Basic Statistics

For a symmetric distribution the kurtosis coefficient is $\beta_2=\mu_4/\sigma^4$, compared with 3 (the normal value). Here $\sigma=7$, so $\sigma^4=7^4=2401$ and $3\sigma^4=7203$.

(i) Mesokurtic ($\beta_2=3$): $\mu_4=3\times2401=7203$.

(ii) Platykurtic ($\beta_2<3$): $\mu_4<7203$.

(iii) Leptokurtic ($\beta_2>3$): $\mu_4>7203$.

For a symmetric distribution, kurtosis is judged by $\beta_2=\mu_4/\sigma^4$ against 3. With $\sigma=9$: $\sigma^4=6561$ and $3\sigma^4=19683$.

(i) Mesokurtic ($\beta_2=3$): $\mu_4=19683$.

(ii) Platykurtic ($\beta_2<3$): $\mu_4<19683$.

(iii) Leptokurtic ($\beta_2>3$): $\mu_4>19683$.

Measures of dispersion

Dispersion is the extent to which observations scatter around a central value. The methods are:

Absolute measures

1. Range: $R = L - S$ (largest minus smallest). Simple but depends only on the two extreme values.
2. Quartile deviation: $QD = \dfrac{Q_3 - Q_1}{2}$. Based on the middle 50% of the data.
3. Mean deviation: $MD = \dfrac{\sum |X-\bar X|}{n}$. Uses all values.
4. Standard deviation: $\sigma = \sqrt{\dfrac{\sum (X-\bar X)^2}{n}}$. The most widely used measure; the square is the variance.

Relative measures: coefficient of range $\dfrac{L-S}{L+S}$, coefficient of quartile deviation $\dfrac{Q_3-Q_1}{Q_3+Q_1}$, coefficient of mean deviation, and coefficient of variation

$CV = \frac{\sigma}{\bar X}\times 100$

Numerical problem

Student A: $\sum X = 610$, so $\bar X_A = 61$.

Deviations: $-17, 19, 15, -13, -9, 11, 7, -5, -1, -7$. Squares: $289, 361, 225, 169, 81, 121, 49, 25, 1, 49$, sum $=1370$.

$\sigma_A = \sqrt{1370/10} = \sqrt{137} = 11.70$

$CV_A = \frac{11.70}{61}\times 100 = 19.19\%$

Student B: $\sum X = 615$, so $\bar X_B = 61.5$.

Deviations: $-13.5, 13.5, -7.5, -1.5, 1.5, 7.5, 10.5, -10.5, -4.5, 4.5$. Squares: $182.25, 182.25, 56.25, 2.25, 2.25, 56.25, 110.25, 110.25, 20.25, 20.25$, sum $=742.5$.

$\sigma_B = \sqrt{742.5/10} = \sqrt{74.25} = 8.62$

$CV_B = \frac{8.62}{61.5}\times 100 = 14.01\%$

(i) Who is better? B has the higher average ($61.5 > 61$), so B is the better performer.

(ii) Consistency: the lower the CV, the more consistent. Since $CV_B = 14.01\% < CV_A = 19.19\%$, B is more consistent and should win the prize.

Regression of performance ($Y$) on experience ($X$): $Y = a + bX$.

$n = 8$, $\sum X = 80$, $\sum Y = 648$, $\sum XY = 6727$, $\sum X^2 = 1018$.

(i) Regression equation

$b = \frac{n\sum XY - \sum X\sum Y}{n\sum X^2 - (\sum X)^2} = \frac{8(6727) - 80(648)}{8(1018) - 6400} = \frac{53816 - 51840}{8144 - 6400} = \frac{1976}{1744} = 1.133$

$a = \bar Y - b\bar X = 81 - 1.133(10) = 69.67$

$\boxed{Y = 69.67 + 1.133X}$

(ii) Estimate for $X = 8$

$Y = 69.67 + 1.133(8) = 78.73$

The estimated performance is about 78.7 good parts per 100 pieces.

Interpretation: the regression coefficient $b = 1.133$ means that each additional year of experience increases performance by about 1.13 good parts per 100 pieces, on average.

Standard normal distribution

The standard normal distribution is the normal distribution with mean 0 and standard deviation 1. Any normal variable $X \sim N(\mu,\sigma^2)$ is converted to it by

$Z = \frac{X-\mu}{\sigma}$

Its density is $f(z) = \dfrac{1}{\sqrt{2\pi}}e^{-z^2/2}$. The curve is bell-shaped, symmetric about 0, and the total area is 1.

Problem

$\mu = 50$, $\sigma = 15$.

(i) $P(50 < X < 70)$

$z_1 = 0$, $z_2 = \dfrac{70-50}{15} = 1.33$

$P = \Phi(1.33) - \Phi(0) = 0.9082 - 0.5 = 0.4082$

(about 0.409 using the more exact value)

(ii) $P(X > 60)$

$z = \dfrac{60-50}{15} = 0.67$

$P = 1 - \Phi(0.67) = 1 - 0.7486 = 0.2514$

(about 0.25)

(iii) $P(X < 45)$

$z = \dfrac{45-50}{15} = -0.33$

$P = \Phi(-0.33) = 1 - 0.6293 = 0.3707$

(about 0.37)

Let $A$ = motherboard problem and $B$ = hard drive problem.

$P(A) = 0.40$, $P(B) = 0.30$, $P(A\cap B) = 0.15$.

$P(A\cup B) = P(A) + P(B) - P(A\cap B) = 0.40 + 0.30 - 0.15 = 0.55$

A computer has fully functioning MB and HD when neither problem occurs:

$P(\text{neither}) = 1 - P(A\cup B) = 1 - 0.55 = 0.45$

The probability is 0.45.

Sampling error

Sampling error is the difference between the value of a sample statistic and the true population parameter, arising because only a part of the population is observed.

• It arises from chance in selecting the sample.
• It decreases as the sample size increases and is absent in a complete census.
• It can be measured, for example by the standard error, and controlled by good sample design and larger samples.

Non-sampling error

Non-sampling errors arise from causes other than the sampling process, and occur in both sample surveys and censuses. Types include:

• Coverage error: incomplete or wrong sampling frame.
• Non-response error: respondents not available or refusing to answer.
• Measurement and response error: faulty questionnaires, interviewer bias, or wrong answers.
• Processing errors: mistakes in data entry, editing, coding and tabulation.

Unlike sampling error, non-sampling error may not fall (and can increase) as the sample size grows, and it is harder to measure. It is reduced by careful planning, training and supervision.

A measurement scale is a system for assigning numbers or labels to the characteristics of objects or observations according to rules. The type of scale decides which statistical operations are valid. Stevens classified four scales.

1. Nominal scale

Classifies data into named categories with no order, for example gender, religion, blood group. Only counting, the mode and chi-square tests are meaningful.

2. Ordinal scale

Categories have a meaningful order or rank, but the differences between ranks are not equal or measurable, for example grades (A, B, C), satisfaction levels, and class rank. The median and percentiles can be used.

3. Interval scale

Has order and equal intervals, but no true zero, so ratios are meaningless, for example temperature in Celsius or Fahrenheit and calendar years. Mean and standard deviation can be used.

4. Ratio scale

Has all properties of the interval scale plus a true zero, so ratios are meaningful, for example height, weight, income, age. All statistical operations, including the geometric mean and the coefficient of variation, are valid.

The scales are in increasing order of information: nominal, ordinal, interval, ratio.

Total hours $= 6+8+2+4+1+3 = 24$. In a pie chart each activity gets a sector angle $\dfrac{\text{hours}}{24}\times 360^\circ = 15^\circ \times \text{hours}$.

Construction: draw a circle, mark a radius, and with a protractor cut successive sectors of $90^\circ, 120^\circ, 30^\circ, 60^\circ, 15^\circ, 45^\circ$ going clockwise. Shade or label each sector with its activity name (and percentage: School 25%, Sleep 33.3%, Playing 8.3%, Study 16.7%, TV 4.2%, Others 12.5%) and give the chart a title, "Distribution of a child's hours on a working day".

Definition: The correlation coefficient (Karl Pearson's $r$) measures the strength and direction of the linear relationship between two variables:

$r=\frac{n\sum XY-\sum X\sum Y}{\sqrt{n\sum X^2-(\sum X)^2}\;\sqrt{n\sum Y^2-(\sum Y)^2}},\qquad -1\le r\le 1$

Calculation: $n=10$, $\sum X=300$, $\sum Y=255$, $\sum XY=7767$, $\sum X^2=9138$, $\sum Y^2=6659$.

$n\sum XY-\sum X\sum Y = 77670-76500 = 1170$ $n\sum X^2-(\sum X)^2 = 91380-90000 = 1380$ $n\sum Y^2-(\sum Y)^2 = 66590-65025 = 1565$

$r=\frac{1170}{\sqrt{1380\times 1565}}=\frac{1170}{1469.6}\approx 0.796$

Interpretation: $r\approx 0.80$ shows a strong positive linear correlation: older husbands tend to have older wives.

(i) Total probability must be 1:

$\int_0^1 k(1-x)^2dx = k\cdot\frac13 = 1 \;\Rightarrow\; k=3.$

(ii) $P(0<X<0.5)=\int_0^{0.5}3(1-x)^2dx=\Big[-(1-x)^3\Big]_0^{0.5}=1-0.125=0.875.$

(iii) $E(X)=\int_0^1 3x(1-x)^2dx=3\int_0^1(x-2x^2+x^3)dx=3\left(\tfrac12-\tfrac23+\tfrac14\right)=3\cdot\tfrac1{12}=0.25.$

(iv) $E(2X+4)=2E(X)+4=2(0.25)+4=4.5.$

Let $X$ = number of adequately coated chips, $X\sim B(n=15,\,p=0.7)$, $q=0.3$, $P(X=x)=\binom{15}{x}(0.7)^x(0.3)^{15-x}$.

(1) $P(X\ge 12)=P(12)+P(13)+P(14)+P(15)$

• $P(12)=455(0.7)^{12}(0.3)^3\approx 0.1700$
• $P(13)=105(0.7)^{13}(0.3)^2\approx 0.0916$
• $P(14)=15(0.7)^{14}(0.3)\approx 0.0305$
• $P(15)=(0.7)^{15}\approx 0.0047$

$P(X\ge12)\approx 0.2969$

(2) $P(X=10)=\binom{15}{10}(0.7)^{10}(0.3)^5=3003\times0.0282475\times0.00243\approx 0.2061.$

(a) Choosing an appropriate measure of central tendency

• Mean: best for quantitative, roughly symmetric data without extreme values; it uses every observation and is needed for further statistics (e.g. variance, regression). Affected by outliers.
• Median: best for skewed data or data with outliers (income, house prices) and for open-ended classes or ordinal data. It is the middle value.
• Mode: best for nominal/categorical data or when the most common value matters (most popular shoe size, brand).
• Geometric mean: for ratios, growth rates and index numbers. Harmonic mean: for averages of rates and speeds.

(b) Parameter and statistic

• A parameter is a numerical measure describing a population (e.g. population mean $\mu$, standard deviation $\sigma$, proportion $P$). It is fixed and usually unknown.
• A statistic is a numerical measure computed from a sample (e.g. $\bar x$, $s$, $p$). It varies from sample to sample and is used to estimate the corresponding parameter.

Absolute vs relative measures:

• An absolute measure (range, quartile deviation, mean deviation, standard deviation) is expressed in the units of the data. It is useful for one data set but cannot compare data in different units or with very different averages.
• A relative measure (coefficient of range, coefficient of quartile deviation, coefficient of variation) is a unit-free ratio of dispersion to an average. It is used to compare variability of different series.

Comparison via the coefficient of variation $CV=\dfrac{\sigma}{\bar x}\times100$.

Tube I: $\bar x=\dfrac{2.94}{7}=0.42$, $\sum (x-\bar x)^2=0.0434$, $\sigma=\sqrt{0.0434/7}=0.0787$

$CV_I=\frac{0.0787}{0.42}\times100\approx18.7\%$

Tube II: $\bar x=\dfrac{3.70}{7}=0.5286$, $\sum (x-\bar x)^2=0.02649$, $\sigma=\sqrt{0.02649/7}=0.0615$

$CV_{II}=\frac{0.0615}{0.5286}\times100\approx11.6\%$

Since $CV_{II}<CV_I$, Tube II is more homogeneous (less variable). (Using the sample SD with $n-1$ gives 20.2% and 12.6%, the same conclusion.)

Let $X$ = circumference (independent), $Y$ = weight (dependent). $n=8$, $\sum X=400$, $\sum Y=153$, $\sum XY=7217$, $\sum X^2=20236$.

(i) $b=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{57736-61200}{161888-160000}=\frac{-3464}{1888}=-1.835$

$\bar X=50,\quad \bar Y=19.125,\quad a=\bar Y-b\bar X=19.125+91.737=110.862$

$\hat Y=110.86-1.835X$

(ii) At $X=40$: $\hat Y=110.86-73.39\approx 37.5$ g.

(iii) At $X=80$: $\hat Y=110.86-146.78=-35.9$ g, an impossible negative weight. $80$ cm lies far outside the observed range ($37$ to $55$ cm), so this is extrapolation and the equation cannot be used reliably there. (Also the negative slope is odd for pumpkins and driven mostly by one point, $X=37$, so the farmer's claim is weakly supported.)

Normal distribution: a continuous, symmetric, bell-shaped distribution with density

$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}},\quad -\infty<x<\infty,$

with parameters mean $\mu$ and standard deviation $\sigma$. Mean = median = mode; total area is 1.

Standard normal distribution: the normal distribution with $\mu=0$, $\sigma=1$; the variable $Z=\dfrac{X-\mu}{\sigma}$ has density $\phi(z)=\dfrac1{\sqrt{2\pi}}e^{-z^2/2}$ and is tabulated.

Problem: $P(X<20)=0.10\Rightarrow z=-1.2816$; $P(X<75)=0.95\Rightarrow z=1.645$.

$\frac{20-\mu}{\sigma}=-1.2816,\qquad \frac{75-\mu}{\sigma}=1.645$

Subtracting: $\dfrac{55}{\sigma}=2.9266\Rightarrow\sigma\approx18.79$.

$\mu=20+1.2816(18.79)\approx 44.09$

Mean $\approx 44.1$ marks, standard deviation $\approx 18.8$ marks.

Secondary data are data already collected by someone else for another purpose, which a researcher reuses. They are cheap and quick to obtain, but may not exactly fit the present need and their accuracy must be checked.

Sources (general): published reports, journals, books, newspapers, websites, and unpublished records of organisations.

Nepalese sources:

• National Statistics Office (formerly Central Bureau of Statistics): Census, Nepal Living Standards Survey, Labour Force Survey, Statistical Pocket Book.
• Nepal Rastra Bank: Economic Report, Quarterly Economic Bulletin, banking statistics.
• Ministry of Finance: Economic Survey and budget documents.
• Ministry of Education: Flash reports, and Ministry of Health: Nepal Demographic and Health Survey (with partners).
• Nepal Telecommunications Authority (MIS reports on internet and mobile users), Department of Customs (trade data), Nepal Stock Exchange (NEPSE).
• National Planning Commission reports, and international sources such as the World Bank, UN agencies and ADB.

The data of Question 2 are not reproduced here, so the method is given for any grouped continuous frequency distribution with class intervals and frequencies $f$.

Histogram:

1. Take the class boundaries on the $X$-axis (make classes continuous if needed) and frequencies on the $Y$-axis.
2. Over each class draw a rectangle with width equal to the class width and height equal to the class frequency (or frequency density if widths are unequal).
3. Rectangles touch each other (no gaps).

Frequency polygon:

1. Find the mid-point of every class: $m=\dfrac{\text{lower}+\text{upper}}{2}$.
2. Plot the points $(m,f)$ and join them by straight lines.
3. Add one empty class at each end (frequency 0) and join the polygon to the $X$-axis. It can equally be drawn by joining the mid-points of the tops of the histogram rectangles.

Frequency curve: smooth the frequency polygon by hand into a continuous curve passing as near as possible to the plotted points, with the same area as the histogram. It shows the shape of the distribution (symmetric, skewed, unimodal, etc.).

All three graphs should share the same axes, with labelled scales and a title.

Spearman's rank correlation (ranks already given):

$r_s=1-\frac{6\sum d^2}{n(n^2-1)}$

$\sum d^2=216$, $n=10$:

$r_s=1-\frac{6(216)}{10(99)}=1-1.309=-0.309$

Interpretation: $r_s\approx-0.31$ indicates a weak negative association: students who rank DELL highly tend to rank HP slightly lower, though the relationship is weak.

(The question calls it Pearson's rank correlation; this is the rank, i.e. Spearman, coefficient.)

Random variable: a variable that takes numerical values determined by the outcome of a random experiment, each value with a certain probability. It may be discrete or continuous.

(i) $\sum P(X)=1$: $0.2+K+0.4+2K+K=1\Rightarrow 4K=0.4\Rightarrow K=0.1$.

So $P(X)=0.2,\,0.1,\,0.4,\,0.2,\,0.1$ for $X=-2,-1,0,1,2$.

(ii) The question's "$P(K\ge0)$" is read as $P(X\ge0)$:

$P(X\ge0)=0.4+0.2+0.1=0.7.$

(iii) $E(X)=(-2)(0.2)+(-1)(0.1)+0+(1)(0.2)+(2)(0.1)=-0.4-0.1+0.2+0.2=-0.1.$

(iv) $E(4X+5)=4(-0.1)+5=4.6.$

Let $X\sim B(15,\,0.7)$, $P(X=x)=\binom{15}{x}(0.7)^x(0.3)^{15-x}$.

(i) $P(X\ge12)=0.1700+0.0916+0.0305+0.0047\approx0.2969.$

(ii) $P(X\le2)=P(0)+P(1)+P(2)$:

• $P(0)=(0.3)^{15}\approx1.4\times10^{-8}$
• $P(1)=15(0.7)(0.3)^{14}\approx5.0\times10^{-7}$
• $P(2)=105(0.7)^2(0.3)^{13}\approx8.2\times10^{-6}$

$P(X\le2)\approx8.7\times10^{-6}\approx0.00001$

(iii) $P(X=10)=\binom{15}{10}(0.7)^{10}(0.3)^5=3003(0.0282475)(0.00243)\approx0.2061.$

Let $E$ = virus enters by e-mail, $I$ = by internet. $P(E)=0.30$, $P(I)=0.40$, $P(E\cap I)=0.15$.

$P(E\cup I)=0.30+0.40-0.15=0.55$

$P(\text{virus does not enter})=1-P(E\cup I)=1-0.55=0.45.$

$\sigma=0.042$ is known and the population is normal, so use the $Z$ interval: $\bar x\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt n}$ with $z_{0.025}=1.96$.

$\text{Margin of error}=1.96\times\frac{0.042}{\sqrt{36}}=1.96\times0.007=0.01372$

$5.12\pm0.0137\;\Rightarrow\;(5.1063,\;5.1337)\text{ mg}$

We are 95% confident that the true mean mass of vitamin E per capsule lies between about $5.106$ mg and $5.134$ mg.

(a) Box and whisker plot

A graphical display of the five-number summary: minimum, first quartile $Q_1$, median $Q_2$, third quartile $Q_3$ and maximum.

• A box is drawn from $Q_1$ to $Q_3$ (length = interquartile range) with a line at the median.
• Whiskers extend from the box to the minimum and maximum (or to the furthest points within $1.5\times IQR$ of the box, with outliers plotted separately).
• Uses: shows centre, spread and skewness (median off-centre in the box, unequal whiskers), identifies outliers, and compares several data sets side by side.

(b) Census survey and sample survey

• Census (complete enumeration): every unit of the population is studied. It gives exact results, but is costly, slow and often impractical (e.g. Nepal's decennial population census).
• Sample survey: only a representative part of the population is studied and results are generalised. It is cheaper, faster and more practical (essential when testing is destructive), but has sampling error and needs a carefully chosen, unbiased sample.

Regression: a statistical method of estimating (predicting) the value of one variable, the dependent variable, from the known value of another, the independent variable, using a fitted relationship (regression line). The slope is the regression coefficient.

(i) $Y$ = blood pressure (dependent), $X$ = age. $n=6$, $\sum X=316$, $\sum Y=827$, $\sum XY=44653$, $\sum X^2=17558$.

$b_{YX}=\frac{6(44653)-316(827)}{6(17558)-316^2}=\frac{267918-261332}{105348-99856}=\frac{6586}{5492}=1.199$

$\bar X=52.67,\;\bar Y=137.83,\;a=137.83-1.199(52.67)=74.68$

$\hat Y=74.68+1.199X$

(ii) At $X=50$: $\hat Y=74.68+59.96\approx134.6$, i.e. about 135 units of blood pressure.

(iii) The regression coefficient $b=1.199$ means that blood pressure rises, on average, by about 1.2 units for each additional year of age.

When the normal distribution is used: for continuous measurements whose values cluster symmetrically about a mean, with few extreme values (heights, marks, errors, salaries in a large group). It also approximates binomial and Poisson distributions for large $n$ (or large mean), and by the central limit theorem models sample means of large samples.

Here $X\sim N(\mu=50000,\;\sigma=20000)$, $Z=\dfrac{X-50000}{20000}$.

(a) $P(X<40000)=P(Z<-0.5)=0.3085$

(b) $P(45000<X<65000)=P(-0.25<Z<0.75)=0.7734-0.4013=0.3721$

(c) $P(X>70000)=P(Z>1)=1-0.8413=0.1587$

Descriptive vs inferential statistics

Consistency is judged by the coefficient of variation $CV=\dfrac{\sigma}{\bar x}\times100$; the smaller the CV, the more consistent.

Basic Statistics: $\bar x=468/10=46.8$, $\sigma^2=107.36$, $\sigma=10.36$

$CV=\frac{10.36}{46.8}\times100\approx22.1\%$

Computer Science: $\bar y=493/10=49.3$, $\sigma^2=464.4$, $\sigma=21.55$

$CV=\frac{21.55}{49.3}\times100\approx43.7\%$

Basic Statistics has the lower CV (22.1% < 43.7%), so marks in Basic Statistics are more consistent.

Statistics is central to information technology:

• Data analysis and data science: summarising and visualising large data sets (mean, variance, correlation, graphs) and extracting patterns in data mining and business analytics.
• Machine learning and AI: models rest on probability, regression, classification and estimation; model evaluation uses statistical measures.
• Quality control and software testing: defect rates, control charts, reliability estimation and sampling inspection of products and code.
• Performance evaluation: analysing response time, throughput, network traffic and queueing using distributions and confidence intervals.
• Database and information retrieval: sampling, query optimisation using data distributions, search ranking.
• Security and networking: anomaly and intrusion detection, error detection and coding use probability.
• Decision making and forecasting: hypothesis testing, A/B testing of designs, time-series forecasting of demand or load.
• Simulation and risk: random numbers, Monte Carlo methods and risk assessment of projects.

Total students $=400$. Percentages: Bus 50%, Bicycle 25%, On foot 20%, By car 5%.

Bar diagram: On the $X$-axis draw four equal-width, equally spaced bars for Bus, Bicycle, On foot, By car. On the $Y$-axis (scale 0 to 200, step 50) draw heights $200,\,100,\,80,\,20$ respectively. Label both axes and give a title.

Pareto diagram: bars are arranged in descending order of frequency and a cumulative percentage line is added on a secondary axis (0 to 100%).

Draw touching or evenly spaced bars in this order (already descending), then plot the points $(\text{Bus},50),(\text{Bicycle},75),(\text{On foot},95),(\text{By car},100)$ at the top of each bar's right edge or centre and join them by a line.

Interpretation: bus and bicycle together account for 75% of the students, so they are the "vital few" modes of transport.

Properties of correlation coefficient $r$:

1. $-1\le r\le1$; $r=+1$ perfect positive, $-1$ perfect negative, $0$ no linear correlation.
2. It is independent of change of origin and scale (for positive scale).
3. It is a pure number, free of units.
4. It is the geometric mean of the two regression coefficients: $r=\pm\sqrt{b_{YX}\,b_{XY}}$, with the same sign.
5. It is symmetric: $r_{XY}=r_{YX}$.
6. It measures only linear relationship and does not prove causation.

Calculation: $n=6$, $\sum X=108$, $\sum Y=169$, $\sum XY=3226$, $\sum X^2=2058$, $\sum Y^2=5267$.

$r=\frac{6(3226)-108(169)}{\sqrt{6(2058)-108^2}\sqrt{6(5267)-169^2}}=\frac{19356-18252}{\sqrt{684}\sqrt{3041}}=\frac{1104}{26.15\times55.15}\approx0.765$

Comment: a fairly strong positive correlation: higher aptitude scores go with higher productivity.

Coefficients of variation:

• $X$: $\bar X=18$, $\sigma_X=\sqrt{2058/6-18^2}=\sqrt{19}=4.36$, $CV_X=\dfrac{4.36}{18}\times100\approx24.2\%$
• $Y$: $\bar Y=28.17$, $\sigma_Y=\sqrt{5267/6-28.17^2}=9.19$, $CV_Y=\dfrac{9.19}{28.17}\times100\approx32.6\%$

Productivity is more variable than aptitude score.

Random variable: a variable whose value is a numerical outcome of a random experiment; each value occurs with some probability.

(i) $0.1+C+0.2+2C+0.3+C=1\Rightarrow 0.6+4C=1\Rightarrow C=0.1$.

Then $P(X)=0.1,\,0.1,\,0.2,\,0.2,\,0.3,\,0.1$ for $X=0,\dots,5$.

(ii) $P(X<3)=0.1+0.1+0.2=0.4$ $P(X\ge1)=1-P(0)=0.9$

(iii) $E(X)=0(0.1)+1(0.1)+2(0.2)+3(0.2)+4(0.3)+5(0.1)=0.1+0.4+0.6+1.2+0.5=2.8$

$E(X^2)=0.1+0.8+1.8+4.8+2.5=10.0$

$\text{Var}(X)=10.0-2.8^2=10-7.84=2.16$

Let $E$ = defective etching, $C$ = crack defect. $P(E)=0.12$, $P(C)=0.29$, $P(E\cap C)=0.07$.

$P(E\cup C)=P(E)+P(C)-P(E\cap C)=0.12+0.29-0.07=0.34.$

The probability of either defect is $0.34$.

Binomial distribution: for $n$ independent trials, each with two outcomes and constant probability of success $p$, the number of successes $X$ has

$P(X=x)=\binom nx p^xq^{n-x},\quad x=0,1,\dots,n,\; q=1-p,$

with mean $np$ and variance $npq$.

Fitting: $n=4$, $N=\sum f=150$.

$\bar x=\frac{\sum fx}{N}=\frac{0+62+92+30+16}{150}=\frac{200}{150}=1.333$

$np=1.333\Rightarrow p=0.3333,\; q=0.6667$

Expected frequency $=150\binom4x(1/3)^x(2/3)^{4-x}$:

The fitted distribution is $B(4,\tfrac13)$ and agrees reasonably with the observed data.

Given $\mu_1=0$, $\mu_2=2.8$, $\mu_3=-2$, $\mu_4=24.5$.

Skewness: $\beta_1=\frac{\mu_3^2}{\mu_2^3}=\frac{4}{21.952}=0.182,\qquad \gamma_1=\sqrt{\beta_1}=-0.427$ (the sign follows $\mu_3<0$).

Kurtosis: $\beta_2=\frac{\mu_4}{\mu_2^2}=\frac{24.5}{7.84}=3.125,\qquad \gamma_2=\beta_2-3=0.125$

Comment: $\mu_3$ is negative so the distribution is moderately (slightly) negatively skewed, with a longer left tail. $\beta_2=3.125$ is just above 3, so it is nearly mesokurtic, very slightly leptokurtic (almost normal peakedness).

$n=50$, $\sum x=525$, $\sum x^2=5625$.

$\bar x=\frac{525}{50}=10.5,\qquad s^2=\frac{5625}{50}-10.5^2=112.5-110.25=2.25,\; s=1.5$

For a large sample ($n=50$) use $\bar x\pm z_{0.005}\dfrac{s}{\sqrt n}$ with $z_{0.005}=2.576$:

$\text{Margin}=2.576\times\frac{1.5}{\sqrt{50}}=2.576\times0.2121=0.546$

$10.5\pm0.546\;\Rightarrow\;(9.95,\;11.05)\text{ km per litre}$

Interpretation: we are 99% confident that the true mean fuel consumption of the new model lies between about 9.95 and 11.05 km per litre. If many such samples were taken, about 99% of the intervals constructed this way would contain the true mean.

(a) Use of the box and whisker plot

It is drawn from the five-number summary (minimum, $Q_1$, median, $Q_3$, maximum). A box spans $Q_1$ to $Q_3$ with the median marked; whiskers reach the extreme values (or limits of $1.5\times IQR$).

Uses: showing the centre, spread and range at a glance; judging symmetry/skewness from the median's position and whisker lengths; detecting outliers; comparing two or more groups on one scale.

(b) Parameter and statistic

• A parameter is a numerical characteristic of a population, such as $\mu$, $\sigma$ or $P$. It is a fixed (usually unknown) constant.
• A statistic is a numerical characteristic of a sample, such as $\bar x$, $s$ or $p$. It is computed from data, varies by sample, and is used to estimate the parameter.

Choosing a measure of central tendency

• Arithmetic mean: for quantitative, fairly symmetric data without extreme values; uses all observations.
• Median: for skewed data, outliers, open-ended classes or ordinal data.
• Mode: for qualitative data or when the most frequent value is wanted.
• Geometric mean: for growth rates, ratios, index numbers; harmonic mean: for averages of speeds and rates.

Computation (mid-points $m=1,3,5,7,9,11$; $N=50$ each):

Computer A: $\sum fm=256$, $\sum fm^2=1706$

$\bar x_A=\frac{256}{50}=5.12\text{ s},\quad \sigma_A=\sqrt{\frac{1706}{50}-5.12^2}=\sqrt{7.9056}=2.81$

$CV_A=\frac{2.81}{5.12}\times100\approx54.9\%$

Computer B: $\sum fm=308$, $\sum fm^2=2146$

$\bar x_B=\frac{308}{50}=6.16\text{ s},\quad \sigma_B=\sqrt{\frac{2146}{50}-6.16^2}=\sqrt{4.9744}=2.23$

$CV_B=\frac{2.23}{6.16}\times100\approx36.2\%$

Conclusion: the average time is 5.12 s for A and 6.16 s for B. B has the smaller CV (36.2% < 54.9%), so Company B's computers are more consistent.

Assumptions of the simple linear regression model $Y=\beta_0+\beta_1X+\varepsilon$:

1. The relationship between $X$ and $Y$ is linear.
2. The errors $\varepsilon$ have mean zero and constant variance (homoscedasticity).
3. The errors are independent of each other (no autocorrelation).
4. The errors are normally distributed.
5. $X$ is measured without error and $\varepsilon$ is independent of $X$.

(a) The response (dependent) variable is processed requests $Y$; the predictor is data size $X$. $n=7$, $\sum X=79$, $\sum Y=238$, $\sum XY=2317$, $\sum X^2=987$.

$b=\frac{7(2317)-79(238)}{7(987)-79^2}=\frac{16219-18802}{6909-6241}=\frac{-2583}{668}=-3.867$

$\bar X=11.286,\;\bar Y=34,\;a=34+3.867(11.286)=77.64$

$\hat Y=77.64-3.867X$

(b) For each additional gigabyte of incoming data, the number of processed requests per hour falls by about 3.87 on average. (Efficiency decreases as data size grows; $r=-0.87$, a strong negative relationship.)

(c) For $X=16$: $\hat Y=77.64-3.867(16)=77.64-61.87\approx15.8$, i.e. about 16 requests per hour.