NOTESAnswer 5 questions(Only the first 5 questions answered will be marked)All questions carry equal marks.

TIME ALLOWED: Three hours and 10 minutes to read the paper.

INSTRUCTIONS: During the reading time you may write notes on the examination paper but you may not commencewriting in your answer book.

Marks for each question are shown. The pass mark required is 50% in total over the whole paper.

Start your answer to each question on a new page.

You are reminded that candidates are expected to pay particular attention to their communication skillsand care must be taken regarding the format and literacy of the solutions. The marking system willtake into account the content of the candidates' answers and the extent to which answers aresupported with relevant legislation, case law or examples where appropriate.

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2005

The Institute of Certified Public Accountants in Ireland, 9 Ely Place, Dublin 2.

THE INSTITUTE OF CERTIFIED PUBLIC ACCOUNTANTS IN IRELAND

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2005

Time Allowed: 3 hours and 10 minutes to read the paper Answer 5 questions Only the first five questions answered will be marked.

All questions carry equal marks.

1. The Superior Products company is purchasing a digital production machine. The cost accountant has received thefollowing quotations from the BIA financial services company. You propose to use the Net Present Value method toassess the alternative proposals which are set out below. The machine is expected to have a life of 8 years.

(i) Rent the machine for €120,000 pa payable in advance.(ii) Hire Purchase of the machine with a deposit of €150,000 and 7 equal annual payments of €100,000. At the

end of 8 years the machine is owned by the company and sold for €50,000(iii) Purchase the machine for €500,000 with a service contract of €40,000 p.a., payable in advance. At the end

of 8 years the machine will be sold for €50,000.

Recommend the most appropriate option to the company using a cost of capital of 14%.

[Total 20 Marks]

2. Trainees in the production department of DIB Ltd. have recently undergone professional examinations. Thecompany has received the distribution of examination results which are set out below.

Marks No. of Trainees

30 and less than 40 240 and less than 50 550 and less than 60 760 and less than 70 1370 and less than 80 1580 and less than 90 590 and less than 100 3

You are asked to provide the following information:

(i) The mean and the median marks. (10 Marks)

(ii) Explain what the values calculated in (i) indicate about the distribution of marks, particularly with regard to thedifference between the mean and median values.

(10 Marks)

[Total 20 Marks]

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3. The New Tyre Company has developed a new radial tyre supported by steel belts that will be sold through a chainof discount garages. Since this is a new tyre NTC considers that a guarantee offered with the tyre will aid itsmarketing. To support this you are asked to provide information on the number of miles the tyre will last. From roadtests, NTC has estimated the mean tyre mileage at 36,500 miles and the standard deviation at 5,000 miles. Thedata collected indicates a Normal distribution.

(i) Estimate the percentage of the tyres that last more than 40,000 miles. (8 Marks)

(ii) Estimate the guaranteed mileage if NTC does not want more than 10% to be eligible for the guarantee.(8 Marks)

(iii) Briefly describe the principles used in the above calculations. (4 Marks)

[Total 20 Marks]

4. A large multinational food company locates its fast-food outlets in areas where it estimates the youth population willprovide adequate sales. The following table provides an estimate of the local population and quarterly sales for 10restaurants.

Restaurant Population (000s) Quarterly Sales (€000)

1 2 58

2 6 105

3 8 88

4 8 118

5 12 117

6 16 137

7 20 157

8 20 169

9 22 149

10 26 202

You are required to:

(i) Draw a scatter diagram as a preliminary step in estimating the form of regression equation.(4 Marks)

(ii) Derive the regression between the variables and position the regression line on the scatter diagram.(8 Marks)

(iii) A further restaurant will be located in an area with a youth population of 21,000. Predict the quarterly salesfor this restaurant.

( 8 Marks)

[Total 20 Marks]

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5. Constructor Ltd. is one of the major house building companies in Leinster. Its records show that it has doubled itsconstruction programme of houses over the past 4 years with the usual peaks and valleys over time. The data from2001 to 2004 is shown in the table below which shows the number of houses (in thousands) in each quarter overthe four year period.

Quarter

Year 1 2 3 4

2001 16 23 28.6 27.1

2002 23.1 29.3 32.6 28.9

2003 23.6 29.1 32.4 29.6

2004 25.5 33.7 39.0 35.0

The MD asks you to prepare a report containing the following analyses.

(i) Plot the trend in the data using centred moving averages. (10 Marks)

(ii) Explain how this process smoothes the data by isolating the trend and cyclical variations. (10 Marks)

[Total 20 Marks]

6. As the investment adviser to CPA accounting you have been asked to deal with a wide range of problems. A numberof the problems are set out below. Provide detailed solutions to the following:

(i) A client wishes to repay a mortgage at €8,000 per year for 5 years at an interest rate of 12%. Calculate thelevel of mortgage required.

(6 Marks)

(ii) DIB Ltd produces batteries at a variable cost of €5 per unit and fixed costs of €20,000. The selling price perunit is €15. Write an equation to represent the company s costs and calculate the number of units requiredto make a profit of €8,000.

(6 Marks)

(iii) In (ii) above the following sample data was extracted from the battery production line. Sample mean = 950kg;sample standard deviation = 15kg; sample size = 36. Calculate the 95% confidence limits.

(8 Marks)

[Total 20 Marks]

END OF PAPER

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Suggested Solutions

BUSINESS MATHEMATICS & QUANTITATIVE METHODS

FORMATION 1 EXAMINATION - APRIL 2005

Solution 1.

Net Present Value of Proposals. In all three cases the payments for years 1 to 7 can be treated as an annuity

(i) Rental. The rent is payable in advance.

Year 0 1 - 7 Total

Cash Flow (€120,000) (€120,000)

Discount Factor 1 4.288

NPV (€120,000) (€514,560) (€634,560)

(ii) Hire Purchase. The deposit is paid in advance.

Year 0 1 - 7 8 Total

Cash Flow (€150,000) (€100,000) €50,000

Discount Factor 1 4.288 0.351

NPV (€150,000) (€428,800) €17,550 (€561,250)

(iii) Purchase. Cash payment and one service fee is paid in advance.

Year 0 1 - 7 8 Total

Cash Flow (€500,000)

(€40,000) (€40,000) €50,000

Discount Factor 1 4.288 0.351

NPV (€540,000) (€171,520) €17,550 (€693,970)

Due to the lower cash out-flow, the hire-purchase option provides the best investment opportunity for thecompany. This reinforces the principle of net present value — if payments can be deferred over a longer period,the discounted cash flow is of greater benefit than paying cash in current terms.

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Solution 2.

(i) The mean and median. Assuming that the data is continuous gives the mid points in the table.

Class f Cum x fx

Boundaries freq

30 — 40 2 2 35 70

40 — 50 5 7 45 225

50 — 60 7 14 55 385

60 — 70 13 27 65 845

70 — 80 15 42 75 1125

80 — 90 5 47 85 425

90 — 100 3 50 95 285

50 3360

Mean = x = ∑fx = 3360 = 67.2∑f 50

Median = Lm + {N/2 - Fm-1} x Cm{ fm }

where

Lm = lower boundary of median class (60 — 70)

Fm-1 = cumulative frequency of class prior to median class (14)

Fm = frequency of median class (13)

Cm = median class width (10), N = ∑f = 50

Therefore, median = 60 + { 51 - 14} x 102

13

= 68.85.

(ii) The median has distinct advantages over the mean particularly in open ended distributions. The mean is generallyused instead of the median especially if there is not much difference between them. The difference between themdepends on the skewness of the distribution. The median splits the area under the frequency curve into two halves.If the distribution is symmetrical, then the mean and median will coincide. With a negatively skewed distribution themedian exceeds the mean and with a positively skewed distribution the mean exceeds the median. In this case themedian is greater than the mean. The distribution therefore has a negative skew. This demonstrates that there is agreater spread of marks amongst the trainees at the lower end of the scale.

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Solution 3.

(i) At x = 40,000, z = x - µ where µ = 36,500, σ = 5,000σ

z = 40,000 - 36,5005000

= 0.70.

A Value of x = 40,000 on the Normal distribution corresponds to a value of z = 0.7 on the standard normal distribution.From the tables the area between the mean and z = 0.70 is 0.2580. Thus 0.5000 - 0.2580 = 0.2420 is the probabilitythat it will exceed 40,000 miles. That is 24.2% of the tyres will exceed 40,000 miles. The appropriate diagram is shownbelow.

(σ) = 5,000

P(x ‡ 40,000)P = 0.2580

Prob µ = 36,500 40,000 Xi

0 0.5

0 0.7 Z

(ii) According to the diagram below, 40% of the area must be between the mean and the guaranteed mileage.

(σ) = 5,00010% of tyres eligible

for discount

Prob µ = 36,500 Xi

Guaranteed mileage

0.4000, from the tables, is approximately 1.28 (0.3997) standard deviations below the mean, that is Z = - 1.28.

To find the corresponding mileage z = x - µ = -1.28 σ

x - µ = - 1.28σ

x = - 1.28σ + µ,

with µ = 36,500 and σ = 5,000

x = 36,500 - 1.28 (5000) = 30,100.

Thus a guarantee of 30,100 miles will meet the requirement that 10% of the tyres will be eligible for the guarantee.

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(iii) The normal distribution plays an important role in business applications since many business generate randomvariables with probability distributions that are well approximated by a normal distribution. It is used to detectdisagreement between a set of data and the assumption of normality. It is perfectly symmetrical about its mean µ.It is defined as a normal distribution with µ = 0 and σ = 1. A random variable with a standard normal distribution isdenoted by z and can be read from tables to find probabilities. To apply the tables to a normal random variable (x)we must first convert the value of x to a z-score. The population z-score for a measurement is defined as thedistance between the measurement and the population mean divided by the population standard deviation. Thusthe z-score gives the distance between a measurement and the mean in units equal to the standard deviation. Thatis z is a standard normal random variable such that z = x - µ

σ

Therefore, all probabilities associated with standard normal random variables can be depicted as areas under thestandard normal curve.

Solution 4.

(i)

Restaurant Population Quarterly Sales

(000s) x (€000) y

1 2 58

2 6 105

3 8 88

4 8 118

5 12 117

6 16 137

7 20 157

8 20 169

9 22 149

10 26 202

The size of the population is shown on the horizontal axis (the independent variable x) and the value of quarterlysales on the vertical axis (the dependent variable y). A scatter diagram enables us to observe the data graphicallyand to draw preliminary conclusions about the possible relationship between the variables. Quarterly sales appearto be higher in locations with larger populations . For this data, the relationship between the population size andsales appears to be approximated by a straight line — there seems to be a positive linear relationship between xand y> We therefore choose the simple linear regression model to represent the relationship between sales andpopulation. The next step is to determine the values of a and b in the estimated regression equation where y = a + bx and

y = estimated value of quarterly sales (€000)

x = the size of the population (000) for each restaurant

a = the y intercept of the regression line

b = the slope of the estimated regression line.

The scatter diagram is shown below (and the regression equation in part (ii) is superimposed on it.

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(ii) The linear regression equation is: y = a + bx where

∑y = na + b∑x∑xy = a ∑ x + b ∑x2 ; and

a = ∑y - b∑xn n

b = n∑xy - ∑x∑y/nn∑x2 - (∑x)2 /n

Restaurant x y x2 xy

1 2 58 4 1162 6 105 36 6303 8 88 64 7044 8 118 64 9445 12 117 144 1,4046 16 137 256 2,1927 20 157 400 3,1408 20 169 400 3,3809 22 149 484 2,27810 26 202 676 5,252

140 1,300 2,528 21,040

Inserting values gives

b = 21,040 - 140 x 1,300/102,528 - 1402/10

= 21,040 - 18,200 = 52528 - 1,960

a = 1300 - 5 x 140 = 6010 10

Therefore, y = 60 + 5x.

(iii) When x = 21, y = 60 + 105 = 165.

Quarterly sales for this restaurant is predicted at €165,000.

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Solution 5.

(i) The trend by means of centred moving averages.

Quarter

Year 1 2 3 4

2001 16 23 28.6 27.1

2002 23.1 29.3 32.6 28.9

2003 23.6 29.1 32.4 29.6

2004 25.5 33.7 39.0 35.0

Year Quarter New Houses Moving Moving Av. Centred (000s) Total Moving Av (TxC)

1 1 16.02 23.03 28.6 94.7 23.675 24.564 27.1 101.8 25.450 26.24

2 1 23.1 108.1 27.025 27.532 29.3 112.1 28.025 28.253 32.6 113.9 28.475 28.544 28.9 114.4 28.600 28.58

3 1 23.6 114.2 28.550 28.532 29.1 114.0 28.500 28.593 32.4 114.7 28.675 28.914 29.6 116.6 29.150 29.73

4 1 25.5 121.2 30.300 31.132 33.7 127.8 31.950 32.633 39.0 133.2 33.3004 35.0

Trend — smoothed data.

(ii) The data given in part (i) is a known time series. By drawing the centred moving average we are decomposing it toisolate the combined trend and cyclical components denoted by TxC. The isolation of TxC is based on the averageexcluding the effects due to seasonal and irregular variations. Because the data are listed by quarters it gives usthe opportunity to isolate the trend and cyclical components by combining four quarters. The peaks and valleys thatoccur within each year are "smoothed". The column of centred moving averages represents estimates of thecombined trend and cyclical components.

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Recognising that the trend (T) is a long-term increasing or decreasing pattern and cyclical variations (C) is apattern that repeats over time periods longer than one year, the combined effect of the TxC component is a longerterm effect, that is, it eliminates the short-term effects from seasonal and irregular variations. The graph of thesmoothed data gives the long term picture. It is possible to further decompose TxC so that the individual trend (T)and cyclical (C) components can be obtained. The trend equation is derived by regression analysis using thedependent variable Y (TxC) and independent X (quarters). The trend value can be derived for each quarter.

The cyclical variation is then derived by C = TxC (from the data in part (i)T (from the regression line).

Values of C are therefore expressed as index numbers that multiply C. In this way the value of C can beseparated.

Solution 6.

(i) This is an annuity of 5 annual payments of €8,000. The interest rate is 12 per year. The present value of an annuity(mortgage) is

PV = 8,000 + + 8,0001.121 1.125

PV = a(1 - rn) where1 - r

a = regular sum received X discount factor,

r = common ratio = 1/(1 + i), i = rate of discount,

n = number of time periods.

Therefore, PV = 8,000 x 1 - (1/1.12)5

1.121 1 - (1/1.12)

= 7,143 x 1 - 0.8935

1 - 0.893

= 7,143 x 0.4420.107

= €29,506.

(ii) Equation to represent company s costs y = ax + b

where y = total costs, x = number of units produced

a = cost per unit, b = fixed costs.

Therefore y = 5x + 20,000.

To generate a profit of €8,000: Profit = total revenue - total costs

8,000 = 15x - (5x + 20,000)

28,000 = 10x

x = 2,800 units.

(iii) 95% confidence interval and limits for µµ = x – 1.96 (s/ˆn)

= 950 – 1.96 (15/ˆ36)

= 950 – 1.96 (2.5)

= 950 – 4.9.

Therefore, µ = 945.1kg to 954.9kg.

We are 95% confident that the mean will lie within these limits

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