(6 points) Client Age 775 / 20 = 38.75 Wait Time 1480 / 20 = 74.45 * What is the median? (6 points) Client Age (39+42)/2 = 40.5 Wait Time (16+16)/2 = 16 * What is the mode? (6 points) Client Age 18.34 Wait Time 197.88 * What is the standard deviation? (6 points) * Create a Bar Graph for each of the 2 out of the 3 questions. What type of skewness is present, if any? (9 points) * Are there any outliers? If so, what are the outliers? Do you believe they were typos? (3 points)
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Logistics Modelling Assessment 1 This assessment is worth 5% of the total unit mark. Submittal details: 1.Submit your report as a softcopy to the submission box on the unit’s Moodle webpage. The logic flow diagram in Problem 2 can be handwritten but it must be attached to the report. 2.Submit the SIMUL8 file from Problem 3 in *.S8 format to the submission box on the unit‘s Moodle webpage. Please use your student ID number as
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true. Step 2: Determine the Characteristics of the Comparison Distribution In the hypothesis-testing process, you compare the actual sample’s score to this comparison distribution. The comparison distribution is the distribution that represents the population situation if the null hypothesis is true. If the null is true both samples will have a normal curve. Step 3: Determine the Cutoff Sample Score on the Comparison Distribution at Which the Null Hypothesis Should Be Rejected You will need Z
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meals sold per month We assume that M obeys a Normal distribution with µ = 3,000 and σ = 1,000 = price of the prix fixe meal We assume that P obeys the following discrete probability distribution Scenario Very healthy market Healthy market Not so healthy market Unhealthy market Price of Prix Fixe Meal $20.00 $18.50 $16.50 $15.00 Probability 0.25 0.35 0.30 0.10 L = labor costs per month We assume that L obeys a uniform distribution with a minimum of $5,040 and maximum of $6,860
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in the text that requires a brief explanation. A key to understanding the t test is to recall an essential component of a Z test; namely, that a Z test is used when the population variance is known. Also, the shape of the population distribution follows a normal, symetrical curve where the left and right side of the curve are the same. Likewise, the Z tables provide exact percentages of the curve for each Z score. Unlike a Z test that is used when the population variance is known, the t test
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Test statistic: Decision: Since tSTAT> 2.0096, reject . There is enough evidence to conclude that the mean number of days is different from 20. (b) The population distribution needs to be normal. (c) The boxplot plot indicates that the distribution is skewed to the right. (d) Even though the population distribution is probably not normally distributed, the result obtained in (a) should still be valid due to the Central Limit Theorem as a result of the relatively large sample size of
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Assessment Why do we assess? Petty describes assessment ‘in the right hands, assessment can inspire, motivate and provide the feedback which is essential for targeting prompt corrective help’ Petty G, (2001) When an assessment is given it’s purpose is to assess either the students understanding of the subject matter or their application of the knowledge they have acquired. Rowntree further describes the framework of assessment under five headings or dimensions. Namely: Why assess
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Distributions, Sampling and Estimation Our main objective in this section is to learn how to estimate some important characteristics of random variables using data. We will concentrate on estimating the mean from a normal distribution, and the proportion of success in a binomial distribution. To talk about estimation, we first need to know something about distributions and sampling. Most data used for decision making exhibit variation. We are interested in drawing conclusions
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calculate the mean, median, variance and standard deviation for all 50 states plus DC. * Next select ten (10) states, any 10 of your choosing, and calculate the mean, median, variance and standard deviation for median home value. Compare and contrast the means and standard deviations for the ten (10) states you selected versus all fifty (50) states plus DC. * How close or how far away from the average is the median home value in your state, i.e. how many standard
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THE NUMBER OF TIMES EVENT A OCCUR AND n IS THE NUMBER OF TIMES THE EXPERIMANT IS REPEATED. 2. IF P(A) = 0, A IS KNOWN TO BE AN IMPOSSIBLE EVENT AND IS P(A) = 1, A IS KNOWN TO BE A SURE EVENT. 3. BINOMIAL DISTRIBUTIONS IS BIPARAMETRIC DISTRIBUTION, WHERE AS POISSION DISTRIBUTION IS UNIPARAMETRIC ONE. 4. THE CONDITIONS FOR THE POISSION MODEL ARE : • THE PROBABILIY OF SUCCESS IN A VERY SMALL INTERAVAL IS CONSTANT. • THE PROBABILITY OF HAVING MORE THAN ONE SUCCESS
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