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# EXAM1 must finish before 10:00 tomorrow 2 / 23 (ca time) (Example)

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Question 1 Possible outcomes 1 2 3 4 5 6 1 (1 1) =2 (1 2) =3 (1 3) =4 (1 4) =5 (1 5) =6 (1 6) =7 2 (2 1) =3 (2 2) =4 (2 3) =5 (2 4) =6 (2 5) =7 (2 6) =8 3 (3 1) =4 (3 2) =5 (3 3) =6 (3 4) =7 (3 5) =8 (3 6) =9 4 (4 1) =5 (4 2) =6 (4 3) =7 (4 4) =8 (4 5) =9 (4 6) =10 5 (5 1) =6 (5 2) =7 (5 3) =8 (5 4) =9 (5 5) =10 (5 6) =11 6 (6 1) =7 (6 2) =8 (6 3) =9 (6 4) =10 (6 5) =11 (6 6) =12 Event A {2 4 6 8 10 12} Event B {9 10 11 12} Union of Event A and event B AUB {2 4 6 mean = x ± t*snMean (x) =12.3 Standard deviation (s) =2.5 n=12 t value associated with 95% confidence interval and 11 degrees of freedom =2.201 95% Confidence Interval for a population mean = 12.3 ± 2.201*2.51295% Confidence Interval for a population mean=12.3 ± 1.588435 Lower bound =12.3-1.588435= 10.7116 Upper bound =12.3+1.588435= 13.8884 95% Confidence Interval for the true population mean weight loss: [10.7116 13.8884] Question 14 Population Mean (µ)=237.6 Standard deviation (σ)=26.3 Sample mean (x) =230 N=55 Z=x-µσnZ=230-237.626.355Z= -7.06 P(X&gt;230) = P(Z&gt; -2.143) From the standard normal table; P(Z&gt; -2.143) = 0.9839 Therefore; The probability that the ATM usage averaged more than 230 students per day = 0.9839 Question 15 Population Mean (µ)=6.8 Standard deviation (σ)=1.8 Sample mean (x) =6 N=36 Z=x-µσnZ=6-6.81.836Z= -2.6667 P(X&lt;6) = P (Z&lt; -2.6667) From the standard normal table; P (Z&lt; -2.6667) = 0.00383 Probability that their mean will be less than 6hrs is 0.00383 [...]

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SEE THE QUESTIONS OF THE ATTACHED FAILED U can handwrite it, must be clear

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This project has already been completed by one of the Studybay experts. The client rated this project:

Project's rating is 5/5

Price \$40

Words 550

Pages 2

Completed in 2 days

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## Client Review

GOOD

Positive
02.25.2017

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