BUS105_Seminar6_Tutor_Copy(Jess)-e.pptx - BUS105 Statistics...

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BUS105 Statistics Seminar 6
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Activity 6.1 The monthly U.S. unemployment rates from January 2003 to January 2013 can be shown as a histogram or as a time series plot. a. What features of the data are clear in the histogram but are not clear in the time series plot? b. What features of the data are clear in the time series plot but are not clear in the histogram? c. Which graphical display seems to be the better way to present the unemployment rate data? Explain. d. Write a brief description of the changes of unemployment rate in the United States over this time period.
Activity 6.1 - Answers a. In histogram, we can observe the shape of the distributions and frequency of unemployment rate. In particular, we noted that this case there is a bimodality distribution. b. In the time series plot, the unemployment rate is plotted against time. As such, we can observe the trend of the data. c. The time series plot is more appropriate because it reveals much more of the structure of the data. The unemployment rates are not stationary. d. Unemployment rate decreased from just over 6% to under 5% from 2004 to 2007. It then increased steadily until 2010 when it reached nearly 10%. Since then it has decreased to about 8%.
Activity 6.2 The number of connections on the Internet during any two-minute period is given by the following distribution: Within a two-minute period, determine the a. probability of having more than 2 connections b. mean number of times a connection is made c. standard deviation of the number of connections made Number of Times Proportion 0 0.1 1 0.2 2 0.1 3 0.4 4 0.2
Activity 6.2 - Answers a. P(more than 2) = 0.4 + 0.2 = 0.6 b. Mean = 0 x 0.1 + 1 x 0.2 + 2 x 0.1 + 3 x 0.4 + 4 x 0.2 = 2.4 c. Variance = (((0 – 2.4) 2 x 0.1 ) + ((1 – 2.4) 2 x 0.2 ) +((2 – 2.4) 2 x 0.1 ) + ((3 – 2.4) 2 x 0.4 ) + ((4 – 2.4) 2 x 0.2 ) = (0 2 x 0.1 + 1 2 x 0.2 + 2 2 x 0.1 + 3 2 x 0.4 + 4 2 x 0.2) 2.4 2 = 1.64 Standard deviation = = 1.28
Activity 6.3 Mr Tan run a small restaurant and recorded his daily revenue for a month as shown in the table. Assume that the daily revenue is normally distributed with mean $4500 and the sample standard deviation is a reasonable substitute for the population standard deviation, calculate the a) Probability that the revenue of a random day is more than $4700.

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