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'1 statistics 2017 mid testpdf , Adobe Acrobat Reader DC
File Edit View Window Help Home Tools statistic52017 mid.. >< (I?) 73113 ' [EJG‘JBEQ ©© 3 I6 9.? Question: 1 2 3 4 5 Total Question 2 [13 marks] "
For this question, use the joint probability distribution of X and Y given in Question 1,
Marks: 13 13 12 Q 13 60 where X was the number of cups of coffee Dirk drinks on any given day and Y was the
number of biscuits Dirk cats on any given day.
Semester 2 , MidsSemest/erv 2017 Page 2 of 6 STAT7055
(a) 3 marks] Find the probability that (X 7 4)Y > 7. (c) 3 marks] Dirk Wants to get a good start to the upcoming 5-day week (Mon . Fri).
so on both Monday and Tuesday, Dirk plans to work productively for twice as long
F Solution: as he normally would on any given day. Find the variance of the total number of 4
0.45 hours that Dirk works productively over the upcoming 5-day week (Mon - Fri).
Dirk has defined a new quantity. Q : X (Y a 4), which he calls his “CC Index“ ((‘offec- 501‘“ng
ceekic index) for that day. 5-28 hours
(bl 3 marks] Fmd ”m “PM“ val“ °f 9' Dirk’s excuse for drinking lots of coffee is that it helps him be more productive at work.
S l t' . Suppose now that the number of hours that Dirk works productively on any given day is
° “ ‘°“' uniformly distributed between a and X hours. Where X is the number of Cups of coffee
’0'3 Dirk drinks on that (lay,
. . (d) 4 marks] Find the probability that Dirk works productively for more than 1.62
(cl 4 marks] Fmd the variance 0f (2' hours today The joint probability distribution of X and Y given in Question 1 is
Solution: dlsplaycd again below:
30-93 31 (Biscuit)
3 4 6
(d) 3 marks] If we assume the values of Q are independent from day to day, find thc
probability that Dirk’s total cumulative CC Index ever the next 60 days is greater 3 0-12 0-25 0.04
than 110. 1: (Coffee) 6 0.14 0.20 0.02 V

2018-04-03 (2).png

'1 statistics 2017 mid testpdf 7 Adobe Acrobat Reader DC — x
File Edit View Window Help Home Tools statistics 2017 mid >< (I?) 73113 '
[31613ng @© 3 /6 9/
p ......... y “is. or... o em. or. W em no. so oayo groom
than 110 1' (Coffee) 6 0.14 0.20 0.02 A
7 0.10 0,05 0.08
Solution:
0.0015
Solution:
Question 3 [12 marks] “-6282
suppose the number of hours that Dirk works productively on any given day is uniformly .
distributed between 0 and 2.4 hours. Assume the number of hours that Dirk works QUGStIOD 4 [9 marks]
productively is independent from day to day. Ron-all from Question 3 that the number of hours that Dirk works productively on any given day is uniformly distributed between 0 and 2.4 hours. Suppose that the num-
ber of hours that Dirk sleeps on any given day is normally distributed with mean 7.2
hours and standard deviation 1.4 hours. Assume that the number of hours that Dirk
works produetively and the number of hours that Dirk sleeps, on any given day, are
independent. (a) 2 marks] If Dirk worked productively for more than an hour today, find the prob-
ability that Dirk worked productively for less than 1.62 hours. Solution:
04429 (a) 3 marks] Find the probability that Dirk sleeps between 8 and 9 hours today, given (b) 3 marks] Over the next 2 days. find the probability that the most amount of time ”1“” 1“ SICCI’S for “10“ than 71‘0““- Dirk worked productively on a day was lnore than 1.62 hours. .
’ Solution: ‘
Solution: 0-3344
0.5444 (b) 3 marks] Find the probability that Dirk sleeps between 7 and 9 hours today or
he works productively for more than 0.75 hours today. Semfiter 2 r MidrSemester, 2017 Page 3 of G STAT7055 Semester 2 r MidrSemester. 2017 Page 4 of 6 STAT7055 day that he works productively for more than 0.75 hours. he gets 5 points. Find
the expected Value of the number of points Dirks gets eaeh day. Solution:

2018-04-03 (3).png

'1 statistics 2017 mid testpdf 7 Adobe Acrobat Reader DC — x
File Edit View Window Help Home Tools statistic52017 mid... >< C?) 73113 ' [EDGJBEIQ ©© 5 I6 9.? tiny that he works productively for more than 015 hours, he gets 5 points. Find A
Solution: the expected value of the number- of points Dirks gets each day.
0.8304
Solution:
(c) 3 marks] If W denotes the number of hours Dirk works productively today and 40051 5 denotes the number of hours Dirk sleeps today, find the varianee of SW + 23. Solution: END OF EXAMINATION 12.16 Question 5 [13 marks] Reeall from Question 4 that the number of hours that Dirk works productively on any
given day, denoted W, is uniformly distributed between o and 2.4 hour-s and the number
of hours that Dirk sleeps on any given day, denoted s, is normally distributed with mean 7.2 hours and standard deviation 1.4 hours. suppose now the number of hours that Dirk
works productively and the number of hours that Dirk sleeps. on airy given day, are not independent. (a) 4 marks] 1i Dirk believes that the variant-e of 3W + 23 is equal to 7, find the expoeted Value of ws. ‘ Solution:
8‘21 For parts (1)) to (d), assume that P({W > 0.75} n {s > 6}) = 0.55. (b) 3 marks] Find the probability tlistDirk sleeps for more tlianslieurs today. given
that he worked produetively for less than 0.75 hours today. Solution:
0.8163 (c) 3 marks] Find the probability that Dirk sleeps for less than 6 hours today and
works productively for less than 0.75 hours today. Solution:
0.0574

2018-04-03 (4).png

“7 statistics 2017 mid testpdf 7 Adobe Acrobat Reader DC File Edit View Window Help Home Tools statistic52017 mid.. >< C?) 73113 '
@QBEQ @© 5 I6 9/
Ker-ell lroni Question 4 tlirtt the number oi hours that Dirk works produr-tively on any A given doy, denoted w, is ullifonlily distributed bulwwu 0 21111124 hours and the number
of hours that Dirk sleeps on ony given day, denoted st is normally distributed with mean
7.2 hours and standard deviation 1.4 hours. suppose now the number of hours that Dirk
works produetivelv and the number of hours that Dirk sleeps. on onv given day. are not
independent. (a) 4 marks] If Dirk believes that the variance of SW + ZS is equal to 7. find the
expected value of WS. Solution:
8.21 For parts (1)) to (d), assume that P({W > 0.75] n {s > 6}) = 0.55. (b) 3 marks] Find the probability that Dirk sleeps for more thonshours today; given
that he worked productively fol' less than 0.75 hours today. Solution:
} 0.8163 ‘ (c) 3 marks] Find the probability that Dirk sleeps for less than 6 hours today and
works productively for less than 0.75 hours today, Solution:
0.0574 (d) 3 marks] Dirk decides to give himself points eoeh day depending on how predue
tively he works and how much he sleeps. If he works productively for less than 0.75
hours end sleeps for less than 6 hol he gets 1 point, If he works produetively for
less than 0.75 hours and sleeps for more then 6 hours, he gets 2 points. For any Screenshot saved Scmcstcx 2 - Mid-Scmwtcr. 2017 Pagc 5 of 6 STAT7055 Somcstcr 2 - Mid-Semester, 2017 ‘ 144 PM
E 0 Wm here to search ' [ll A [3 2704mm E4

2018-04-03 (5).png

': statistics 2017 mid testpdf 7 Adobe Acrobat Reader DC — x
File Edit View Window Help Home Tools statistics 2017 mid >< C?) 73113 '
[33$ng ®© 1/6 9/
HUSIrallan ' Dirk likes- having terror and biseui‘ts everyday. Let X denote the nurnlrer oleups ofeoffee a
National Dirk drinks on any given day. The number of cups of coffee Dirk drink.» is independent
University from day to day. Let Y denote the number of biscuits Dirk oats on any given day. The number of biseuits Dirk Cats is independent from day to day. The relationship between
X and Y is described by the following joint probability distribution:
Research School of Finance, Actuarial Studies and Statistics
y (Biscuit)
EXAMINATION SOLUTIONS 3 4 6
Semester 2 - Mid-Semester. 2017 —
3 0.12 0.25 0.04
STAT7055 Introductory Statistics for Business and Finance 1’ ((70500) 6 0'14 0'20 0-02
7 0.10 0.05 0.08 Writing Time: 90 minutts _ , _ _ _ _ .
Reading Time: 15 minutes (a) 3 marks] Find the probability that Dirk drank 3 cups oi col-lea today, given that he ate more than 3 biocuiis.
Exam Conditions:
Central examination. Solution:
Students must return the examination paper at the end of the examination. This examination paper is not available to the ANU Lihrary archives. “-4531
’ Materials permitted in the Exam Venue: (b) 3 marks] Find the probability that Dirk ate 3 biscuits today. given that he drank ‘
(No electronic aids are permitted, e.g., laptops, phones). mm ”m 3 cups of mm,“
Calculator (non»pmgra.mm£tble).
One A4 page with notes on both sides. .
Unannetated paper—based dictionary (no approval required). 501mm“:
0.4068 Materials to be Supplied to Students:
Script beck. (C)
Scribble paper. 4 marks] over the next 10 days, find the probability that Dirk drank more than
3 cups of tattoo on more than 7 days. Instructions to Students:
Please write your student number in the spa/Ce provided on the front of the script hook. Solution:
Attempt till 5 questions. (11517
Sturt your solution to each question on a new page and clearly label each solution with the corresponding
question number. _ _ _
To ensure full marks show all the steps in working out your solutions. Marks may be dedueted for failure (<1) 3 malrksl Over the next 30 daysr find the Probability that Dirk (hank more than
to show working or formulae. 3 cups of eoifee on more than 20 days. Selected statistical tzthles are attached to the hawk of the examination paper.
Round all numeric answers to 4 decimal places. n 0 Type here to search Solution:
0. 1977 museum.-

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