Test 1 - Marking Scheme

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Tutorial (Day and Time):

UNIVERSITI TUNKU ABDUL RAHMAN

ACADEMIC YEAR 2015 / 2016

TEST 1

UDBB1243 BIOSTATISTICS / UDPG1643 ANCILLARY STATISTICS /UDDD1244 BIOSTATISTICS AND BASIC EPIDEMIOLOGY

9 JULY 2015 TIME: 8.15 AM – 9.45 AM

BACHELOR OF SCIENCE (HONS) BIOTECHNOLOGY

BACHELOR OF SCIENCE (HONS) MICROBIOLOGYBACHELOR OF SCIENCE (HONS) BIOMEDICAL SCIENCE

Instruction to Candidates:

There are total of SIX (6) questions.

You are to answer ALL questions.

Marks allocated for each part of the questions are indicated in brackets.

The total marks for this paper is 100 marks.

Marking Scheme Questions Marks

1

2

3

4

5

6

Total

-------------------------------------------- Attendance Slip -----------------------------------------------

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This question paper consists of 6 questions on 10 printed pages.

2

Q1. The following table lists the number of strikeouts per game (K / game) for each of the

30 Major League baseball teams during the 2010 regular season. Note that FloridaMarlins are now Miami Marlins.

Team K / game Team K / gameArizona Diamondbacks 9.44 Milwaukee Brewers 7.51Atlanta Braves 7.04 Minnesota Twins 5.97Baltimore Orioles 6.52 New York Mets 6.76Boston Red Sox 7.04 New York Yankees 7.01Chicago Cubs 7.63 Oakland Athletics 6.55Chicago White Sox 5.69 Philadelphia Phillies 6.57Cincinnati Reds 7.52 Pittsburgh Pirates 7.45Cleveland Indians 7.31 San Diego Padres 7.30Colorado Rockies 7.86 San Francisco Giants 6.78

Detroit Tigers 7.08 Seattle Mariners 7.31Florida Marlins 8.49 St. Louis Cardinals 6.34Houston Astros 6.33 Tampa Bay Rays 7.98Kansas City Royals 5.59 Texas Rangers 6.09Los Angeles Angels 6.60 Toronto Blue Jays 7.19Los Angeles Dodgers 7.31 Washington Nationals 7.53

(a)

Construct a frequency distribution table. Take 5.50 as the lower boundary ofthe first class and 0.8 as the width of each class. (10 marks)

(b) Prepare the relative frequency and percentage distribution columns for the

frequency distribution table of part a. (10 marks)[Total : 20 marks]

Solution:

30 1 100(5 marks) (5 marks) (5 marks) (5 marks)





3

Q2. (a) The test scores for a large statistics class have an unknown distribution with a

mean of 70 and a standard deviation of 10.(i)

Find k so that at least 50% of the scores are within k standard

deviations of the mean. (3 marks)(ii)

Find k so that at most 10% of the scores are more than k standarddeviations above the mean. (4 marks)

(b) Environmental Department of Perak monitors the air quality of the state everyyear. The quality indices for 30 randomly selected weeks in year 2014 werelisted as follows:20 25 78 75 65 50 45 30 21 32 33 40 35 85 7165 64 57 59 40 33 21 20 22 22 34 35 21 20 19(i) Estimate the average and standard deviation of the air quality index.

(6 marks)

(ii)

Comment on the symmetry of the data by using appropriate values.(7 marks)

[Total : 20 marks]

Solution:

(a) 70 , 10

(i)

2

11 0.50

k (1 mark)

2

2

10.50

2

1.414

k

k

k

(1 mark)

(1 mark)

(ii) 2

11 0.90

k (2 marks)

2

10.10

3.162

k

k

(1 mark)

(1 mark)

(b) (i)

(4 marks)

(2 marks)





4

(ii)

Q3. (a) A small country bought oil from three different sources in one week, as shownin the following table.

Source Barrels Purchased Price per Barrel ($)Mexico 1000 95Kuwait 200 92Spot Market 100 99

Find the mean price per barrel for all the barrels of oil purchased in that week. (5 marks)

(b) On a 300-mile auto trip, Lisa averaged 52 miles per hour (mph) for the first100 miles, 65 mph for the second 100 miles and 58 mph for the last 100 miles.(i) How long did the 300-mile trip take? (5 marks)(ii) Could you find Lisa’s average speed for the 300-mile trip by

calculating (52 + 65 + 58) / 3? If not, find the correct average speed forthe trip. (5 marks)

(c)

Jeffrey is serving on a six-person jury for a personal-injury lawsuit. All six jurors want to award damages to the plaintiff but cannot agree on the amountof the award. The jurors have decided that each of them will suggest an

amount that he or she thinks should be awarded; then they will use the meanof these six numbers as the award to recommend to the plaintiff. Jeffrey thinksthe plaintiff should receive $20,000, but he thinks the mean of the other five

jurors’ recommendations will be about $12,000. He decides to suggest aninflated amount so that the mean for all six jurors is $20,000. What amountwould Jeffrey have to suggest? (5 marks)

[Total : 20 marks]

(1 mark)

(3 marks)

(1 mark) (1 mark)

(1 mark)

0.9867





5

Solution:(a)

(b) (i)

(ii)

(c)

(3 marks)

(2 mark)

(3 marks)

(1 mark)(1 mark)

(2 marks)

(3 marks)

(3 marks)

(1 mark)

(1 mark)





6

Q4. (a) A furniture store sells living room chairs which are available in 5 styles, 10

types of fabrics and 8 colors. In how many ways can a customer buy a livingroom chair? (1 mark)

(b) A quiz consists of 6 multiple choice questions, each with five choices. In howmany ways can a student mark the answers to all the questions by marking allof them incorrectly? (1 mark)

(c) A shopper wants to visit three of five department stores and wants to decidewhich stores she will visit first, second and third. She will not visit the samestore twice. In how many ways can she select the three stores? (2 marks)

(d) Suppose you are to choose a basketball team (five players) from eightavailable athletes. How many ways can you choose a team composed of two

guards, two forwards and a centre? (Assume each position was distinct.)(2 marks)

[Total : 6 marks]

Solution:(a) No. of ways = 5 10 8 = 400 (1 mark)

(b)

No. of ways = 64 4 4 4 4 4 4 4096 (1 mark)

(c)

No. of ways =5

3 3 2 1 60C or5

3 60P (2 marks)

(d) No. of ways = 8 6 4

2 2 1 1680C C C or8

5 16802!2!

P (2 marks)





7

Q5. (a) Suppose the events A and B are mutually exclusive and complementary events,

such that 0.75P A and 0.25P B . Consider another event C such that

0.3P C A

and 0.5P C B

. Find:(i)

P A C , (3 marks)

(ii) P B C and (2 marks)

(iii) P C . (4 marks)

(b) If 0.9P D and 0.8P E , show that 0.7P D E . (5 marks)

[Total : 14 marks]Solution:

(a) 0P A B , 0.75P A , 0.25P B

(i) 0.3P C A

0.3

0.3 0.75

0.225

P C A

P A

P C A

P A C

(1 mark)

(1 mark)

(1 mark)

(ii) 0.5P C B

0.5

0.5 0.25

0.125

P C B

P B

P C B

P B C

(1 mark)

(1 mark)

(iii) C

A

C

C B

C (1 mark)

0.225 0.125

0.35

P C P A P C A P B P C B

P A C P B C

(1 mark)

(1 mark)

(1 mark)

(b) 0 1P D E (1 mark)

1

0.9 0.8 1

1 0.9 0.8

0.7

P D P E P D E

P D E

P D E

P D E

(1 mark)

(1 mark)

(1 mark)

(1 mark)

0.5

0.3

0.25

0.75





8

Q6. (a) The probability that a certain door is locked is 0.6. The key to the door is one

of five unidentified keys hanging on a key rack. You randomly select two keys before approaching the door. What is the probability that you can open the

door without returning for another key? (10 marks)

(b) A laboratory blood test is 95% effective in detecting a certain disease whenthe disease is, in fact, present. However, the test also yields a ‘false positive’result for 1% of the healthy person. If 0.5% of the population actually has thedisease, what is the probability a person has the disease given that the testresult is positive. (10 marks)

[Total : 20 marks]Solution:(a) Let L = the door is locked

K = the key to unlock the door

open door1 4 1

0.4 0.6 0.65 5 4

0.64

P P L P L K P L K K

(3 marks)

(3 marks)

(1 mark)

(b) Let D = the event that the tested person has the disease. E = the event that the test result is positive.

.

. .

0.005 0.95 95

0.005 0.95 0.995 0.01 294

P D E P D E

P E

P D P E D

P D P E D P D P E D

(1 mark) (2 mark)

(3 marks) (1 mark)

0.9950.01

0.95

0.005

3

4

1

4

4

5

1

5

0.4

0.6 L

L

K

K

K

K

(3 marks)

D

D

E

E

E

E

Given 0.95P E D (1 mark)

0.01P E D (1 mark)

0.005P D (1 mark)





9

AppendixSome Useful Formula

∆∆ ∆

∆ ∆

L lower boundary od median class N= Total number of dataF Cumulative frequency before median class f = frequency of median class C= Median class size

² 1

²

² 1

²

1 1 ² 1 1 1

² 1

² ² ∑ ∑ ²

² ∑ ² 1 1 1

!

(1-α)100% CI for μ √

(1-α)100% CI for μ

√

(1-α)100% CI for μ , √

(1-α)100% CI for p

(1-α)100% CI for μ1-μ2

1 2 2 1 2 1 2

2 1 2

1 2 11 22 1 2 11 22

(1-α)100% CI for

2





10

√ √

1 2 1 21 2

1 2 11 22

1 2 1 21 2

1 2 11 22

∑

∑ ² = ∑ ²∑ ²

1 2

1 21 2

² 1 (1-α)100% CI for σ²

1

1

²

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Test 1 - Marking Scheme