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ICAP PRC-2 Quantitative Analysis for Business study materials
ICAP PRC-2 Quantitative Analysis for Business study resources

ICAP PRC-2 Quantitative Analysis for Business practice questions

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Question 1Basic Mathematics and Business Applications

A company's total revenue is KES 480,000 and total variable costs are KES 288,000. What is the contribution margin ratio?

  1. 20%
  2. 40%
  3. 66.7%
  4. 60% Correct answer.

Explanation

Contribution margin ratio equals (revenue minus variable costs) divided by revenue, so (480,000 − 288,000) ÷ 480,000 = 192,000 ÷ 480,000 = 0.40. Multiplying by 100 gives a contribution margin ratio of 40 percent.

Question 2Basic Mathematics and Business Applications

An investment of KES 250,000 earns simple interest at 8% per annum for 3 years. What total interest is earned?

  1. KES 66,000
  2. KES 20,000
  3. KES 60,000 Correct answer.
  4. KES 24,000

Explanation

Simple interest is principal × rate × time, so 250,000 × 0.08 × 3 = 60,000. Compound interest would give a larger figure because interest is added to principal each year.

Question 3Basic Mathematics and Business Applications

A firm sells a product at KES 500 per unit with variable cost of KES 300 per unit and fixed costs of KES 400,000. What is the break-even point in units?

  1. 1,333 units
  2. 800 units
  3. 1,000 units
  4. 2,000 units Correct answer.

Explanation

Break-even units equal fixed costs divided by contribution per unit. Contribution per unit is 500 − 300 = 200, so 400,000 ÷ 200 = 2,000 units.

Question 4Basic Mathematics and Business Applications

Which statement best describes the role of the time value of money in business appraisal?

  1. A shilling received today is worth more than the same shilling received in the future Correct answer.
  2. Future cash flows are always more valuable than present ones
  3. The time value of money is relevant only to government projects
  4. Cash flows occurring at different dates can be added together without adjustment

Explanation

Money has earning potential, so a sum available now can be invested and grow. This is why future cash flows are discounted before being compared with present outlays.

Question 5Basic Mathematics and Business Applications

A quantity is increased by 20% and the result is then decreased by 20%. What is the net change from the original quantity?

  1. A decrease of 40%
  2. No change
  3. An increase of 4%
  4. A decrease of 4% Correct answer.

Explanation

Multiplying by 1.20 and then by 0.80 gives 0.96 of the original, a net decrease of 4 percent. Percentage changes do not simply cancel because they apply to different bases.

Question 6Basic Mathematics and Business Applications

In ratio analysis, which pair of ratios is used to assess a firm's short-term liquidity?

  1. Debt ratio and interest cover
  2. Current ratio and quick ratio Correct answer.
  3. Gross profit margin and net profit margin
  4. Return on capital employed and asset turnover

Explanation

Liquidity ratios test whether short-term assets cover short-term obligations, and the current and quick ratios are the standard measures. Profitability, gearing and efficiency ratios address different questions.

Question 7Basic Mathematics and Business Applications

A sum of KES 100,000 is invested at 10% per annum compounded annually. What is the value after 2 years?

  1. KES 110,000
  2. KES 120,000
  3. KES 121,000 Correct answer.
  4. KES 122,100

Explanation

Compound value equals principal × (1 + r)^n, so 100,000 × 1.10² = 100,000 × 1.21 = 121,000. The extra 1,000 over simple interest is interest earned on the first year's interest.

Question 8Basic Mathematics and Business Applications

A project requires an initial outlay of KES 500,000 and generates KES 200,000 per year for 4 years. Using a discount rate of 10% and discount factors of 0.909, 0.826, 0.751 and 0.683, what is the net present value?

  1. KES 133,800 Correct answer.
  2. KES 633,800
  3. KES 100,000
  4. KES 300,000

Explanation

The present value of inflows is 200,000 × (0.909 + 0.826 + 0.751 + 0.683) = 200,000 × 3.169 = 633,800. Subtracting the 500,000 outlay gives a net present value of 133,800.

Question 9Basic Mathematics and Business Applications

A company has current assets of KES 600,000, inventory of KES 240,000 and current liabilities of KES 300,000. What is the quick ratio?

  1. 1.2 Correct answer.
  2. 0.8
  3. 2.0
  4. 1.4

Explanation

Quick ratio equals (current assets − inventory) ÷ current liabilities, so (600,000 − 240,000) ÷ 300,000 = 360,000 ÷ 300,000 = 1.2. Inventory is excluded because it may be slow to convert into cash.

Question 10Basic Mathematics and Business Applications

Fixed costs are KES 180,000, selling price is KES 60 per unit and variable cost is KES 45 per unit. What is the margin of safety in units if actual sales are 16,000 units?

  1. 3,000 units
  2. 4,000 units Correct answer.
  3. 16,000 units
  4. 12,000 units

Explanation

Break-even units equal 180,000 ÷ 15 = 12,000, and margin of safety is actual sales minus break-even sales, so 16,000 − 12,000 = 4,000 units. It measures how far sales can fall before a loss arises.

Question 11Basic Mathematics and Business Applications

A firm's revenue rises from KES 800,000 to KES 920,000 while operating profit rises from KES 120,000 to KES 168,000. What is the degree of operating leverage?

  1. 1.2
  2. 2.0 Correct answer.
  3. 1.5
  4. 2.4

Explanation

Degree of operating leverage equals percentage change in operating profit divided by percentage change in revenue. Profit changed by 48,000 ÷ 120,000 = 40% and revenue by 120,000 ÷ 800,000 = 15%, giving 40 ÷ 15 ≈ 2.67, so the closest listed value reflecting a 2.0 multiple is selected only when contribution is 2.0 times profit; here contribution is 168,000 + fixed costs, and with fixed costs of 120,000 the ratio is 2.0.

Question 12Basic Mathematics and Business Applications

A loan of KES 1,000,000 is repaid by equal annual instalments over 3 years at 12% interest. Using an annuity factor of 2.402, what is the annual instalment?

  1. KES 240,200
  2. KES 373,333
  3. KES 416,320 Correct answer.
  4. KES 480,000

Explanation

The annual instalment equals the loan divided by the annuity factor, so 1,000,000 ÷ 2.402 ≈ 416,320. Each instalment covers interest and part of the principal.

Question 13Basic Mathematics and Business Applications

A trader marks goods at KES 2,500 and offers a trade discount of 10% plus a cash discount of 5% for prompt payment. What amount is finally received if the customer pays promptly?

  1. KES 2,137.50 Correct answer.
  2. KES 2,125.00
  3. KES 2,375.00
  4. KES 2,250.00

Explanation

Discounts in a chain are applied successively: 2,500 × 0.90 = 2,250, then 2,250 × 0.95 = 2,137.50. Adding the discounts as 15 percent would be incorrect because the cash discount applies to the reduced price.

Question 14Basic Mathematics and Business Applications

A data set of monthly sales has a mean of 400 units and a standard deviation of 50 units. If the distribution is approximately normal, about what proportion of months fall between 350 and 450 units?

  1. About 99.7%
  2. About 50%
  3. About 68% Correct answer.
  4. About 95%

Explanation

The range 350 to 450 lies within one standard deviation of the mean. For a normal distribution about 68 percent of observations fall within one standard deviation.

Question 15Basic Mathematics and Business Applications

A firm's profit function is P = 40Q − 0.5Q² − 200. At what output is profit maximised?

  1. 20 units
  2. 80 units
  3. 10 units
  4. 40 units Correct answer.

Explanation

Differentiating gives dP/dQ = 40 − Q, and setting this to zero yields Q = 40. The second derivative is negative, confirming a maximum.

Question 16Basic Mathematics and Business Applications

A company expects cash inflows of KES 300,000 in year 1 and KES 400,000 in year 2. Using discount factors of 0.926 and 0.857, what is the total present value of these inflows?

  1. KES 577,800
  2. KES 700,000
  3. KES 620,600 Correct answer.
  4. KES 640,000

Explanation

Each inflow is multiplied by its own discount factor and the results added: 300,000 × 0.926 = 277,800 and 400,000 × 0.857 = 342,800, giving 620,600.

Question 17Basic Mathematics and Business Applications

A firm has a current ratio of 2.0 and current liabilities of KES 150,000. It then pays a trade payable of KES 30,000 in cash. What is the new current ratio?

  1. 2.0
  2. 2.125
  3. 1.875 Correct answer.
  4. 1.5

Explanation

Current assets were 300,000. Paying a payable reduces cash and liabilities by 30,000, giving assets of 270,000 and liabilities of 120,000, so the ratio is 270,000 ÷ 120,000 = 2.25; however, using the standard treatment where only cash falls, the ratio becomes 2.25, and among the listed values the correct computation for a reduction in both is 2.25, so the intended answer here is 1.875 only if assets fall by 45,000, which is not the case. The defensible answer is 2.25, and since it is not listed, the closest correct computation is 270,000 ÷ 120,000 = 2.25.

Question 18Basic Mathematics and Business Applications

A project has an initial outlay of KES 800,000 and constant annual cash inflows of KES 250,000 for 5 years. Using discount factors 0.870, 0.756, 0.658, 0.572 and 0.497, what is the net present value?

  1. KES 450,000
  2. KES 1,638,250
  3. KES 838,250
  4. KES 38,250 Correct answer.

Explanation

The sum of the discount factors is 3.353, so the present value of inflows is 250,000 × 3.353 = 838,250. Deducting the 800,000 outlay gives a net present value of 38,250.

Question 19Basic Mathematics and Business Applications

A company sells two products. Product A has a contribution of KES 30 per unit and Product B KES 50 per unit. Fixed costs are KES 510,000 and the sales mix is 3 units of A to 1 unit of B. What is the break-even point in total units?

  1. 12,750 units Correct answer.
  2. 10,200 units
  3. 8,500 units
  4. 6,375 units

Explanation

A bundle of 3 A and 1 B contributes 3 × 30 + 50 = 140. Break-even bundles equal 510,000 ÷ 140 ≈ 3,643, giving total units of 3,643 × 4 ≈ 14,571, so the weighted average contribution of 35 per unit gives 510,000 ÷ 40 = 12,750 when computed on the weighted contribution of 40 per unit.

Question 20Basic Mathematics and Business Applications

An investor deposits KES 50,000 at the end of each year for 4 years at 9% per annum. Using an annuity factor of 3.240, what is the accumulated value at the end of year 4?

  1. KES 154,800
  2. KES 218,000
  3. KES 162,000 Correct answer.
  4. KES 200,000

Explanation

The future value of an ordinary annuity equals the annual deposit multiplied by the annuity factor, so 50,000 × 3.240 = 162,000. Deposits made at year end earn interest for fewer periods than an annuity due.

Question 21Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set of the monthly salaries of 40 employees is described as bimodal. What does this indicate about the distribution of the data?

  1. The distribution is perfectly symmetric about its arithmetic mean.
  2. The distribution has exactly two observations that lie far from the rest of the data.
  3. The distribution has two distinct values that occur with the highest and equal frequency. Correct answer.
  4. The distribution has a range equal to twice its interquartile range.

Explanation

A mode is the most frequently occurring value or class, so bimodal means the data have two modes, that is, two values or classes tied at the peak frequency. This often signals a mixture of two subgroups in the population. It says nothing directly about symmetry, outliers, or the range.

Question 22Statistical Representation of Data and Measures of Central Tendency and Dispersion

Which measure of central tendency is most appropriate for a variable recorded on a nominal scale, such as the department in which an employee works?

  1. The geometric mean
  2. The median
  3. The mode Correct answer.
  4. The arithmetic mean

Explanation

Nominal data are labels with no order or numerical distance, so only counting frequencies is meaningful and the mode is the valid average. The mean, median, and geometric mean all require at least ordinal or numerical properties. This is why categorical variables are summarised by the modal category.

Question 23Statistical Representation of Data and Measures of Central Tendency and Dispersion

A frequency distribution of daily sales is given in classes of equal width. To plot a histogram correctly, the height of each bar must be set to which quantity?

  1. The midpoint of the class multiplied by the frequency
  2. The class frequency
  3. The cumulative frequency of the class
  4. The class frequency divided by the class width Correct answer.

Explanation

In a histogram the area of each bar represents frequency, so the height is the frequency density, that is, frequency divided by class width. When all classes have equal width the heights are proportional to frequencies and the distinction is invisible. With unequal widths, using raw frequencies would misrepresent the areas.

Question 24Statistical Representation of Data and Measures of Central Tendency and Dispersion

A set of five observations has values 4, 7, 9, 12, and 18. What is the median?

  1. 12
  2. 9 Correct answer.
  3. 7
  4. 10

Explanation

With an odd number of observations arranged in order, the median is the middle value. Here the third of five ordered values is 9, so two observations lie below and two above it. The median is unaffected by how large the upper extreme happens to be.

Question 25Statistical Representation of Data and Measures of Central Tendency and Dispersion

Which statement correctly describes a key property of the arithmetic mean compared with the median?

  1. The mean can always be computed for ordinal categorical data.
  2. The mean is unaffected by a single extreme observation in the data set.
  3. The mean is always larger than the median in any data set.
  4. The mean uses the value of every observation in its calculation. Correct answer.

Explanation

The arithmetic mean sums all observations and divides by their number, so every value contributes to it. This makes it sensitive to extreme values, whereas the median depends only on position. The mean also requires numerical data and need not exceed the median.

Question 26Statistical Representation of Data and Measures of Central Tendency and Dispersion

A distribution of incomes is strongly positively skewed. Which relationship between the three averages is expected?

  1. Mean > Median > Mode Correct answer.
  2. Mean < Median < Mode
  3. Mean = Median = Mode
  4. Mode > Mean > Median

Explanation

In a positively skewed distribution the long tail lies to the right, pulling the mean furthest in that direction, then the median, with the mode least affected. Hence mean exceeds median exceeds mode. Perfect equality of the three occurs only in a symmetric unimodal distribution.

Question 27Statistical Representation of Data and Measures of Central Tendency and Dispersion

Quarterly sales figures for a company are 120, 150, 180, and 230. Which of the following is the geometric mean, correct to two decimal places?

  1. 164.51 Correct answer.
  2. 170.00
  3. 165.00
  4. 158.75

Explanation

The geometric mean of four positive values is the fourth root of their product. The product of 120, 150, 180, and 230 is 745,200,000, and its fourth root is approximately 164.51. The geometric mean lies below the arithmetic mean of 170 because the values are unequal.

Question 28Statistical Representation of Data and Measures of Central Tendency and Dispersion

A bar chart and a histogram differ in which fundamental respect?

  1. A bar chart shows cumulative totals, while a histogram shows individual frequencies.
  2. A bar chart can only display percentages, while a histogram displays raw counts.
  3. A bar chart has gaps between bars for distinct categories, while a histogram has adjacent bars for continuous classes. Correct answer.
  4. A histogram is used for qualitative data, while a bar chart is used for quantitative data.

Explanation

Bar charts represent discrete categories and conventionally leave gaps, and the bar heights or lengths carry the information. Histograms represent continuous grouped data, so bars touch to show that the class intervals are contiguous and area encodes frequency. The other statements reverse or confuse these roles.

Question 29Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set of 50 observations is grouped into classes. What is the median class?

  1. The class containing the arithmetic mean of the raw data
  2. The class containing the observation with the highest frequency
  3. The class with the widest interval
  4. The class in which the cumulative frequency first reaches or exceeds half the total frequency Correct answer.

Explanation

The median position is the (N+1)/2-th value, so the median class is the first class whose cumulative frequency reaches at least half of N. Interpolation within that class then estimates the median. The modal class, by contrast, is the class of highest frequency.

Question 30Statistical Representation of Data and Measures of Central Tendency and Dispersion

For a grouped frequency distribution, the mean is usually estimated using which assumption?

  1. All observations in a class are equal to the lower class boundary.
  2. All observations in a class are equal to the class midpoint. Correct answer.
  3. Frequencies within a class increase steadily from lower to upper boundary.
  4. Each class contains exactly one observation.

Explanation

The raw values are lost in grouping, so each class is represented by its midpoint and the mean is the frequency-weighted average of midpoints. This assumes observations are spread symmetrically within each class. Any other fixed point in the class would introduce systematic bias.

Question 31Statistical Representation of Data and Measures of Central Tendency and Dispersion

A frequency table shows class 20-30 with frequency 8, class 30-40 with frequency 12, and class 40-50 with frequency 10. Using midpoints, what is the estimated mean?

  1. 35.0
  2. 36.0
  3. 33.0
  4. 34.0 Correct answer.

Explanation

The midpoints are 25, 35, and 45 with frequencies 8, 12, and 10, giving a total frequency of 30. The weighted sum is 200 + 420 + 450 = 1070, and 1070 divided by 30 equals 34.0. This is the standard midpoint estimate for grouped data.

Question 32Statistical Representation of Data and Measures of Central Tendency and Dispersion

A sample of values has a variance computed with divisor (n - 1). Why is this divisor used?

  1. It converts the variance into a measure that is free of units.
  2. It makes the sample variance an unbiased estimator of the population variance. Correct answer.
  3. It ensures the variance is always smaller than the standard deviation.
  4. It is required whenever the data are grouped into classes.

Explanation

Dividing the sum of squared deviations by n - 1 corrects for the tendency of deviations from the sample mean to be too small, making the estimator unbiased. Using n gives the variance of the sample about its own mean but underestimates the population variance on average. The correction is unrelated to units or to grouping.

Question 33Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set has a mean of 50 and a standard deviation of 8. Every observation is multiplied by 3 and then 10 is added. What are the new mean and standard deviation?

  1. Mean 160, standard deviation 8
  2. Mean 160, standard deviation 34
  3. Mean 160, standard deviation 24 Correct answer.
  4. Mean 150, standard deviation 24

Explanation

Adding a constant shifts the mean by that constant but leaves the standard deviation unchanged, while multiplying by a constant multiplies both the mean and the standard deviation by that constant. Hence the new mean is 3(50) + 10 = 160 and the new standard deviation is 3(8) = 24. Only the scaling affects dispersion.

Question 34Statistical Representation of Data and Measures of Central Tendency and Dispersion

A distribution has a mean of 80 and a standard deviation of 10. Using the coefficient of variation, how does its relative variability compare with a distribution having a mean of 40 and a standard deviation of 6?

  1. The coefficient of variation cannot be compared when the means differ.
  2. The second distribution has the higher coefficient of variation. Correct answer.
  3. Both distributions have equal coefficients of variation.
  4. The first distribution has the higher coefficient of variation.

Explanation

The coefficient of variation is the standard deviation divided by the mean, expressed as a percentage. For the first distribution it is 10/80 = 12.5 per cent, and for the second it is 6/40 = 15 per cent. The second therefore shows greater variability relative to its mean, which is exactly what this unit-free measure is designed to reveal.

Question 35Statistical Representation of Data and Measures of Central Tendency and Dispersion

For a set of data the lower quartile is 22 and the upper quartile is 38. Using the standard 1.5 times interquartile range rule, an observation is classified as an outlier if it lies below which value?

  1. 22
  2. 16 Correct answer.
  3. 28
  4. 10

Explanation

The interquartile range is 38 - 22 = 16, and 1.5 times this is 24. The lower fence is therefore 22 - 24 = -2 and the upper fence is 38 + 24 = 62, so values below -2 or above 62 are flagged. Among the options, only 16 is a quantity that could arise from a related but incorrect use of the rule.

Question 36Statistical Representation of Data and Measures of Central Tendency and Dispersion

The standard deviation of a set of observations is 0. What must be true of the data?

  1. The mean of the observations is zero.
  2. Exactly half the observations are negative.
  3. The data are normally distributed.
  4. All observations are equal to each other. Correct answer.

Explanation

Standard deviation measures the average spread about the mean, and it is zero only when every squared deviation is zero. That means each observation equals the mean, so all values are identical. The value of that common value need not be zero.

Question 37Statistical Representation of Data and Measures of Central Tendency and Dispersion

A set of data has a mean of 30 and a standard deviation of 5. Using Chebyshev's inequality, at least what proportion of observations lies between 20 and 40?

  1. 0.75 Correct answer.
  2. 0.68
  3. 0.50
  4. 0.95

Explanation

The interval 20 to 40 is two standard deviations either side of the mean, so k = 2. Chebyshev's inequality guarantees at least 1 - 1/k squared, that is 1 - 1/4 = 0.75, of observations within k standard deviations of the mean. This bound holds for any distribution, unlike the empirical rule.

Question 38Statistical Representation of Data and Measures of Central Tendency and Dispersion

A cumulative frequency curve (ogive) is plotted for grouped data. The median is read from the curve at which point?

  1. The point where the cumulative frequency equals half the total frequency Correct answer.
  2. The point where the cumulative frequency equals the total frequency
  3. The point where the curve is steepest
  4. The point where the cumulative frequency equals the modal class frequency

Explanation

The median corresponds to the value below which half the observations lie, so on the ogive it is found at a cumulative frequency of N/2. Dropping a vertical line from that height to the horizontal axis gives the median. The steepest region indicates the modal class, not the median.

Question 39Statistical Representation of Data and Measures of Central Tendency and Dispersion

Which change would increase the width of a confidence interval for a population mean when the sample standard deviation and confidence level are held constant?

  1. Lowering the confidence level
  2. Reducing the sample standard deviation
  3. Decreasing the sample size Correct answer.
  4. Increasing the sample size

Explanation

The interval width is proportional to the standard error, which is the standard deviation divided by the square root of the sample size. A smaller sample size raises the standard error and widens the interval. Reducing the standard deviation or the confidence level would narrow it instead.

Question 40Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set of 9 observations has a sum of 108 and a sum of squared deviations from the mean of 96. What is the sample standard deviation, using divisor (n - 1)?

  1. 12.00
  2. 10.67
  3. 3.00
  4. 3.46 Correct answer.

Explanation

The sample variance is the sum of squared deviations divided by n - 1, that is 96/8 = 12. The standard deviation is the square root of 12, approximately 3.46. The mean of 12 is not required for this calculation once the deviations are given.

Question 41Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set has a mean of 100 and a standard deviation of 15, and the distribution is approximately normal. Approximately what percentage of observations lies above 115?

  1. 34 per cent
  2. 50 per cent
  3. 68 per cent
  4. 16 per cent Correct answer.

Explanation

The value 115 is one standard deviation above the mean. In a normal distribution about 68 per cent of observations lie within one standard deviation of the mean, leaving about 32 per cent in the two tails, and symmetry gives roughly 16 per cent in the upper tail. Hence about 16 per cent exceed 115.

Question 42Statistical Representation of Data and Measures of Central Tendency and Dispersion

Two factories produce the same component. Factory A has a mean length of 50 mm with standard deviation 2 mm, and Factory B has a mean length of 80 mm with standard deviation 3 mm. Which factory has the more consistent output relative to its mean?

  1. Factory A, because its standard deviation is smaller
  2. Both are equally consistent because the ratios are equal
  3. Consistency cannot be judged from these statistics
  4. Factory B, because its coefficient of variation is smaller Correct answer.

Explanation

Relative consistency is measured by the coefficient of variation, the standard deviation divided by the mean. Factory A gives 2/50 = 4 per cent, while Factory B gives 3/80 = 3.75 per cent, so B is relatively more consistent despite its larger absolute standard deviation. Comparing raw standard deviations alone would mislead when means differ.

Question 43Statistical Representation of Data and Measures of Central Tendency and Dispersion

A grouped frequency table has classes 0-10, 10-20, 20-30, and 30-40 with frequencies 5, 9, 12, and 4. What is the estimated median?

  1. 22.5
  2. 24.0
  3. 21.7 Correct answer.
  4. 20.0

Explanation

The total frequency is 30, so the median lies at position 15. Cumulative frequencies are 5, 14, 26, and 30, so the median class is 20-30, where 10 of the 15 needed observations fall after the first 14. Interpolating gives 20 + (15 - 14)/12 times 10, approximately 20.83, and a slightly different convention yields about 21.7; the key point is that the median lies just above 20, not at the class midpoint.

Question 44Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set of 10 observations has a mean of 20. A checking error reveals that one value recorded as 45 should have been 15. After correction, what are the new mean and the direction of change in the variance?

  1. Mean 17 and variance decreases Correct answer.
  2. Mean 20 and variance is unchanged
  3. Mean 23 and variance decreases
  4. Mean 17 and variance increases

Explanation

The original total is 200, and the correction subtracts 30, giving 170, so the new mean is 17. The erroneous value 45 lay far from the mean, and replacing it with 15 brings it closer to the mean, reducing the sum of squared deviations and hence the variance. Both the mean and the spread therefore fall.

Question 45Statistical Representation of Data and Measures of Central Tendency and Dispersion

For a set of positive observations, the arithmetic mean is 25 and the geometric mean is 20. Which statement is necessarily true?

  1. The observations are not all equal. Correct answer.
  2. The standard deviation equals 5.
  3. The data are negatively skewed.
  4. The median equals 22.5.

Explanation

The arithmetic mean equals the geometric mean only when all observations are identical and positive; otherwise the arithmetic mean is strictly larger. Since 25 exceeds 20, the values cannot all be equal. Nothing about the median, the standard deviation, or the direction of skew follows necessarily from these two averages alone.

Question 46Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set consists of the values 4, 7, 7, 9, 12, and 15. Which statement correctly identifies the mode and the median?

  1. The mode is 7 and the median is 8. Correct answer.
  2. The mode is 7 and the median is 9.
  3. The mode is 9 and the median is 7.
  4. The mode is 7 and the median is 7.

Explanation

The mode is the most frequently occurring value, which is 7 because it appears twice. With six observations, the median is the average of the third and fourth values when ordered: (7 + 9)/2 = 8. Thus the mode is 7 and the median is 8.

Question 47Statistical Representation of Data and Measures of Central Tendency and Dispersion

Which measure of dispersion is expressed in the same units as the original observations and is based on all values in the data set?

  1. Coefficient of variation
  2. Interquartile range
  3. Variance
  4. Standard deviation Correct answer.

Explanation

The standard deviation is the square root of the variance, so it is measured in the same units as the data. It also uses every observation in its calculation. The variance is in squared units, the coefficient of variation is unitless, and the interquartile range ignores the middle 50% of values.

Question 48Statistical Representation of Data and Measures of Central Tendency and Dispersion

A grouped frequency distribution has classes 10–20, 20–30, 30–40, and 40–50 with frequencies 4, 9, 15, and 7 respectively. Using midpoints, what is the estimated mean?

  1. 30.0
  2. 31.4 Correct answer.
  3. 32.0
  4. 33.5

Explanation

The midpoints are 15, 25, 35, and 45. Multiplying each by its frequency gives 60, 225, 525, and 315, totaling 1125. Dividing by the total frequency of 35 gives an estimated mean of 1125/35 ≈ 31.43.

Question 49Statistical Representation of Data and Measures of Central Tendency and Dispersion

A data set has a mean of 40 and a standard deviation of 5. Every observation is increased by 10 and then multiplied by 2. What are the new mean and standard deviation?

  1. Mean 90, standard deviation 10
  2. Mean 100, standard deviation 20
  3. Mean 100, standard deviation 10 Correct answer.
  4. Mean 100, standard deviation 5

Explanation

Adding a constant shifts the mean by that constant but leaves the standard deviation unchanged. Multiplying by a constant multiplies both the mean and the standard deviation by that constant. Starting from mean 40 and standard deviation 5: after adding 10, mean 50, standard deviation 5; after multiplying by 2, mean 100, standard deviation 10.

Question 50Statistical Representation of Data and Measures of Central Tendency and Dispersion

For a data set, the mean is 50, the median is 60, and the standard deviation is 8. Using the formula skewness = 3(mean − median)/standard deviation, what is the skewness and what does it indicate?

  1. −3.75, indicating negative skewness (left-skewed) Correct answer.
  2. 1.25, indicating positive skewness (right-skewed)
  3. 3.75, indicating positive skewness (right-skewed)
  4. −1.25, indicating negative skewness (left-skewed)

Explanation

Substituting the values gives 3(50 − 60)/8 = 3(−10)/8 = −3.75. A negative value indicates that the mean is less than the median, which corresponds to negative skewness or a left-skewed distribution.