Category: statistics

Statistics – Individual Series Arithmetic Mode ”; Previous Next When data is given on individual basis. Following is an example of individual series − Items 5 10 20 30 40 50 60 70 In case of individual items, the number of times each value occurs is counted and the value which is repeated maximum number of times is the modal value. Example Problem Statement − Calculate Arithmetic Mode for the following individual data − Items 14 36 45 36 105 36 Solution − The Arithmetic Mode of the given numbers is 36 as it is repeated maximum number of times,3. Calculator Print Page Previous Next Advertisements ”;

Statistics – Negative Binomial Distribution ”; Previous Next Negative binomial distribution is a probability distribution of number of occurences of successes and failures in a sequence of independent trails before a specific number of success occurs. Following are the key points to be noted about a negative binomial experiment. The experiment should be of x repeated trials. Each trail have two possible outcome, one for success, another for failure. Probability of success is same on every trial. Output of one trial is independent of output of another trail. Experiment should be carried out until r successes are observed, where r is mentioned beforehand. Negative binomial distribution probability can be computed using following: Formula ${ f(x; r, P) = ^{x-1}C_{r-1} times P^r times (1-P)^{x-r} }$ Where − ${x}$ = Total number of trials. ${r}$ = Number of occurences of success. ${P}$ = Probability of success on each occurence. ${1-P}$ = Probability of failure on each occurence. ${f(x; r, P)}$ = Negative binomial probability, the probability that an x-trial negative binomial experiment results in the rth success on the xth trial, when the probability of success on each trial is P. ${^{n}C_{r}}$ = Combination of n items taken r at a time. Example Robert is a football player. His success rate of goal hitting is 70%. What is the probability that Robert hits his third goal on his fifth attempt? Solution: Here probability of success, P is 0.70. Number of trials, x is 5 and number of successes, r is 3. Using negative binomial distribution formula, let”s compute the probability of hitting third goal in fifth attempt. ${ f(x; r, P) = ^{x-1}C_{r-1} times P^r times (1-P)^{x-r} \[7pt] implies f(5; 3, 0.7) = ^4C_2 times 0.7^3 times 0.3^2 \[7pt] , = 6 times 0.343 times 0.09 \[7pt] , = 0.18522 }$ Thus probability of hitting third goal in fifth attempt is $ { 0.18522 }$. Print Page Previous Next Advertisements ”;

Statistics – Histograms ”; Previous Next A histogram is a graphical representation of the distribution of numerical data. It is an estimate of the probability distribution of a continuous variable (quantitative variable). Problem Statement: Every month one measure the amount of weight one”s dog has picked up and get these outcomes: 0.5 0.5 0.3 -0.2 1.6 0 0.1 0.1 0.6 0.4 Draw the histogram demonstrating how much is that dog developing. Solution: monthly development vary from -0.2 (the fox lost weight that month) to 1.6. Putting them in order from lowest to highest weight gain. -0.2 0 0.1 0.1 0.3 0.4 0.5 0.5 0.6 1.6 We decide to put the results into groups of 0.5: The -0.5 to just below 0 range. The 0 to just below 0.5 range, etc. And here is the result: There are no values from 1 to just below 1.5, but we still show the space. Print Page Previous Next Advertisements ”;

Statistics – Individual Series Arithmetic Median ”; Previous Next When data is given on individual basis. Following is an example of individual series − Items 5 10 20 30 40 50 60 70 In case of a group having even number of distribution, Arithmetic Median is found out by taking out the Arithmetic Mean of two middle values after arranging the numbers in ascending order. Formula Median = Value of ($frac{N+1}{2})^{th} item$. Where − ${N}$ = Number of observations Example Problem Statement − Let”s calculate Arithmetic Median for the following individual data − Items 14 36 45 70 105 145 Solution − Based on the above mentioned formula, Arithmetic Median M will be − $M = Value of (frac{N+1}{2})^{th} item. \[7pt] , = Value of (frac{6+1}{2})^{th} item. \[7pt] , = Value of 3.5^{th} item. \[7pt] , = Value of (frac{3^{rd} item + 4^{th} item}{2})\[7pt] , = (frac{45 + 70}{2}) , = {57.5}$ The Arithmetic Median of the given numbers is 57.5. In case of a group having odd number of distribution, Arithmetic Median is the middle number after arranging the numbers in ascending order. Example Let”s calculate Arithmetic Median for the following individual data − Items 14 36 45 70 105 Given numbers are 5, an odd number thus middle number is the Arithmetic Median. ∴ The Arithmetic Median of the given numbers is 45. Calculator Print Page Previous Next Advertisements ”;

Statistics – Inverse Gamma Distribution ”; Previous Next Inverse Gamma Distribution is a reciprocal of gamma probability density function with positive shape parameters $ {alpha, beta } $ and location parameter $ { mu } $. $ {alpha } $ controls the height. Higher the $ {alpha } $, taller is the probability density function (PDF). $ {beta } $ controls the speed. It is defined by following formula. Formula ${ f(x) = frac{x^{-(alpha+1)}e^{frac{-1}{beta x}}}{ Gamma(alpha) beta^alpha} \[7pt] , where x gt 0 }$ Where − ${alpha}$ = positive shape parameter. ${beta}$ = positive shape parameter. ${x}$ = random variable. Following diagram shows the probability density function with different parameter combinations. Print Page Previous Next Advertisements ”;

Statistics – Relative Standard Deviation ”; Previous Next In probability theory and statistics, the coefficient of variation (CV), also known as relative standard deviation (RSD), is a standardized measure of dispersion of a probability distribution or frequency distribution. Relative Standard Deviation, RSD is defined and given by the following probability function: Formula ${100 times frac{s}{bar x}}$ Where − ${s}$ = the sample standard deviation ${bar x}$ = sample mean Example Problem Statement: Find the RSD for the following set of numbers: 49, 51.3, 52.7, 55.8 and the standard deviation are 2.8437065. Solution: Step 1 – Standard deviation of sample: 2.8437065 (or 2.84 rounded to 2 decimal places). Step 2 – Multiply Step 1 by 100. Set this number aside for a moment. ${2.84 times 100 = 284}$ Step 3 – Find the sample mean, ${bar x}$. The sample mean is: ${frac{(49 + 51.3 + 52.7 + 55.8)}{4} = frac{208.8}{4} = 52.2.}$ Step 4Divide Step 2 by the absolute value of Step 3. ${frac{284}{|52.2|} = 5.44.}$ The RSD is: ${52.2 pm 5.4}$% Note that the RSD is expressed as a percentage. Print Page Previous Next Advertisements ”;

Statistics – Scatterplots ”; Previous Next A scatterplot is a graphical way to display the relationship between two quantitative sample variables. It consists of an X axis, a Y axis and a series of dots where each dot represents one observation from a data set. The position of the dot refers to its X and Y values. Patterns of Data in Scatterplots Scatterplots are used to analyze patterns which generally varies on the basis of linearity, slope, and strength. Linearity – data pattern is either linear/straight or nonlinear/curved. Slope – direction of change in variable Y with respect to increase in value of variable X. If Y increases with increase in X, slope is positive otherwise slope is negative. Strength – Degree of spreadness of scatter in the plot. If dots are widely dispersed, the relationship is consider weak. If dot are densed around a line then the relationship is said to be strong. Print Page Previous Next Advertisements ”;

Statistics – Process Sigma ”; Previous Next Process sigma can be defined using following four steps: Measure opportunities, Measure defects, Calculate yield, Look-up process sigma. Formulae Used ${DPMO = frac{Total defect}{Total Opportunities} times 1000000}$ ${Defect (%) = frac{Total defect}{Total Opportunities} times 100}$ ${Yield (%) = 100 – Defect (%) }$ ${Process Sigma = 0.8406+sqrt{29.37}-2.221 times (log (DPMO)) }$ Where − ${Opportunities}$ = Lowest defect noticeable by customer. ${DPMO}$ = Defects per Million Opportunities. Example Problem Statement: In equipment organization hard plate produced is 10000 and the defects is 5. Discover the process sigma. Solution: Given: Opportunities = 10000 and Defects = 5. Substitute the given qualities in the recipe, Step 1: Compute DPMO $ {DPMO = frac{Total defect}{Total Opportunities} times 1000000 \[7pt] , = (10000/5) times 1000000 , \[7pt] , = 500}$ Step 2: Compute Defect(%) $ {Defect (%) = frac{Total defect}{Total Opportunities} times 100 \[7pt] , = frac{10000}{5} times 100 , \[7pt] , = 0.05}$ Step 3: Compute Yield(%) $ {Yield (%) = 100 – Defect (%) \[7pt] , = 100 – 0.05 , \[7pt] , = 99.95}$ Step 3: Compute Process Sigma $ {Process Sigma = 0.8406+sqrt{29.37}-2.221 times (log (DPMO)) \[7pt] , = 0.8406 + sqrt {29.37} – 2.221 times (log (DPMO)) , \[7pt] , = 0.8406+sqrt(29.37) – 2.221*(log (500)) , \[7pt] , = 4.79 }$ Print Page Previous Next Advertisements ”;

Statistics – Harmonic Resonance Frequency ”; Previous Next Harmonic Resonance Frequency represents a signal or wave whose frequency is an integral multiple of the frequency of a reference signal or wave. Formula ${ f = frac{1}{2 pi sqrt{LC}} } $ Where − ${f}$ = Harmonic resonance frequency. ${L}$ = inductance of the load. ${C}$ = capacitanc of the load. Example Calculate the harmonic resonance frequency of a power system with the capcitance 5F, Inductance 6H and frequency 200Hz. Solution: Here capacitance, C is 5F. Inductance, L is 6H. Frequency, f is 200Hz. Using harmonic resonance frequency formula, let”s compute the resonance frequency as: ${ f = frac{1}{2 pi sqrt{LC}} \[7pt] implies f = frac{1}{2 pi sqrt{6 times 5}} \[7pt] , = frac{1}{2 times 3.14 times sqrt{30}} \[7pt] , = frac{1}{ 6.28 times 5.4772 } \[7pt] , = frac{1}{ 34.3968 } \[7pt] , = 0.0291 }$ Thus harmonic resonance frequency is $ { 0.0291 }$. Print Page Previous Next Advertisements ”;

Statistics – Grand Mean ”; Previous Next When sample sizes are equal, in other words, there could be five values in each sample, or n values in each sample. The grand mean is the same as the mean of sample means. Formula ${X_{GM} = frac{sum x}{N}}$ Where − ${N}$ = Total number of sets. ${sum x}$ = sum of the mean of all sets. Example Problem Statement: Determine the mean of each group or set”s samples. Use the following data as a sample to determine the mean and grand mean. Jackson 1 6 7 10 4 Thomas 5 2 8 14 6 Garrard 8 2 9 12 7 Solution: Step 1: Compute all means $ {M_1 = frac{1+6+7+10+4}{5} = frac{28}{5} = 5.6 \[7pt] , M_2 = frac{5+2+8+14+6}{5} = frac{35}{5} = 7 \[7pt] , M_3 = frac{8+2+9+12+7}{5} = frac{38}{5} = 7.6 }$ Step 2: Divide the total by the number of groups to determine the grand mean. In the sample, there are three groups. $ {X_{GM} = frac{5.6+7+7.6}{3} = frac{20.2}{3} \[7pt] , = 6.73 }$ Print Page Previous Next Advertisements ”;