–15 15–20 20–25 25–30 30–35Frequency30 Kr 0.1% Hz765Class interval462136135–40 40–45 45–50 50–55Frequency30 Kr 0.1% Hz000117. Following is the information on the distribution obtained for the characternumber of days to maturity in castor for different treatments. CalculateMean, Median, Mode.(a)CIFrequency30 Kr125–130 130–135 135–140 140–145 145–150 150–1556314147187(b)CIFrequency0.1% Hz125–130 130–135 135–140 140–145 145–150 150–15573246302212
114Fundamentals of Biostatistics24. Effect of temperature on Chitinase activity was tested in TMV2 groundnut by incubating the enzymatic reaction mixture in a water bath rangingfrom 10 –60 and measuring the absorbance.Calculate Mean Absorbance for Mean Temperature.Temperature (0 )102030405060Absorbance (A-420)0.20.30.30.350.370.3925. Distribution of persons as per HbGrams of Hbper 100 CC Grams of Hbper 100 CC �12.01214101Calculate Mean, Standard Deviation, S.E.26. Draw frequency polygonTotal SerumProtein g/100mlNo. of Females6–77–88–99–1010–11241210227. The number of E. coli ( 104 ) observed every 40 minutes is given below:60, 80, 100, 125, 145Calculate Geometric Mean.28. Serum creatine kinase levels in the blood is providedCI0-1010-2020-30Data Frequency6412CI30-4040-5050-60Calculate Standard Deviation and C.V.Data Frequency1424
Frequency Distribution — Measures of Central Values and Dispersion11529. Trigonella leaflet data is givenLength (mm)Breadth .20.91.41.51.3Calculate mean length, mean breadth, data and mean for l/b.30. Consider the following data:Tumour size (cm)No. of patients1–22–33–44–55684Calculate Mean.31. After ingestion of drug, the excretion % of the same in urine samples isgiven after certain length of time.% excreted 256Calculate Mean excretion.32. IQ score are given below:Mid value727680889298102Calculate variance.Frequency46830354010Mid value114116118120122126Frequency14108421
Frequency Distribution — Measures of Central Values and DispersionHeight in ��671518352010Height in 15862Calculate S.D., Variance and Standard error.39. Dysentery is caused by the virus family. The double stranded RNA mass(kbp Kilobase pairs) are given below:18, 20, 22, 24, 26, 28, 30Calculate Harmonic mean.40. AIDS is due to retroviridae. The size (nm) is given below. CalculateGeometric mean 4610441. Out of some of the main virus families infecting humans, the familyadenoviridae causing common cold has the following size measurements(nm) collected among 20 samples. Prepare frequency 0.072.074.078.079.080.084.085.086.088.090.042. Hepatitis B is caused by the family of viruses. The genome as DNAshows following kilobase pairs (kbp) of double stranded DNA among 10samples.1.7, 1.9, 2.0, 2.4, 2.4, 2.1, 2.2, 2.3, 2.6, 2.8Calculate Standard Deviation, Range, Variance, Standard error and C.V.43. Influenza is caused by the family of virus whose frequency distribution ofsize (nm) is given below. Calculate Mean, Median and 110110–115115–1202212321
Frequency Distribution — Measures of Central Values and DispersionSamples no.Under stressUnder relaxed 32.0128.0131.0132.012345611949. Compute Mean and Mode for the given data:Diameter (in mm)0-44-88-1212-1616-20461464No. of bearings50. Compute Standard error for the following data:6, 9, 11, 8, 14, 3, 20, 1251. Six bushes of flowering plant gave the following data with reference tono. of buds:60, 40, 80, 50, 30, 64Calculate Standard deviation ‘S’ by short cut method.][2/n2 (Σx)ΣxiiHint : S 2 n 152. Protein in gms/litre is giving below. Calculate Variance, Coefficient 13.014.616.653. Haemoglobin (g/dl) data is given below:3.8, 2.6, 3.2, 4.8, 3.9, 2.8, 3.6, 6.2Find Mean deviation.54. Albumin (g/dl) data is given Calculate Mean deviation.Frequency56843212.412.611.214.516.8
120Fundamentals of Biostatistics55. Creatinine levels mg/kg body weight are given below:23, 24, 25, 26, 28, 30, 32, 34, 36.Calculate Standard error.56. Phosphatase levels (mg/litre) are 5Draw histogram and frequency polygon.57. Total aflatoxin levels are given below in (mg/l). On days 6, 9 and 12after infection with Aspergillus in groundnut.No. of days6912Amount8.29.410.6Draw Frequency polygon.ReviewI. The temperaturesIn degrees Fahr
Length of leaf (cm) 9.5–14.5 14.5–19.5 19.5–24.5 24.5–29.5 Frequency 3 8 12 7 12 10 8 6 4 2 0 9.5 14.5 19.5 24.5 29.5 Length of leaf Frequency FREQUENCY POLYGONS A frequency distribution may be displayed as a frequency polygon. A frequency polygon may be superimposed on a histogra
1 3 5 7 9 11 13 15 17 (oW ‐ WF) and GH WindFarmer, gross and net (with ‐0.15% ‐0.10% ‐0.05% e nt Deviation Gross Net Modified Park) model F ‐0.25% Perc ‐0.20% Plant Number requency distribution of predicte
Education 5 0:1667 Law 3 0:1000 Literature 4 0:1333 Medicine 7 0:2333 Science 5 0:1667 Technology 6 0:2000 (b) Relativ e F requency Ed Law Lit Med Sci Tech 0 0.05 0.1 0.15 0.2 0.25 Book type PSfrag replacements Education Law Literature Medicine Science Technology (c) Perhaps medicine. H
The Poor Man’s Electrically Tuned 6-band HF Beam Caesium Clock By Dave McQue, G4NJU* F REQUENCY COUNTERS with lots of digits are cheap today. However, you . in 108 - way beyond the achievable setting accuracy of most hand-held frequency counters. SETTING UP TO SET
Dec 01, 2013 · Job Description : new product design of medium voltage (MV) variable f requency drives for general industrial applications, power rating from 1 -6 MVA, voltage rating 6.6 kV with low cost, high perf
DraganFly Innovations Inc or Vibration Isolators & Mounts by Lord Corporation). Various tests show excellent compensation results within certain amplitude and f requency ranges, but stabilizing vibrations over the entire amplitude and frequency spectr
Intro high resolution broad bandwidth or low temporal coherence Interferometry plane wave or high spatial coherence Basics Super Luminescend Diode SLD or SLED Center wavelength 840 nm – 1310 nm Bandwidth 30 – 100 nm e s Fs Laser (Ti:S) Center wavelength 820 nm Sourc Bandwidth 120 – 2
18 proved with the use of mathematical programming. The automated model of 19 agricultural manufacture development and easures to increase the level of its 20 ecologization were provided. Pr
Use the English phonemic alphabet page, which you find at the beginning of good dictionaries, as a guide to pronouncing new words. Effective English Learning ELTC self-study materials Tony Lynch and Kenneth Anderson, English Language Teaching Centre, University of Edinburgh 2012 9 3. Don't forget to learn the word stress of a new word. Every English word has its own normal stress pattern. For .
