Badarinarayanan_Charanya _First Exam W2

Badarinarayanan_Charanya _First Exam W2 - NAME STD ID...

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NAME: CHARANYA BADARINARAYANAN STD ID: 11365498 Problem Value Earned 1 20 2 20 3 20 4 20 5 20 Total 100 0 1.) A marketing firm is intrested in sampling public opinion about a new produ in the downtown area at which to conduct random interviews of pedestrian Use the the methods discussed in this course to randomly pick the sites. Avenue with a street. 1st 2nd 3rd 4th 5th 6th Avenue Street Street Street Street Street Street A B C D E F G H I J (Hint: Pick an avenue at random and independently a street at random.) Avenue random values street random values J 0.14 10 0.02 B 0.31 9 0.12 C 0.37 7 0.3 E 0.42 2 0.32 H 0.46 5 0.35 D 0.52 3 0.48 I 0.52 1 0.57 A 0.61 8 0.66 G 0.78 6 0.73 F 0.95 4 0.8
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so the random sites would be the intersection of the first 5 avenues and the first 5 str they are J10,B9,C7,E2,H5
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uct. They wish to randomly pick five sites ns. Below is a map of the downton area. A site is described by the intersection of a 7th 8th 9th 10th Street Street Street Street
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reets.
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2) The following data represents a random sample of the daily "annualized" rates of return o (The "anualized rate" is the rate that would occur in a one year period if the daily rate were h for an entire year). 2.01 2.47 2.71 3.06 3.52 4.06 4.51 2.05 2.49 2.71 3.12 3.53 4.08 4.70 2.07 2.50 2.75 3.16 3.60 4.14 4.78 2.11 2.53 2.86 3.22 3.65 4.19 4.83 2.15 2.53 2.88 3.24 3.75 4.21 4.96 2.17 2.56 2.88 3.31 3.79 4.24 5.05 2.21 2.58 2.92 3.37 3.86 4.36 5.10 2.35 2.63 2.92 3.43 3.88 4.37 5.34 2.37 2.64 3.00 3.45 3.95 4.42 5.45 2.45 2.67 3.02 3.47 3.95 4.51 5.52 a) Find the arithmetic mean of this data: 3.80 b) Find the median of this data: 3.49 c) Find the standard deviation of this data: 1.37 d) Construct a Histogram of this Data using intervals starting at 1.5 and increasing by 1 until th (more sections below) bins Bin Frequency 1.5 1.5 0 2.5 2.5 12 3.5 3.5 28 4.5 4.5 19 5.5 5.5 10 6.5 6.5 7 7.5 7.5 3 8.5 8.5 1 More 0 lower limit upper limit 1.5 2.5 2.5 3.5 3.5 4.5 4.5 5.5 Frequ
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5.5 6.5 6.5 7.5 7.5 8.5 e) Find the arithmetic mean using the grouped data only: (I.e. pretend you don't have the raw data, just the histogram) 3.81 f) Find the median using the grouped data only: n=80 so median is the 40th position in the 3.5 g) Find the standard deviaiton using the grouped data only: 1.4 h) Comment on how the values compare. The mean and median values are almost the same. There is a very slight difference of .02 in the values of standard deviation
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on an investment. held constant 5.61 5.79 5.86 6.08 6.31 6.45 6.68 6.87 7.29 7.88 he value of 8.5. midpoint f f*m F f*m*m 2 12 24 12 48 3 28 84 40 252 4 19 76 59 304 5 10 50 69 250 1.5 3.5 5.5 7.5 More 0 5 10 15 20 25 30 Histogram for rates of return Column F Bin(rates of return) uency
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6 7 42 76 252 7 3 21 79 147 8 1 8 80 64 80 305 1317 e intervals ie., last member in the range 2.5 to 3.5
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3) The following data was collected and it was found that a regression analysis showed that the higher the divorce rate, the lower was the death rate seeming to indicate that "divorce" was healthy. Rate per Rate per 1000 of 1000 of Region Death Divorce States: Mountain 6.9 7.0 MO, ID,WY, CO, NM,AZ,UT,NE Pacific 7.7 5.4 WA, OR,CA, AL, HW West South Central 7.9 7.0 AR,LA,OK TX East North Central 8.9 4.6 OH, IN,IL,MI,WI South Atlantic 9.0 5.2 DE,MD,DC,VI,WV,NC,SC,GE,FL West North Central 9.2 4.2 MN,IW,MS,ND,SD,NB,KA East South Central 9.3 5.7 KY,TN,AL,MS New England 9.3 3.9 ME,NH,VT,MA,RI,CN Mid Atlantic 9.9 3.8 NY,NJ, PA a) Using regression analysis, verify that if y is the death rate and x is the divorce rate, there is a negative relationship.
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