Category: statistics

Statistics – Permutation ”; Previous Next A permutation is an arrangement of all or part of a set of objects, with regard to the order of the arrangement. For example, suppose we have a set of three letters: A, B, and C. we might ask how many ways we can arrange 2 letters from that set. Permutation is defined and given by the following function: Formula ${^nP_r = frac{n!}{(n-r)!} }$ Where − ${n}$ = of the set from which elements are permuted. ${r}$ = size of each permutation. ${n,r}$ are non negative integers. Example Problem Statement: A computer scientist is trying to discover the keyword for a financial account. If the keyword consists only of 10 lower case characters (e.g., 10 characters from among the set: a, b, c… w, x, y, z) and no character can be repeated, how many different unique arrangements of characters exist? Solution: Step 1: Determine whether the question pertains to permutations or combinations. Since changing the order of the potential keywords (e.g., ajk vs. kja) would create a new possibility, this is a permutations problem. Step 2: Determine n and r n = 26 since the computer scientist is choosing from 26 possibilities (e.g., a, b, c… x, y, z). r = 10 since the computer scientist is choosing 10 characters. Step 2: Apply the formula ${^{26}P_{10} = frac{26!}{(26-10)!} \[7pt] = frac{26!}{16!} \[7pt] = frac{26(25)(24)…(11)(10)(9)…(1)}{(16)(15)…(1)} \[7pt] = 26(25)(24)…(17) \[7pt] = 19275223968000 }$ Print Page Previous Next Advertisements ”;

Statistics – Combination ”; Previous Next A combination is a selection of all or part of a set of objects, without regard to the order in which objects are selected. For example, suppose we have a set of three letters: A, B, and C. we might ask how many ways we can select 2 letters from that set. Combination is defined and given by the following function − Formula ${C(n,r) = frac{n!}{r!(n-r)!}}$ Where − ${n}$ = the number of objects to choose from. ${r}$ = the number of objects selected. Example Problem Statement − How many different groups of 10 students can a teacher select from her classroom of 15 students? Solution − Step 1 − Determine whether the question pertains to permutations or combinations. Since changing the order of the selected students would not create a new group, this is a combinations problem. Step 2 − Determine n and r n = 15 since the teacher is choosing from 15 students. r = 10 since the teacher is selecting 10 students. Step 3 − Apply the formula ${^{15}C_{10} = frac{15!}{(15-10)!10!} \[7pt] = frac{15!}{5!10!} \[7pt] = frac{15(14)(13)(12)(11)(10!)}{5!10!} \[7pt] = frac{15(14)(13)(12)(11)}{5!} \[7pt] = frac{15(14)(13)(12)(11)}{5(4)(3)(2)(1)} \[7pt] = frac{(14)(13)(3)(11)}{(2)(1)} \[7pt] = (7)(13)(3)(11) \[7pt] = 3003}$ Calculator Print Page Previous Next Advertisements ”;

Discrete Series Arithmetic Mean ”; Previous Next When data is given along with their frequencies. Following is an example of discrete series − Items 5 10 20 30 40 50 60 70 Frequency 2 5 1 3 12 0 5 7 For discrete series, the Arithmetic Mean can be calculated using the following formula. Formula $bar{x} = frac{f_1x_1 + f_2x_2 + f_3x_3……..+ f_nx_n}{N}$ Alternatively, we can write same formula as follows − $bar{x} = frac{sum fx}{sum f}$ Where − ${N}$ = Number of observations ${f_1,f_2,f_3,…,f_n}$ = Different values of frequency f. ${x_1,x_2,x_3,…,x_n}$ = Different values of variable x. Example Problem Statement − Calculate Arithmetic Mean for the following discrete data − Items 14 36 45 70 Frequency 2 5 1 3 Solution − Based on the given data, we have − Items Frequencyf ${fx}$ 14 2 28 36 5 180 45 1 45 70 3 210   ${N=11}$ ${sum fx=463}$ Based on the above mentioned formula, Arithmetic Mean $bar{x}$ will be − $bar{x} = frac{463}{11} \[7pt] , = {42.09}$ The Arithmetic Mean of the given numbers is 42.09. Calculator Print Page Previous Next Advertisements ”;

Statistics – Interval Estimation ”; Previous Next Interval estimation is the use of sample data to calculate an interval of possible (or probable) values of an unknown population parameter, in contrast to point estimation, which is a single number. Formula ${mu = bar x pm Z_{frac{alpha}{2}}frac{sigma}{sqrt n}}$ Where − ${bar x}$ = mean ${Z_{frac{alpha}{2}}}$ = the confidence coefficient ${alpha}$ = confidence level ${sigma}$ = standard deviation ${n}$ = sample size Example Problem Statement: Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the interval estimation for the population mean at a 95% confidence level? Solution: The student calculated the sample mean of the boiling temperatures to be 101.82, with standard deviation ${sigma = 0.49}$. The critical value for a 95% confidence interval is 1.96, where ${frac{1-0.95}{2} = 0.025}$. A 95% confidence interval for the unknown mean. ${ = ((101.82 – (1.96 times 0.49)), (101.82 + (1.96 times 0.49))) \[7pt] = (101.82 – 0.96, 101.82 + 0.96) \[7pt] = (100.86, 102.78) }$ As the level of confidence decreases, the size of the corresponding interval will decrease. Suppose the student was interested in a 90% confidence interval for the boiling temperature. In this case, ${sigma = 0.90}$, and ${frac{1-0.90}{2} = 0.05}$. The critical value for this level is equal to 1.645, so the 90% confidence interval is ${ = ((101.82 – (1.645 times 0.49)), (101.82 + (1.645 times 0.49))) \[7pt] = (101.82 – 0.81, 101.82 + 0.81) \[7pt] = (101.01, 102.63)}$ An increase in sample size will decrease the length of the confidence interval without reducing the level of confidence. This is because the standard deviation decreases as n increases. Margin of Error The margin of error ${m}$ of interval estimation is defined to be the value added or subtracted from the sample mean which determines the length of the interval: ${Z_{frac{alpha}{2}}frac{sigma}{sqrt n}}$ Suppose in the example above, the student wishes to have a margin of error equal to 0.5 with 95% confidence. Substituting the appropriate values into the expression for ${m}$ and solving for n gives the calculation. ${ n = {(1.96 times frac{1.2}{0.5})}^2 \[7pt] = {frac{2.35}{0.5}^2} \[7pt] = {(4.7)}^2 = 22.09 }$ To achieve 95% interval estimation for the mean boiling point with total length less than 1 degree, the student will have to take 23 measurements. Print Page Previous Next Advertisements ”;

Statistics – Outlier Function ”; Previous Next An outlier in a probability distribution function is a number that is more than 1.5 times the length of the data set away from either the lower or upper quartiles. Specifically, if a number is less than ${Q_1 – 1.5 times IQR}$ or greater than ${Q_3 + 1.5 times IQR}$, then it is an outlier. Outlier is defined and given by the following probability function: Formula ${Outlier datas are, lt Q_1 – 1.5 times IQR (or) gt Q_3 + 1.5 times IQR }$ Where − ${Q_1}$ = First Quartile ${Q_2}$ = Third Quartile ${IQR}$ = Inter Quartile Range Example Problem Statement: Consider a data set that represents the 8 different students periodic task count. The task count information set is, 11, 13, 15, 3, 16, 25, 12 and 14. Discover the outlier data from the students periodic task counts. Solution: Given data set is: 11 13 15 3 16 25 12 14 Arrange it in ascending order: 3 11 12 13 14 15 16 25 First Quartile Value() ${Q_1}$ ${ Q_1 = frac{(11 + 12)}{2} \[7pt] = 11.5 }$ Third Quartile Value() ${Q_3}$ ${ Q_3 = frac{(15 + 16)}{2} \[7pt] = 15.5 }$ Lower Outlier Range (L) ${ Q_1 – 1.5 times IQR \[7pt] = 11.5 – (1.5 times 4) \[7pt] = 11.5 – 6 \[7pt] = 5.5 }$ Upper Outlier Range (L) ${ Q_3 + 1.5 times IQR \[7pt] = 15.5 + (1.5 times 4) \[7pt] = 15.5 + 6 \[7pt] = 21.5 }$ In the given information, 5.5 and 21.5 is more greater than the other values in the given data set i.e. except from 3 and 25 since 3 is greater than 5.5 and 25 is lesser than 21.5. In this way, we utilize 3 and 25 as the outlier values. Print Page Previous Next Advertisements ”;

Statistics – Continuous Series Arithmetic Median ”; Previous Next When data is given based on ranges along with their frequencies. Following is an example of continous series − Items 0-5 5-10 10-20 20-30 30-40 Frequency 2 5 1 3 12 Formula $Median = {L} + frac{(frac{n}{2} – c.f.)}{f} times {i}$ Where − ${L}$ = Lower limit of median class, median class is that class where $frac{n}{2}^{th}$ item is lying. ${c.f.}$ = Cumulative frequency of the class preceding the median class. ${f}$ = Frequency of median class. ${i}$ = Class interval of median class. Arithmetic Median is a useful measure of central tendency in case the data type is nominal data. Since it is a positional average, it does not get affected by extreme values. Example Problem Statement − In a study conducted in an organization, the distribution of income across the workers is observed. Find the the median wage of the workers of the organization. 06 men get less than Rs. 500 13 men get less than Rs. 1000 22 men get less than Rs. 1500 30 men get less than Rs. 2000 34 men get less than Rs. 2500 40 men get less than Rs. 3000 Solution − Given are the cumulative frequencies of the workers. Hence we first find the simple frequency and present the data in tabular form. Income (rs.) M.P. m Frequency f (m-1250)/500 d fd c.f 0 – 500 250 6 -2 -12 6 500 – 1000 750 7 -1 -7 13 1000 – 1500 1250 9 0 0 22 1500 – 2000 1750 8 1 8 30 2000 – 2500 2250 4 2 8 34 2500 – 3000 2750 6 3 18 40     N = 40   ∑ fd = 15   In order to simplify the calculation, a common factor i = 500 has been taken. Using the following formula for calculating median wage. $Median = {L} + frac{(frac{n}{2} – c.f.)}{f} times {i}$ Where − ${L}$ = 1000 $frac{n}{2}$ = 20 ${c.f.}$ = 13 ${f}$ = 9 ${i}$ = 500 Thus $Median = {1000} + frac{(20 – 13)}{9} times {500} \[7pt] , = {1000 + 388.9} \[7pt] , = {1388.9}$ As 1388.9 ≃ 1389. The median wage is Rs. 1389. Calculator Print Page Previous Next Advertisements ”;

Statistics – Hypergeometric Distribution ”; Previous Next A hypergeometric random variable is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a hypergeometric distribution. Hypergeometric distribution is defined and given by the following probability function: Formula ${h(x;N,n,K) = frac{[C(k,x)][C(N-k,n-x)]}{C(N,n)}}$ Where − ${N}$ = items in the population ${k}$ = successes in the population. ${n}$ = items in the random sample drawn from that population. ${x}$ = successes in the random sample. Example Problem Statement: Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)? Solution: This is a hypergeometric experiment in which we know the following: N = 52; since there are 52 cards in a deck. k = 26; since there are 26 red cards in a deck. n = 5; since we randomly select 5 cards from the deck. x = 2; since 2 of the cards we select are red. We plug these values into the hypergeometric formula as follows: ${h(x;N,n,k) = frac{[C(k,x)][C(N-k,n-x)]}{C(N,n)} \[7pt] h(2; 52, 5, 26) = frac{[C(26,2)][C(52-26,5-2)]}{C(52,5)} \[7pt] = frac{[325][2600]}{2598960} \[7pt] = 0.32513 }$ Thus, the probability of randomly selecting 2 red cards is 0.32513. Print Page Previous Next Advertisements ”;

Statistics – Central limit theorem ”; Previous Next If the population from which the sample has a been drawn is a normal population then the sample means would be equal to population mean and the sampling distribution would be normal. When the more population is skewed, as is the case illustrated in Figure, then the sampling distribution would tend to move closer to the normal distribution, provided the sample is large (i.e. greater then 30). According to Central Limit Theorem, for sufficiently large samples with size greater than 30, the shape of the sampling distribution will become more and more like a normal distribution, irrespective of the shape of the parent population. This theorem explains the relationship between the population distribution and sampling distribution. It highlights the fact that if there are large enough set of samples then the sampling distribution of mean approaches normal distribution. The importance of central limit theorem has been summed up by Richard. I. Levin in the following words: The significance of the central limit theorem lies in the fact that it permits us to use sample statistics to make inferences about population parameters without knowing anything about the shape of the frequency distribution of that population other than what we can get from the sample. Print Page Previous Next Advertisements ”;

Statistics – Deciles Statistics ”; Previous Next A system of dividing the given random distribution of the data or values in a series into ten groups of similar frequency is known as deciles. Formula ${D_i = l + frac{h}{f}(frac{iN}{10} – c); i = 1,2,3…,9}$ Where − ${l}$ = lower boundry of deciles group. ${h}$ = width of deciles group. ${f}$ = frequency of deciles group. ${N}$ = total number of observations. ${c}$ = comulative frequency preceding deciles group. Example Problem Statement: Calculate the deciles of the distribution for the following table:   fi Fi [50-60] 8 8 [60-60] 10 18 [70-60] 16 34 [80-60] 14 48 [90-60] 10 58 [100-60] 5 63 [110-60] 2 65   65   Solution: Calculation of First Decile $ {frac{65 times 1}{10} = 6.5 \[7pt] , D_1= 50 + frac{6.5 – 0}{8} times 10 , \[7pt] , = 58.12}$ Calculation of Second Decile $ {frac{65 times 2}{10} = 13 \[7pt] , D_2= 60 + frac{13 – 8}{10} times 10 , \[7pt] , = 65}$ Calculation of Third Decile $ {frac{65 times 3}{10} = 19.5 \[7pt] , D_3= 70 + frac{19.5 – 18}{16} times 10 , \[7pt] , = 70.94}$ Calculation of Fourth Decile $ {frac{65 times 4}{10} = 26 \[7pt] , D_4= 70 + frac{26 – 18}{16} times 10 , \[7pt] , = 75}$ Calculation of Fifth Decile $ {frac{65 times 5}{10} = 32.5 \[7pt] , D_5= 70 + frac{32.5 – 18}{16} times 10 , \[7pt] , = 79.06}$ Calculation of Sixth Decile $ {frac{65 times 6}{10} = 39 \[7pt] , D_6= 70 + frac{39 – 34}{14} times 10 , \[7pt] , = 83.57}$ Calculation of Seventh Decile $ {frac{65 times 7}{10} = 45.5 \[7pt] , D_7= 80 + frac{45.5 – 34}{14} times 10 , \[7pt] , = 88.21}$ Calculation of Eighth Decile $ {frac{65 times 8}{10} = 52 \[7pt] , D_8= 90 + frac{52 – 48}{10} times 10 , \[7pt] , = 94}$ Calculation of Nineth Decile $ {frac{65 times 9}{10} = 58.5 \[7pt] , D_9= 100 + frac{58.5 – 58}{5} times 10 , \[7pt] , = 101}$ Print Page Previous Next Advertisements ”;

Statistics – Harmonic Number ”; Previous Next Harmonic Number is the sum of the reciprocals of the first n natural numbers. It represents the phenomenon when the inductive reactance and the capacitive reactance of the power system becomes equal. Formula ${ H = frac{W_r}{W} \[7pt] , where W_r = sqrt{ frac{1}{LC}} } \[7pt] , and W = 2 pi f $ Where − ${f}$ = Harmonic resonance frequency. ${L}$ = inductance of the load. ${C}$ = capacitanc of the load. Example Calculate the harmonic number 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 number formula, let”s compute the number as: ${ H = frac{sqrt{ frac{1}{LC}}}{2 pi f} \[7pt] implies H = frac{sqrt{ frac{1}{6 times 5}} }{2 times 3.14 times 200} \[7pt] , = frac{0.18257}{1256} \[7pt] , = 0.0001 }$ Thus harmonic number is $ { 0.0001 }$. Print Page Previous Next Advertisements ”;