DEPARTMENT OF STUDIES \& RESEARCH IN MANAGEMENT KARNATAKA STATE OPEN UNIVERSITY Mukthagangotri, Mysuru - 570006 MBA 1st SEMESTER INTERNAL ASSIGNMENT Academic Year 2023-24 January Cycle MBHC-1.4: Statistics and Optimization Techniques (Answer any two) \[ 10 \times 2=20 \] 1. Prove that median lies between mean and mode from following data. \begin{tabular}{|c|c|c|c|c|c|c|} \hline Heights in cms & \( 120-125 \) & \( 125-130 \) & \( 130-135 \) & \( 135-140 \) & \( 140-145 \) & \( 145-150 \) \\ \hline No. of Children & 7 & 10 & 18 & 25 & 13 & 7 \\ \hline \end{tabular} 2. The following data represents the marks in Algebra (x) and Geometry (y) of a group of 10 students. Find both the regression equations and hence estimate \( \mathbf{y} \) if \( \mathbf{x}=\mathbf{7 8} \) and \( \mathbf{x} \) if \( \mathbf{y}=\mathbf{9 4} \). \begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|} \hline \( \mathbf{x} \) & 75 & 80 & 93 & 65 & 87 & 71 & 98 & 68 & 89 & 77 \\ \hline \( \mathbf{y} \) & 82 & 78 & 86 & 72 & 91 & 80 & 95 & 72 & 89 & 74 \\ \hline \end{tabular} 3. Mean and Standard deviation of chest measurements of 1200 soldiers are \( 85 \mathrm{cms} \) and \( 5 \mathrm{cms} \) respectively. How many of them are expected to have chest measurements exceeding \( 95 \mathrm{cms} \), assuming the measurements follow the normal distribution. How many soldiers have their chest measurements between \( 80 \mathrm{cms} \) and \( 90 \mathrm{cms} \). 4. Find the optimum schedule for the given projects. The overhead cost is ?75 per day. \begin{tabular}{|c|c|c|c|c|} \hline \multirow[b]{2}{*}{ Activity } & \multirow[b]{2}{*}{ Predecessor } & \multicolumn{2}{|c|}{ Duration (days) } & \multirow{2}{*}{\begin{tabular}{c} Increase in cost for \\ crashing by one day \\ ? \end{tabular}} \\ \hline & & \begin{tabular}{c} Normal \\ Time \end{tabular} & \begin{tabular}{l} Crash \\ Time \end{tabular} & \\ \hline \( \mathrm{A} \) & - & 3 & 2 & 150 \\ \hline B & - & 4 & 3 & 100 \\ \hline \( \mathrm{C} \) & \( \mathrm{A} \) & 5 & 4 & 200 \\ \hline \( \mathrm{D} \) & \( \mathrm{A} \) & 7 & 5 & 300 \\ \hline \( \mathrm{E} \) & \( \mathrm{B}, \mathrm{C} \) & 3 & 3 & 0 \\ \hline \( \mathrm{F} \) & \( \mathrm{B}, \mathrm{C}, \mathrm{D} \) & 6 & 2 & 75 \\ \hline \end{tabular} a) Draw the Project using normal duration. b) Find the path and projects duration for above. c) Find the optimal schedule and project duration. \( * * * * * * \) Comment Highlight Draw Text Fill \& Sign More tools
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Need Solutions for 31-39 in Statistics Exam
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Deviation, and standard whose mean and = mean rc population standard. The mean and the random samples of size 200 and 18 are taken from an infinite population. The population has mean 200 and standard deviation of 18. The probability that the sample mean will be between 297 to 303 for a sample of 144 observations taken from an infinite population with a mean of 300 and standard deviation of 18 is 0.9544 0.4332 0.8664 0.668 Doubling the size of the sample will reduce the standard error of the mean to one-half its current value have no effect on the standard error of the mean double the standard error of the mean reduce the standard error of the mean to approximately 70% of its current value The finite population correction factor is needed in computing the standard deviation of the sampling distribution of sample means whenever the population is infinite sample size is less than 5% of the population size population has normal distribution sample size is more than 5% of the population size In a local university, 40% of the students live in the dormitories. A random sample of 80 students is selected for a particular study. (assuming infinite population) The standard deviation known as the standard error of the proportion is approximately 0.5477 0.0547 5.477 54.77 Refer to the information in Q19. The probability that the sample proportion (the proportion living in the dormitories) is between 0.30 and 0.50 is 0.9328 0.4664 0.0336 0.0672
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