Statistics Essays Histogram

Statistics Essays HistogramA histogram is often used for representing entropy from a continuous variable which argon summarised as a assort frequency distri furtherion.We use Excel to generate a Box to represents both the original and the corrected sets of data. The issuing is the following diagramThe different methods of diagrammatic representation of statistical data argon bar chat, histogram, go and leaf, and lineplots. The bar chart is more appropriate to data from a discrete distri providedion that atomic number 18 summarised apply a frequency distri bution. A histogram is often used for representing data from a continuous variable which are summarised as a grouped frequency distribution. A histogram is therefore similar to a bar chat, but is used to present continuous data. Steam and leaf gives a visual representation similar to the histogram but has the advantage that it does not lose the detail of the individual data point in the grouping. All these diagrams assist t o examine the general shape of the distribution of data and help in making conjecture closely values of quantities such as the median, the mean or the interquartile range. The last one, the lineplot, is often appropriate for sm entirelyer data sets, and tramp be useful for example to check whether toe data sets sustain a parking area variance.We denote by and the mean of the original set and the corrected set respectively. Then we aim i.e. . i.e. .Since we have an even number of observation, the median in this case will be the midpoint of the two philia observations. ThatsFor the original set the median is For the corrected set the median is .The standard deviation of each(prenominal) data set is given by , where , are the different values in each data set. thuslyFor the original set, , and for the corrected set .The lower quartile is defined to be the th observation enumeration from below, and the upper quartile is the resembling but counting from above. The interquarti le is only if the difference between the upper and the lower quartile. We have the results in the following table.Original setCorrected setLower quartile3.8153.7475Upper quartile3.39253.3925Interquartile0.42250.355Question 2Theoretically, the fact that 9 and 12 can be made up in as galore(postnominal) ways as 10 and eleven 11 means that both sets of numbers should have the same prospect to appear. The first thing that should be noted here is the fact that this is true if and only if when we virgule a dice, all the numbers have the same probability of appearance, which if not always the case in practice when if when we need to allow consideration such as the on uniformity of the surface on which the dice is thrown, the angle and the velocity at which the is thrown, and even any deformation on the dice which all have an effect on the number that we will get. This problem thus highlights the impossibility of the probability to be an absolutely precise science as oppose to the other branches of mathematics.Question 3The probability that a film processed on machine X is . Also, the note of a film is independent of the quality of all the films processed before it. Thus the probability that three films randomly chosen from a portion coming from machine X is simply .Lets denote bythe position the pile came from machine X, the event the three film are all of nigh quality. Clearly, what we are asking for is the probability that and occur at the same time, which is the probability that the three films are all of good quality and the batch came from machine X. Using the theory of conditional probabilities, we have.Since all of all films are processed on machine X, consequently . is simply the probability the probability that we calculated above. Thus . Hence.Question 4At each question only two things can happen 1-the scholarly person can answer the correctly, and we denote by the probability that this does happen 2-or the student can choose the aggrieve outcom es among the v possible, and we denote by the probability that this does happen.Obviously we must have . Given that only five outcomes are available at each question, only one of which being correct, we have , and .The sample that consists in answering a single question can therefore be viewed as a Bernoulli experimentation with parameter . Hence, Taking all the multiple-choice examination can be viewed as Binomial experiment with parameter , where . Lets be the random variable representing the number of correct answer achieved by the student. Clearly, the distribution of Binomial with parameter . The probability that the student passes the test is the , which is equivalent to . But,where for each , .Hence,.This gives us , and thus the probability that the student passes the test is .Question 5Bayes Formula Let E, F be subsets of some sample space S, and let Fc be the complement of F in S. We can express E as because in order for a point to be in E it must be either in E and F or in E but not in F. As EF and EFc are mutually exclusive we can write Applying this to the conditional probability equation gives . Consider the following problemWe have three boxes labelled U1, U2 and U3. Each of them contains a rumple of white and red balls. The proportion of white balls is each of them is as follows 30% for U1, 60% for U2, 40% for U3. We delineate one ball from U1 if it is a white ball then we get out a ball in U2, otherwise we magnet a ball in U3. We would like to find the probability that the first draw gives a red ball knowing that the second draw has given a given a white ball.We denote by the event the second draw is made in the box Ui, the event the second draw gives a white ball.Clearly, if the first draw gives a red ball, then the second can be made only in U3. Thus the probability that the first draw gives a red ball knowing that the second draw has given a given a white one is exactly the same as the probability that the second ball comes from U3 knowing that it is a white ball, which is nothing else than . Using the Bayes formula, we have. (1)It can be easily seen that and are mutually exclusive as a the second draw can not happen in both U2 and U3 simultaneously. Also since the second draw can happen only either in U2 or U3, then gives all the possibility on where the second draw can happen. That is why .The top of the fraction (1) is simply application of the conditional probability.Hence