Stat200 Introduction to Statistics

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Page1of14Stat200Introduction to StatisticsName______________________________Answer SheetRecord your answers andwork.ProblemNumberSolution1(25 pts)Answers:(a)True(b)True(c)False(d)True(e)FalseWork for (a), (b), (c), (d) and (e):2(5 pts)Answer:70%Work:CheckoutTimeFrequency(in minutes)1.0-1.922.0-2.9123.0-3.924.0-4.94Total20percentage of the checkout times was less than 3 minutes = 14/20 = 0.7= 70%3(5 pts)Answer:2ndclass:2.0-2.9Work:Here, n = 20,Median = n/2th observation when n = even,Here, n = 20/2th obs = 10thobservation. Thus median is expected to lie in 2ndclasswhich is 2.0–2.9.

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Page2of144(5 pts)Answer:2.85Work:CheckoutTimeFrequency(in minutes)fimid point(xi(fixi1.0-1.921.452.92.0-2.9122.4529.43.0-3.923.456.94.0-4.944.4517.8Total2057Mean = ∑fixi / ∑fi= 57 / 20 = 2.855(10 pts)Answer:0.92Work:CheckoutTimeFrequency(in minutes)Fimid point(xi(fixifi*xi^21.0-1.921.452.94.2052.0-2.9122.4529.472.033.0-3.923.456.923.8054.0-4.944.4517.879.212057179.25Variance = (1/n)*{ ∑ (fi*xi^2)–n*x-bar^2 }= (1/20)*( 179.25–20*2.85^2)= (1/20)*(179.25–162.45)= (1/20)*(16.8)= 0.84Standard deviation = sqrt(variance)= sqrt(0.84)= 0.9165= 0.92 (approximately)

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Page3of146(10 pts)Answer:13.77Work:xi(xi-xbar)^2225615915918040484total90758Mean == 90/5=18Sample Variance == 758 / 4=189.5Standard deviation = sqrt(variance)= sqrt(189.5)= 13.7659= 13.77 (approximately)7(5 pts)Answer:0.765Work:Coefficient of variation ===0.7658(5 pts)Answer:noWork:

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Page4of149(10 pts)Answer:1/221Work:Probability (1stcard is ace and 2ndcard is also an ace) = ?Probability of drawing 1stcard as ace = 4/52Probability of drawing 2ndcard as ace without replacement = 3/51Thus required probability = (4/52)*(3/51) = (1/13)*(1/17) = 1/22110(10 pts)Answer:1/169Work:Probability (1stcard is ace and 2ndcard is also an ace) = ?Probability of drawing 1stcard as ace = 4/52Probability of drawing 2ndcard as ace with replacement = 4/52Thus required probability = (4/52)*(4/52) = (1/13)*(1/13) = 1/16911(5 pts)Answer:yes, events are independentin case of with replacement because P(A) * P(B)= P(A and B)Work:two events are said to be independent if P(AandB) = P(A) * P(B)In case of with replacement,LetA denotethe event of drawing 1stace and event B denote the event of drawing 2ndace.Now, P(A) = Probability of drawing 1stcard as ace = 4/52= 1/13P(B) = Probability of drawing 2ndcard as ace(with replacement)= 4/52= 1/13P(A) * P(B) = 1/13 * 1/13 =1/169P(A and B) =P(drawing 1stcard as ace and 2ndcard as ace without replacement) =(4/52 * 4/52) =1/169Since, P(A and B) = P(A)*P(B), thus events are independent12(10 pts)Answer:0.25Work:Let event A denoteSTAT200And event B denote PHY300Here totalnumber of juniorsin college is 1000

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