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Stat200 Introduction to Statistics

Fundamental concepts and methods in statistical analysis

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    Page 1 of 14 Stat200Introduction to Statistics Nam e______________________________ Answer Sheet Record your answers and work. Problem Number Solution 1 (25 pts) Answers: (a) True (b) True (c) False (d) True (e) False Work for (a), (b), (c), (d) and (e): 2 (5 pts) Answer: 70% Work: Checkout Time Frequency (in minutes) 1.0 - 1.9 2 2.0 - 2.9 12 3.0 - 3.9 2 4.0 - 4.9 4 T otal 20 percentage of the checkout times was less than 3 minutes = 14/20 = 0.7 = 70% 3 (5 pts) Answer: 2 nd class: 2.0 - 2.9 Work: Here, n = 20, Median = n/2th observation when n = even, Here, n = 20/2th obs = 10 th observation . Thus median is expected to lie in 2 nd class which is 2.0 – 2.9.

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    Page 2 of 14 4 (5 pts) Answer: 2.85 Work: Checkout Time Frequency (in minutes) fi mid point (xi( fixi 1.0 - 1.9 2 1.45 2.9 2.0 - 2.9 12 2.45 29.4 3.0 - 3.9 2 3.45 6.9 4.0 - 4.9 4 4.45 17.8 T otal 20 57 Mean = ∑fixi / ∑fi = 57 / 20 = 2.85 5 (10 pts) Answer: 0.92 Work: Checkout Time Frequency (in minutes) F i mid point (xi( fixi fi*xi^2 1.0 - 1.9 2 1.45 2.9 4.205 2.0 - 2.9 12 2.45 29.4 72.03 3.0 - 3.9 2 3.45 6.9 23.805 4.0 - 4.9 4 4.45 17.8 79.21 20 57 179.25 Variance = (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) = 0.84 Standard deviation = sqrt(variance) = sqrt(0.84) = 0.9165 = 0.92 (approximately)

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    Page 3 of 14 6 (10 pts) Ans wer: 13.77 Work: xi (xi - xbar)^2 2 256 15 9 15 9 18 0 40 484 total 90 758 Mean = = 90/5 =18 Sample Variance = = 758 / 4 =189.5 Standard deviation = sqrt(variance) = sqrt(189.5) = 13.7659 = 13.77 (approximately) 7 (5 pts) Answer: 0.765 Work: Coefficient of variation = = =0.765 8 (5 pts) Answer: no Work:

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    Page 4 of 14 9 (10 pts) Answer: 1/221 Work: Probability (1 st card is ace and 2 nd card is also an ace) = ? Probability of drawing 1 st card as ace = 4/52 Probability of drawing 2 nd card as ace without replacement = 3/51 Thus required probability = (4/52)*(3/51) = (1/13)*(1/17) = 1/221 10 (10 pts) Answer: 1/169 Work: Probability (1 st card is ace and 2 nd card is also an ace) = ? Probability of drawing 1 st card as ace = 4/52 Probability of drawing 2 nd card as ace with replacement = 4/52 Thus required probability = (4/52)*(4/52) = (1/13)*(1/13) = 1/169 11 (5 pts) Answer: yes, events are independent in 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 denote the event of drawing 1 st ace and event B denote the event of drawing 2 nd ace. Now, P(A) = Probability of drawing 1 st card as ace = 4/52 = 1/13 P(B) = Probability of drawing 2 nd card as ace (with replacement) = 4/52 = 1/13 P(A) * P(B) = 1/13 * 1/13 = 1/169 P(A and B) = P(drawing 1 st card as ace and 2 nd card as ace without replacement) = (4/52 * 4/52) = 1/169 Since, P(A and B) = P(A)*P(B), thus events are independent 12 (10 pts) Answer: 0.25 Work: Let event A denote STAT200 And event B denote PHY300 Here total number of juniors in college is 1000

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