Exam 2011, Questions And Answers PDF

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ANSWERS

MGT 137 & 138 Multiple Choice Exam January 2012 Instructions:  2 hours  Graph paper to be provided  Answer all 35 questions (Total marks 50)  There are no penalty marks for wrong answers  While it is intended that the correct answer for each question should be one of the choices (a) to (e), there is a slight chance that a mistake has been made and that this is not the case. In this situation you should answer (f) and then give your numerical answer  Normal distribution tables are provided at the end of the exam paper; t distribution tables are not needed.  Calculators are allowed  An A4 sheet of personal notes is allowed  You need to ensure that your answer to each question is clear

ANSWERS

1

ANSWERS

PRICE INDICES Average Salary

Average salary

Number of

Number of

2000 (£k)

2009 (£k)

employees, 2000

employees, 2009

Administrative

10.5

14.6

10

15

Professional

30.7

38.8

25

19

Managerial

38.0

55.9

8

17

Table 1:

The change in the salary costs for a company between 2000 and 2009.

Q.1 For the data in Table 1, determine the simple (price) index for professional salaries. a) 1.26 b) 7.91 c) 79.12 d) 96.05 e) 126.38 Correct:

e) (38.8 ÷ 30.7) × 100 = 126.38

Incorrect:

(30.7 ÷ 38.8) × 100 = 79.12

Incorrect:

(38.8 ÷ 30.7) = 1.26

Incorrect:

(30.7 ÷ 38.8) = 7.91

Incorrect:

((38.8 × 19) ÷ (30.7 × 25)) × 100 = 96.05 [MARKS: 1]

Q.2 For the salary costs in Table 1, determine the Laspeyres (base period) index. a) 61.71 b) 75.26 c) 132.87 d) 137.47 e) 162.05 C: c) ((14.6 × 10 + 38.8 × 25 + 55.9 × 8) ÷ (10.5 × 10 + 30.7 × 25 + 38.0 × 8)) × 100 = 132.87 I: ((14.6 × 15 + 38.8 × 19 + 55.9 × 17) ÷ (10.5 × 15 + 30.7 × 19 + 38.0 × 17)) × 100 = 137.47 I: ((14.6 × 15 + 38.8 × 19 + 55.9 × 17) ÷ (10.5 × 10 + 30.7 × 25 + 38.0 × 8)) × 100 = 162.05 I: ((10.5 × 10 + 30.7 × 25 + 38.0 × 8) ÷ (14.6 × 15 + 38.8 × 19 + 55.9 × 17)) × 100 = 61.71

2

ANSWERS I: ((10.5 × 10 + 30.7 × 25 + 38.0 × 8) ÷ (14.6 × 10 + 38.8 × 25 + 55.9 × 8)) × 100 = 75.26 [MARKS: 2] Q.3 For the salary costs in Table 1, determine the Paasche (current period) index. a) 61.71 b) 75.26 c) 132.87 d) 137.47 e) 162.05 C:d) ((14.6 × 15 + 38.8 × 19 + 55.9 × 17) ÷ (10.5 × 15 + 30.7 × 19 + 38.0 × 17)) × 100 =137.47 I: ((14.6 × 10 + 38.8 × 25 + 55.9 × 8) ÷ (10.5 × 10 + 30.7 × 25 + 38.0 × 8)) × 100 = 132.87 I: ((14.6 × 15 + 38.8 × 19 + 55.9 × 17) ÷ (10.5 × 10 + 30.7 × 25 + 38.0 × 8)) × 100 = 162.05 I: ((10.5 × 10 + 30.7 × 25 + 38.0 × 8) ÷ (14.6 × 15 + 38.8 × 19 + 55.9 × 17)) × 100 = 61.71 I: ((10.5 × 10 + 30.7 × 25 + 38.0 × 8) ÷ (14.6 × 10 + 38.8 × 25 + 55.9 × 8)) × 100 = 75.26 [MARKS: 2] Q.4 If the Retail Price Index (RPI) was 181.3 in the middle of 2003 and 186.7 in the middle of 2004, then what was the rate of inflation between the middle of 2003 and the middle of 2004? a) 0.0289% b) 0.0298% c) 2.89% d) 2.98% e) 5.40% Correct:

d) ((186.7 – 181.3) ÷ 181.3) × 100 = 2.98

Incorrect:

(186.7 – 181.3) = 5.40

Incorrect:

((186.7 – 181.3) ÷ 186.7) × 100 = 2.89

Incorrect:

((186.7 – 181.3) ÷ 181.3) = 0.0298

Incorrect:

((186.7 – 181.3) ÷ 186.7) = 0.0289 [MARKS: 1]

Q.5 If the local House Price Index was 204.0 in January of 2003 when a house sold for £159,995 and the index is 311.2 in January 2011, then what is the best estimate of the house’s value in January 2011? 3

ANSWERS a) £160,048, b) £160,102 c) £215,109 d) £244,071 e) £264,876 Correct:

d) (311.2  204)  159,995 = 244,071

Incorrect:

((311.2 – 204)  204)  100 + 159,995 = 160,048,

Incorrect:

159,995 + (311.2 – 204) = 160,102

Incorrect:

(204  311.2)  159,995 + 159,995 = 264,876

Incorrect:

(311.2-204)  311.2  159,995 + 159,995 = 215,109 [MARKS: 1]

INVESTMENT APPRAISAL Q.6 £3,500 is invested in a bond paying an annual simple interest rate of 2.0%. What is the value of the bond after 3 years? a) £3,301.89 b) £3,710.00 c) £3,714.23 d) £5,600.00 e) £10,710.00 Correct:

b) 3,500  (1 + 3  0.02) = 3,710.00

Incorrect:

3,500  (1 + 0.02)3 = 3,714.23

Incorrect:

3,500  (1 + 3  0.2) = 5,600.00

Incorrect:

3,500  (1 + 3  0.02) = 3,301.89

Incorrect:

3,500  3  (1 + 0.02) = 10,710.00 [MARKS: 1]

Q.7 £3,500 is invested in a bond paying an annual compound interest rate of 1.5%. What is the value of the bond after 3 years? a) £3,500.04 b) £3,657.50

4

ANSWERS c) £3,659.87 d) £5,075.00 e) £5,323.06 Correct:

c) 3,500  (1 + 0.015)3 = 3,659.87

Incorrect:

3,500  (1 + 0.15)3 = 5,323.06

Incorrect:

3,500  (1 + 3  0.015) = 3,657.50

Incorrect:

3,500  (1 + 3  0. 15) = 5,075.00

Incorrect:

3,500  (1 + 0.0153 )= 3,500.04 [MARKS: 1]

Year

0

1

2

3

Expenditure /

–100

20

40

60

income (£)

Table 2:

Returns from an investment opportunity

Q.8 For the data in Table 2, determine the Net Present Value of the project if the interest rate is 2%. a) -£20.83 b) £14.49 c) £14.59 d) £16.12 e) £25.69 Correct:

c) –100 + (20  1.02) + (40  1.022) + (60  1.023) = 14.59

Incorrect:

–100 + (20  1.02) + (40  1.022) + (60  1.023) = 25.69

Incorrect:

–100 + (20  1. 2) + (40  1.22) + (60  1. 23) = -20.83

Incorrect:

–100 + (20  1.023) + (40  1.022) + (60  1.021) = 16.12

Incorrect:

–100 + (20  0.98) + (40  0.982) + (60  0.983) = 14.49 [MARKS: 2]

Q.9 For the data in Table 2, estimate the Internal Rate of Return of the project. a) 7% b) 8% 5

ANSWERS c) 9% d) 10% e) 11% Correct:

b) 8%

(Excel gives 8.2%) [MARKS: 2]

Q.10 You take out a £1,000 loan at a compound annual interest rate of 10%. If you clear the loan by making two equal payments, one of them after 1 year and the other after 2 years, how large would each repayment be? a) 432.90 b) 454.55 c) 550.00 d) 576.19 e) 605.00 Correct:

d) 1.12  1000  2.1 = 576.19

Incorrect:

1.12  1000  2 = 605.00

Incorrect:

1000  (1.1 + 1.12) = 432.90

Incorrect:

500  1.1 = 454.55

Incorrect:

1.1  1000  2 = 550.00 [MARKS: 2]

LOCATION AND SPREAD 10

Table 3:

20

6

2

8

32

4

14

14

45

Data for location and spread

Q.11 For the data in Table 3, determine the median. a) 10 b) 12 c) 13 d) 14 e) 20 6

ANSWERS ORDERED: 2

4

6

8

10

Correct:

b) (10 + 14)  2 = 12

Incorrect:

(8+32)  2 = 20 (Data not sorted)

14

14

20

32

45

[MARKS: 1] Q.12 For the data in Table 3, determine the mean. a) 14.0 b) 15.5 c) 16.1 d) 16.2 e) 17.2 Correct:

b) 15.5

Incorrect:

155  9 = 17.2 [MARKS: 1]

Q.13 For the data in Table 3, determine the 20% trimmed mean. a) 11.0 b) 12.0 c) 12.5 d) 13.0 e) 13.5 Correct:

b) 12.0 (mean of 6 8

10

14

14

Incorrect:

13.5 (10% trimmed mean)

Incorrect:

11.0 (20% trimmed mean without ordering the data)

Incorrect:

12.5 (10% trimmed mean without ordering the data)

20)

[MARKS: 1] Q.14 For the data in Table 3, determine the sample standard deviation. a) 12.1 b) 12.9 c) 13.4 d) 13.6 e) 184.3 7

ANSWERS Correct:

d) 13.6

Incorrect:

12.9 (population standard deviation)

Incorrect:

184.3 (sample variance) [MARKS: 1]

Q.15 For the data in Table 3, determine the range. a) 2 and 45 b) 33 c) 41 d) 43 e) 45 Correct:

d) 43

Incorrect:

2 and 45

Incorrect:

33 (top – median) [MARKS: 1] Days off work

Number of employees

0

80

1

30

2

25

3

20

4

10

5

10

6 or more

25

Table 4: Data for lower and upper quartile values.

Q.16 For the data in Table 4, which of the following are the lower and upper quartile values? a) 0 & 3 b) 0 & 5 c) 1 & 3 d) 1 & 4 e) 1 & 5

Correct: a) 0 & 3

[MARKS: 2] 200 employees so 50th and 150th position. 8

ANSWERS

PROBABILITY An advertising campaign using television commercials and magazine advertisements was carried out to promote a new chocolate bar. A month after the advertising campaign finished, a survey was carried out to determine people’s recall of the different adverts. 35% of people said that they could recall the television adverts and 45% could recall the magazine adverts. However, 45% of people said they could recall neither the television nor the magazine adverts. Q.17 What is the probability that someone can recall both the television and the magazine adverts? a) 0.20 b) 0.25 c) 0.30 d) 0.35 e) 0.40

10% TV[35%]

Correct:

25%

20%

45% M[45%]

b) 0.25 [MARKS: 2]

Q.18 If someone can recall the television adverts, what is the probability that they cannot recall the magazine adverts? a) 0.29 b) 0.34 c) 0.37 d) 0.40 e) 0.47 Correct:

a) 10  35 = 0.29 [MARKS: 2]

9

ANSWERS Less than 5 days

More than 5 days

sick leave

sick leave

Administrative

5

10

Professional

9

10

Managerial

12

5

Table 5:

The amount of sick leave taken by staff grades in the last 12 months.

Q.19 For the data in Table 5, what is the probability that someone selected at random is in the professional grade? a) 0.19 b) 0.27 c) 0.29 d) 0.37 e) 0.47 Less than 5 days

More than 5 days

sick leave

sick leave

Administrative

5

10

15

Professional

9

10

19

Managerial

12

5

17

26

25

51

Correct:

d) 19  51 = 0.37 [MARKS: 1]

Q.20 For the data in Table 5, what is the probability that someone selected at random is in the professional grade and took less than 5 days sick leave? a) 0.18 b) 0.19 c) 0.35 d) 0.47 e) 0.51 Correct:

a) 9  51 = 0.18 10

ANSWERS Incorrect:

9  19 = 0.47

Incorrect:

9  26 = 0.35

Incorrect:

26  51 = 0.51 [MARKS: 1]

Q.21 For the data in Table 5, if the person took less than 5 days sick leave, what is the probability that they are in the professional grade? a) 0.09 b) 0.18 c) 0.19 d) 0.35 e) 0.47 Correct:

d) 9  26 = 0.35

Incorrect:

9  51 = 0.18

Incorrect:

9  19 = 0.47 [MARKS: 1]

NORMAL DISTRIBUTION Using the normal distribution tables at the end of the exam paper: Q.22 For the standard normal distribution, i.e. mean zero and standard deviation 1, what is the probability that Z is greater than or equal to 0.75? a) 0.20 b) 0.22 c) 0.23 d) 0.24 e) 0.67 Correct:

c) 0.22663  0.23

Incorrect:

0.24196  0.24 [MARKS: 1]

Q.23 For the standard normal distribution, i.e. mean zero and standard deviation 1, what is the probability that Z is greater than or equal to –0.55? 11

ANSWERS a) 0.21 b) 0.29 c) 0.58 d) 0.71 e) 0.79 Correct:

d) 1 – 0.29116 = 0.71

Incorrect:

0.5 – 0.29116 = 0.21

Incorrect:

0.5 + 0.29116 = 0.79

Incorrect:

0.29116 = 0.29

Incorrect:

2  0.29116 = 0.58 [MARKS: 1]

Q.24 For the standard normal distribution, i.e. mean zero and standard deviation 1, what is the probability that Z is between 0.31 and 0.52? a) 0.07 b) 0.08 c) 0.21 d) 0.32 e) 0.68 Correct: Incorrect:

b) 0.37828 – 0.30153 = 0.07675 = 0.08 1 – (0.37828 + 0.30153) = 0.32

Incorrect:

0.37828 + 0.30153 = 0.68

Incorrect:

0.52-0,31 = 0.21 [MARKS: 1]

Area

z

Figure 1:

The standard normal distribution

Q.25 If the shaded region in Figure 1 is 0.80, then what approximate value does z take?

12

ANSWERS a) -0.84 b) -0.42 c) 0.42 d) 0.84 e) 2.86 Correct:

a) Look up Area = 1.00 – 0.80 = 0.20  Z = 0.84  Z = – 0.84

Incorrect:

Look up Z = 0.20  Z = 0.42074  Z = – 0.42 [MARKS: 2]

Q.26 If X is distributed as a normal distribution with mean 4 and standard deviation 2, then what is the probability that X is greater than 3? a) 0.00 b) 0.31 c) 0.62 d) 0.69 e) 0.81 Correct:

d) Z=(X – 4)  2 = – 0.5  0.30854  1 – 0.30854 = 0.69146 = 0.69

Incorrect:

Z = 3  0.00135 = 0.00

Incorrect:

0.30854 = 0.31

Incorrect:

0.30854 + 0.5 = 0.80854 = 0.81

Incorrect:

2  0.30854 = 0.62 [MARKS: 2]

Q.27 If X is distributed as a normal distribution with mean 4 and standard deviation 2 and the shaded area in Figure 1 has size 0.2, then what is the value of X? a) 0.84 b) 4.84 c) 5.68 d) 8.84 e) 9.68 Correct:

c) 0.2  Z=0.84  X = (Z 2) + 4 = 5.68

Incorrect:

0.2  Z=0.84  X = 0.84

Incorrect:

Z=0.20  P = 0.42074  X = (P 2) + 4 = 4.84 13

ANSWERS Incorrect:

0.2  Z=0.84  X = (Z + 8)  2= 9.68

Incorrect:

P = 0.42074  X = (P 2) + 4 = 8.84 [MARKS: 2]

DECISION MAKING

0.3

5

0.7 9

A 0.5

3 4

0.6

3 -2

B 0.1

11 -2

Figure 2:

Partially completed decision tree

Q.28 For the partially completed decision tree shown in Figure 2, what value should replace A? a) 3.84 b) 5.25 c) 6.51 d) 7.80 e) 11.85 Correct:

c) 0.7  (0.3  5 + 0.7  9) + 0.3  (0.5  3 + 0.5 4) = 6.51

Incorrect:

Maximum((0.3  5 + 0.7  9) , (0.5  3 + 0.5  4) ) = 7.80

Incorrect:

0.7  0.3  (5 + 9) + 0.3  (0.5  3 + 0.5  3) = 3.84

Incorrect:

.7  (0.3  5 + 1  9) + 1  (0.5  3 + 1  3) = 11.85 14

ANSWERS [MARKS: 2] Q.29 For the partially completed decision tree shown in Figure 2, what value should replace B? a) 0.15 b) 1 c) 2 d) 5 e) 11 Correct:

b) Maximum ((0.6  3 + 0.4  (-2)) , (0.1  11 + 0.9  (-2)) ) = 1

Incorrect:

0.5  (0.6  3 + 0.4  (-2)) + 0.5  (0.1  11 + 0.9  (-2)) = 0.15 [MARKS: 2]

Q.30 You are considering holding a firework display to raise funds for your playgroup. If it is mild and dry you estimate that your profits will be £9,000, if it is cold and dry you estimate that the profits will be £7,000 but if it is wet you think that you will make a loss of £1,000 no matter whether it is mild or cold. Records show that the chance of it being mild and dry is 0.1, cold and dry is 0.6, and wet is 0.3. What is the expected monetary value of holding the event? a) £1,020 b) £1,320 c) £4,800 d) £5,100 e) £5,400 Correct:

c) EMV = 9  0.1 + 7  0.6 + (-1)  0.3 = 0.90 + 4.2 – 0.30 = 4.8

Incorrect:

EMV = 9  0.1 + 7  0.6 + (-1)  0.3 = 0.90 + .42 - 0.30 = 1.02 EMV = 9  0.1 + 7  0.6 + 1  0.3 = 0.90 + 4.2 + 0.30 = 5.4 [MARKS: 2]

Q.31 For the firework display situation in the previous question, what is the expected value of knowing what the weather will be (Expected Value of Perfect Information)? a) £0 b) £30 15

ANSWERS c) £300 d) £798 e) £7,980 Correct:

c) New EMV = 9  0.1 + 7  0.6 + 0  0.3 = 0.90 + 4.2 = 5.1 So EVPI = 0.3 [MARKS: 1]

EXCEL Q.32 Using the values in the picture what would be the result in cell B5 if you entered the formula: =(B2*B3)+B4

a) 130 b) 800 c) 840 d) 4000 e) none of these Correct: c) 840 [MARKS: 1] Q.33 What result will you get in cell D8 if you enter the formula: =SUM(B3:D6)

16

ANSWERS a) 12 b) 20 c) 31 d) 37 e) 80 Correct: e) 80 [MARKS: 1]

Q.34 You highlight cells B2 and B3 and drag the AutoFill handle two cells down to B5. What will appear in cells B4 and B5?

a) 7 in B4, 8 in B5 b) 7 in B4, 10 in B5 c) 9 in B4, 12 in B5 d) 3 in B4, 6 in B5 e) 6 in B4, 6 in B5 Correct: c) 9 in B4, 12 in B5 [MARKS: 1]

Q.35 Which formula needs to be entered into cell D13 to automatically calculate the correct Postage and packing amount?

17

ANSWERS

a) =IF(D11$C$4,$E$5,IF(D11$C$3,$E$4,IF(D11>$C$5,$E$5,$E$3)) d) =IF(D11>=$C$3,$E$4,IF(D11>$C$5,$E$5,$E$3)) e) =IF(D11$C$5,$E$5,$E$4)) Correct: e) =IF(D11$C$5,$E$5,$E$4)) [MARKS: 2]

18

ANSWERS

Normal Distribution (Areas) f(z)

Area in the tail of the standardised Normal curve, N(0,1), for different positive values of z

Area

0

z

z

0.00

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.0 0.1 0.2 0.3 0.4

.50000 .46017 .42074 .38209 .34458

.49601 .45620 .41683 .37828 .34090

.49202 .45224 .41294 .37448 .33724

.48803 .44828 .40905 .37070 .33360

.48405 .44433 .40517 .36693 .32997

.48006 .44038 .40129 .36317 .32636

.47608 .43644 .39743 .35942 .32276

.47210 .43251 .39358 .35569 .31918

.46812 .42858 .38974 .35197 .31561

.46414 .42465 .38591 .34827 .31207

0.5 0.6 0.7 0.8 0.9

.30854 .27425 .24196 .21186 .18406

.30503 .27093 .23885 .20897 .18141

.30153 .26763 .23576 .20611 .17879

.29806 .26435 .23270 .20327 .17619

.29460 .26109 .22965 .20045 .17361

.29116 .25785 .22663 .19766 .17106

.28774 .25463 .22363 .19489 .16853

.28434 .25143 .22065 .19215 .16602

.28096 .24825 .21770 .18943 .16354

.27760 .24510 .21476 .18673 .16109

1.0 1.1 1.2 1.3 1.4

.15866 .13567 .11507 .09680 .08076

.15625 .13350 .11314 .09510 .07927

.15386 .13136 .11123 .09342 .07780

.15151 .12924 .10935 .09176 .07636

...