Business Statistics

$r = \frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}$

$\sigma_X = \sqrt{16} = 4$, so

$0.8 = \frac{20}{4\,\sigma_Y} \Rightarrow \sigma_Y = \frac{20}{3.2} = 6.25$

$r=\frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}$

$\sigma_X=\sqrt{16}=4$, so

$0.9=\frac{30}{4\,\sigma_Y}\;\Rightarrow\;\sigma_Y=\frac{30}{4\times0.9}=\frac{30}{3.6}=8.33$

The standard deviation of $Y$ is about $8.33$.

For a moderately asymmetric distribution, $\text{Mode} = 3\,\text{Median} - 2\,\text{Mean}$.

$\text{Mode} = 3(20) - 2(16) = 60 - 32 = 28$

$Q.D. = \frac{Q_3 - Q_1}{2} = \frac{70 - 40}{2} = 15$

$\bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} = \frac{20(120)+15(130)}{35} = \frac{4350}{35} = 124.29$

The percentile coefficient of kurtosis is

$K = \frac{Q.D.}{P_{90}-P_{10}} = \frac{2}{24-16} = 0.25$

Since $0.25 < 0.263$ (the value for a normal curve), the distribution is platykurtic.

$Sk = \frac{\text{Mean} - \text{Mode}}{\sigma} = \frac{65-62}{5} = 0.6$

The distribution is moderately positively skewed.

$P(A\cup B) = P(A)+P(B)-P(A\cap B) = 0.5+0.4-0.2 = 0.7$

$P(x) = \frac{e^{-\lambda}\lambda^x}{x!}$

$P(x=2) = \frac{e^{-2}\,2^2}{2!} = 2e^{-2} = 2(0.1353) = 0.2707$

Since the sample is more than 5% of the population ($50/500 = 10\%$), the finite population correction is used.

$S.E.(\bar{x}) = \frac{\sigma}{\sqrt{n}}\sqrt{\frac{N-n}{N-1}} = \frac{5}{\sqrt{50}}\sqrt{\frac{450}{499}} = 0.7071\times 0.9496 = 0.671$

The types of random (probability) sampling techniques are:

1. Simple random sampling
2. Systematic sampling
3. Stratified random sampling
4. Cluster sampling
5. Multistage sampling

Let the missing frequencies be $x$ (20-30) and $y$ (40-50). Mid-values are 5, 15, 25, 35, 45, 55.

Total: $4+6+x+20+y+10 = 60 \Rightarrow x+y = 20$ ... (i)

$\sum fm = 20+90+25x+700+45y+550 = 1360+25x+45y$

... (ii)

From (i), $x = 20-y$. Substituting in (ii): $100-5y+9y = 148 \Rightarrow y = 12$, so $x = 8$.

The missing frequencies are 8 (for 20-30) and 12 (for 40-50).

$n=400$, $\bar{x}=128$, $s=13$. Because $n$ is large, the normal distribution is used with $z = 1.96$ at 95%.

$S.E. = \frac{s}{\sqrt{n}} = \frac{13}{20} = 0.65$

$\mu = \bar{x} \pm 1.96\,S.E. = 128 \pm 1.96(0.65) = 128 \pm 1.274$

The 95% confidence limits are 126.73 mm to 129.27 mm.

The classes are made continuous by subtracting 0.5 from the lower limits and adding 0.5 to the upper limits.

$N = 1520$.

$Q_1$ lies at $N/4 = 380$, in the class 999.5-1999.5:

$Q_1 = 999.5 + \frac{380-50}{500}\times 1000 = 1659.5$

$Q_3$ lies at $3N/4 = 1140$, in the class 2999.5-3999.5:

$Q_3 = 2999.5 + \frac{1140-1105}{100}\times 1000 = 3349.5$

$\text{Coefficient of Q.D.} = \frac{Q_3-Q_1}{Q_3+Q_1} = \frac{1690}{5009} = 0.337$

Consistency is compared by the coefficient of variation.

$C.V._A = \frac{5}{11}\times 100 = 45.45\%$

$C.V._B = \frac{6}{12}\times 100 = 50\%$

Since $C.V._A < C.V._B$, brand A shows greater consistency in the life of its cars.

The class with the highest frequency (45) is the modal class $20\text{-}30$.

$Z = L + \frac{\Delta_1}{\Delta_1+\Delta_2}\times h$

Here $L=20$, $\Delta_1 = 45-20 = 25$, $\Delta_2 = 45-15 = 30$, $h=10$.

$Z = 20 + \frac{25}{25+30}\times 10 = 20 + 4.545 = 24.55$

Mode $\approx 24.55$ marks.

The lowest mark of the top 10% students is the 90th percentile, $P_{90}$.

Position of $P_{90}$: $\dfrac{90N}{100} = \dfrac{90\times 500}{100} = 450$. The c.f. first reaches 450 in the class $60\text{-}80$, so $L=60$, $c.f.=390$, $f=60$, $h=20$.

$P_{90} = 60 + \frac{450-390}{60}\times 20 = 60 + 20 = 80$

The lowest marks of the top 10% students is 80.

Use Karl Pearson's coefficient of skewness, $Sk = \dfrac{\bar X - Z}{\sigma}$.

Mean. With mid-values $10, 30, 45, 55, 70, 90$:

$\sum fm = 500+3000+6750+4950+4200+4500 = 23900,\quad \bar X = \frac{23900}{500} = 47.8$

Standard deviation.

$\sum fm^2 = 5000+90000+303750+272250+294000+405000 = 1370000$

$\sigma = \sqrt{\frac{1370000}{500} - 47.8^2} = \sqrt{2740 - 2284.84} = \sqrt{455.16} = 21.33$

Mode. The class widths are unequal, so convert the frequencies to a common width of 10: $25, 50, 150, 90, 30, 25$. The modal class is $40\text{-}50$, with $\Delta_1 = 150-50 = 100$ and $\Delta_2 = 150-90 = 60$.

$Z = 40 + \frac{100}{100+60}\times 10 = 46.25$

Coefficient of skewness.

$Sk = \frac{47.8-46.25}{21.33} = 0.073$

Interpretation. $Sk$ is positive but very close to zero, so the distribution is almost symmetrical with a very slight positive skew (the tail extends marginally to the higher marks).

Percentile coefficient of kurtosis:

$K = \frac{Q.D.}{P_{90}-P_{10}} = \frac{(P_{75}-P_{25})/2}{P_{90}-P_{10}}$

Convert to true class limits (subtract 0.5 from the lower limit and add 0.5 to the upper limit). Cumulative frequencies: $5, 55, 110, 120, 150, 160$, so $N=160$.

• $P_{10}$: position $16$, class $99.5\text{-}199.5$: $99.5 + \dfrac{16-5}{50}\times 100 = 121.5$
• $P_{25}$: position $40$, same class: $99.5 + \dfrac{40-5}{50}\times 100 = 169.5$
• $P_{75}$: position $120$, class $299.5\text{-}399.5$: $299.5 + \dfrac{120-110}{10}\times 100 = 399.5$
• $P_{90}$: position $144$, class $399.5\text{-}499.5$: $399.5 + \dfrac{144-120}{30}\times 100 = 479.5$

$K = \frac{(399.5-169.5)/2}{479.5-121.5} = \frac{115}{358} = 0.321$

Comment. For a normal (mesokurtic) curve $K = 0.263$. Since $0.321 > 0.263$, the distribution is platykurtic (flatter than the normal curve).

Let $X\sim N(\mu=120,\ \sigma=20)$ and $Z=\dfrac{X-120}{20}$.

(a) For $X=105$: $Z=-0.75$. For $X=140$: $Z=1$.

$P(105<X<140) = P(-0.75<Z<1) = 0.2734 + 0.3413 = 0.6147$

The probability is about $0.615$, i.e. about $400\times 0.6147\approx 246$ employees.

(b) For $X=125$: $Z=0.25$. For $X=150$: $Z=1.5$.

$P(125<X<150) = P(0.25<Z<1.5) = 0.4332 - 0.0987 = 0.3345$

The probability is about $0.334$, i.e. about $400\times 0.3345\approx 134$ employees.

Given $n=100$, $\bar x=65$, $s=20$, $\mu_0=60$ (large sample, so a Z-test is used).

Hypotheses. $H_0:\mu=60$ against $H_1:\mu\ne 60$ (two-tailed), $\alpha=0.05$.

Test statistic.

$Z=\frac{\bar x-\mu_0}{s/\sqrt n} = \frac{65-60}{20/\sqrt{100}} = \frac{5}{2} = 2.5$

Critical value. At the 5% level (two-tailed), $Z_{0.05}=1.96$.

Decision. $|Z|=2.5>1.96$, so $H_0$ is rejected. The mean weight of the population cannot be taken as 60 kg.

Let $X$ = income and $Y$ = expenditure. Use mid-values $X: 250, 750, 1250, 1750, 2250$ and $Y: 200, 600, 1000, 1400, 1800$.

Marginal frequencies. Income: $24, 33, 33, 11, 9$. Expenditure: $26, 40, 24, 14, 6$. Total $N=110$.

Means and standard deviations.

$\bar X=\frac{\sum fX}{N}=\frac{111500}{110}=1013.64,\qquad \bar Y=\frac{\sum fY}{N}=\frac{83600}{110}=760$

$\sigma_X=\sqrt{\frac{\sum fX^2}{N}-\bar X^2}=586.63,\qquad \sigma_Y=\sqrt{\frac{\sum fY^2}{N}-\bar Y^2}=455.11$

Covariance.

$\text{Cov}(X,Y)=\frac{\sum fXY}{N}-\bar X\bar Y = 150545.45$

(b) Correlation coefficient.

$r=\frac{\text{Cov}(X,Y)}{\sigma_X\sigma_Y}=\frac{150545.45}{586.63\times 455.11}=0.564$

There is a moderate positive correlation between income and expenditure.

(a) Regression lines.

$b_{YX}=r\frac{\sigma_Y}{\sigma_X}=0.564\times\frac{455.11}{586.63}=0.4375$

$b_{XY}=r\frac{\sigma_X}{\sigma_Y}=0.564\times\frac{586.63}{455.11}=0.7268$

Regression of $Y$ on $X$:

$Y-760=0.4375(X-1013.64)\ \Rightarrow\ Y=316.57+0.4375X$

Regression of $X$ on $Y$:

$X-1013.64=0.7268(Y-760)\ \Rightarrow\ X=461.25+0.7268Y$

(c) Probable error.

$P.E.=0.6745\frac{1-r^2}{\sqrt N}=0.6745\times\frac{1-0.318}{\sqrt{110}}=0.0439$

$6\,P.E.=0.263$. Since $r=0.564>6\,P.E.$, the correlation coefficient is significant: the correlation is real and not due to sampling fluctuation.

(d) For income $X=4000$:

$Y=316.57+0.4375\times 4000=316.57+1750=\text{Rs } 2066.43$

The estimated expenditure is about Rs 2066 (this is an extrapolation beyond the observed income range, so it should be used with caution).

Favourable cases are 3 girls out of 6 and 2 boys out of 4. Total cases are 5 members out of 10.

$P = \frac{\binom{6}{3}\binom{4}{2}}{\binom{10}{5}} = \frac{20 \times 6}{252} = \frac{120}{252} = \frac{10}{21} \approx 0.476$

For 99% confidence, $z = 2.576$. Given $\sigma = 12$ and $E = 3$:

$n = \left(\frac{z\sigma}{E}\right)^2 = \left(\frac{2.576 \times 12}{3}\right)^2 = (10.304)^2 \approx 106.2$

The sample size must be at least 107.

$P(r=3) = \binom{10}{3}(0.6)^3(0.4)^7 = 120 \times 0.216 \times 0.0016384 \approx 0.0425$

$N = 30$.

$Q_1$ is the size of the $\frac{N+1}{4} = 7.75$th item. The c.f. first exceeds 7.75 at 8, so $Q_1 = 7$.

$Q_3$ is the size of the $3\cdot\frac{N+1}{4} = 23.25$th item. The c.f. first exceeds 23.25 at 26, so $Q_3 = 18$.

$Q.D. = \frac{Q_3 - Q_1}{2} = \frac{18 - 7}{2} = 5.5$

The coefficient of quartile deviation is $\dfrac{Q_3-Q_1}{Q_3+Q_1} = \dfrac{11}{25} = 0.44$.

Given $n = 50$, $\bar x = 300$, $s = 11$, $\mu_0 = 250$, $\alpha = 0.05$.

$H_0: \mu = 250$ against $H_1: \mu > 250$ (right-tailed test).

Since $n$ is large, use the $z$ test:

$z = \frac{\bar x - \mu_0}{s/\sqrt n} = \frac{300-250}{11/\sqrt{50}} = \frac{50}{1.556} \approx 32.1$

The critical value is $z_{0.05} = 1.645$. Since $32.1 > 1.645$, $H_0$ is rejected.

The sample indicates that the current average monthly wage is higher than Rs 250.

Population: $N=5$, values 3, 5, 8, 10, 14.

$\mu = \frac{40}{5} = 8, \qquad \sigma^2 = \frac{(-5)^2+(-3)^2+0^2+2^2+6^2}{5} = \frac{74}{5} = 14.8,\quad \sigma = 3.85$

There are $\binom{5}{2} = 10$ samples of size 2 without replacement:

Mean of sample means.

$\mu_{\bar x} = \frac{4+5.5+6.5+8.5+6.5+7.5+9.5+9+11+12}{10} = \frac{80}{10} = 8 = \mu$

Standard deviation of the sampling distribution.

$\sum(\bar x-8)^2 = 16+6.25+2.25+0.25+2.25+0.25+2.25+1+9+16 = 55.5$

$\sigma_{\bar x}^2 = \frac{55.5}{10} = 5.55, \qquad \sigma_{\bar x} = 2.356$

This agrees with $\sqrt{\dfrac{\sigma^2}{n}\cdot\dfrac{N-n}{N-1}} = \sqrt{7.4 \times 0.75} = \sqrt{5.55}$.

Since $\sigma_{\bar x} = 2.36 < \sigma = 3.85$, the standard deviation of the sampling distribution is less than the population standard deviation.

The middle 70% of families lie between $P_{15}$ and $P_{85}$. Here $N = 150$.

$P_{15}$. The position is $\frac{15\times150}{100} = 22.5$, which lies in the class 20-40.

$P_{15} = 20 + \frac{22.5-15}{25}\times 20 = 26$

$P_{85}$. The position is $\frac{85\times150}{100} = 127.5$, which lies in the class 60-80.

$P_{85} = 60 + \frac{127.5-100}{30}\times 20 = 78.33$

The middle 70% of families pay between Rs 2,600 and Rs 7,833, so the range is $78.33 - 26 = 52.33$ (in '00 Rs), that is about Rs 5,233.

Sampling technique. It is the procedure by which a part (sample) of the population is selected so that it represents the whole population, and conclusions about the population can be drawn from it.

A. Probability (random) sampling. Every unit has a known chance of selection.

1. Simple random sampling. Each unit has an equal chance, using a lottery or random number table.
2. Systematic sampling. Every $k$th unit is selected after a random start, where $k = N/n$.
3. Stratified sampling. The population is divided into homogeneous strata, and random samples are drawn from each stratum, proportionally or equally.
4. Cluster sampling. The population is divided into clusters, and whole randomly chosen clusters are surveyed.
5. Multistage sampling. Sampling is done in stages, for example district, then ward, then household.

B. Non-probability sampling. Selection depends on the researcher's judgement.

1. Convenience sampling. Units that are easily available are chosen.
2. Judgement (purposive) sampling. The researcher picks units that seem typical.
3. Quota sampling. Fixed numbers are taken from each category.
4. Snowball sampling. Respondents refer further respondents.

Let $X$ be advertisement expenditure and $Y$ be sales (both in '000). Here $n=7$.

$\sum X = 56$, $\sum Y = 54$, $\sum XY = 459$, $\sum X^2 = 466$, $\sum Y^2 = 464$.

$\bar X = 8$, $\bar Y = 7.714$.

$S_{xy} = 459 - \frac{56\times54}{7} = 27,\quad S_{xx} = 466 - \frac{56^2}{7} = 18,\quad S_{yy} = 464 - \frac{54^2}{7} = 47.43$

Regression of $Y$ on $X$.

$b_{yx} = \frac{27}{18} = 1.5$

$Y - 7.714 = 1.5(X - 8) \;\Rightarrow\; Y = -4.286 + 1.5X$

For $X = 15$: $Y = -4.286 + 22.5 = 18.21$. Estimated sales are about Rs 18,214 (18.21 thousand).

Regression of $X$ on $Y$.

$b_{xy} = \frac{27}{47.43} = 0.569$

$X - 8 = 0.569(Y - 7.714) \;\Rightarrow\; X = 3.608 + 0.569Y$

For $Y = 20$: $X = 3.608 + 11.385 = 14.99$. The estimated advertisement expenditure is about Rs 14,994 (about Rs 15,000).

Correlation coefficient.

$r = \sqrt{b_{yx}\, b_{xy}} = \sqrt{1.5\times0.569} = 0.924$

(or $r = \dfrac{27}{\sqrt{18\times47.43}} = 0.924$). There is a strong positive correlation.

Use class mid-points $m = 2, 7, 12, 17, 22, 27$ ($N = 100$ for each type).

Type A. $\sum fm = 1280$ and $\sum fm^2 = 19120$.

$\bar x = \frac{1280}{100} = 12.8,\qquad \sigma = \sqrt{\frac{19120}{100} - 12.8^2} = \sqrt{27.36} = 5.23$

$C.V. = \frac{5.23}{12.8}\times100 = 40.9\%$

Type B. $\sum fm = 1310$ and $\sum fm^2 = 21040$.

$\bar x = 13.1,\qquad \sigma = \sqrt{210.4 - 171.61} = \sqrt{38.79} = 6.23$

$C.V. = \frac{6.23}{13.1}\times100 = 47.5\%$

Conclusion. Type A has the smaller C.V. ($40.9\% < 47.5\%$), so its profit is more uniform (consistent).

I would invest in Type A. Its profit is more stable and less risky, while its average profit (Rs 1,280) is almost as high as that of Type B (Rs 1,310), and the small difference does not compensate for B's greater variability.

Sampling is the process of selecting a part (sample) of the units from the whole population so that conclusions about the population can be drawn from the sample.

Qualities of a good questionnaire:

• It is short and contains only the questions necessary for the objective.
• Questions are simple, clear, unambiguous and in easy language.
• Questions are arranged in a logical order, from easy and general to difficult and personal.
• Questions are not leading, biased, or embarrassing, and avoid offending the respondent.
• Each question deals with one idea only.
• Questions are, where possible, answerable by "yes/no" or by choosing from alternatives, which makes tabulation easy.
• Instructions are given clearly, with a covering letter explaining the purpose and assuring confidentiality.
• It has been pre-tested on a small group, and questions can be verified by cross-checks.

Total income of all residents $=250\times 5000=\text{Rs }12{,}50{,}000$

Total income of 100 females $=100\times 3000=\text{Rs }3{,}00{,}000$

Number of males $=250-100=150$

Total income of males $=12{,}50{,}000-3{,}00{,}000=\text{Rs }9{,}50{,}000$

$\text{Average income of males}=\frac{9{,}50{,}000}{150}=\text{Rs }6{,}333.33$

Dispersion is the extent to which the observations of a distribution are scattered or spread about a central value (average). Examples: range, quartile deviation, mean deviation, standard deviation.

Total outcomes $=2^3=8$: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

Favourable outcomes (two heads, one tail): HHT, HTH, THH $=3$.

$P(\text{2 heads and 1 tail})=\frac{3}{8}=0.375$

Given $\mu_2=3$ and $\mu_3=20$ (central moments).

$\beta_1=\frac{\mu_3^2}{\mu_2^3}=\frac{20^2}{3^3}=\frac{400}{27}=14.81$

$\gamma_1=\sqrt{\beta_1}=\frac{\mu_3}{\mu_2^{3/2}}=\frac{20}{5.196}=3.85$

The coefficient of skewness is $\gamma_1\approx 3.85$ (with $\beta_1\approx14.81$), indicating strong positive skewness.

The correlation coefficient is the geometric mean of the two regression coefficients:

$r=\pm\sqrt{b_{yx}\,b_{xy}}$

with the sign of $r$ the same as the sign of the regression coefficients.

Here $b_{yx}\,b_{xy}=0.7\times(-0.8)=-0.56$, which is negative. Since $r^2$ cannot be negative, the two regression coefficients must have the same sign, so the given values are inconsistent and no valid $r$ exists.

If both coefficients are taken with the same sign (as $-0.7$ and $-0.8$):

$r=-\sqrt{0.7\times0.8}=-\sqrt{0.56}=-0.748$

Trend (the long-term movement of a time series) is measured by four methods:

1. Freehand (graphic) method: plot the data on a graph and draw a smooth line or curve through the points by eye. It is simple but subjective.
2. Semi-average method: divide the data into two equal parts, find the average of each part and plot them against the mid-points of their periods. Joining the points gives the trend line.
3. Moving average method: compute the averages of successive groups of $m$ observations (3-yearly, 5-yearly, etc.), which smooth out short-term fluctuations. The period of the moving average should match the cycle length.
4. Method of least squares: fit the straight line $Y=a+bX$ by minimising $\sum (Y-Y_c)^2$. The normal equations are $\sum Y=na+b\sum X$ and $\sum XY=a\sum X+b\sum X^2$. With the origin at the middle year, $\sum X=0$, so $a=\bar Y$ and $b=\sum XY/\sum X^2$. It is the most objective method, and can be extended to parabolic and exponential trends.

Tests of an ideal index number (adequacy tests):

1. Time reversal test: if the time subscripts are interchanged, the two index numbers should be reciprocals: $P_{01}\times P_{10}=1$.
2. Factor reversal test: interchanging price and quantity should give a result such that the product of the price index and quantity index equals the true value ratio: $P_{01}\times Q_{01}=\dfrac{\sum p_1q_1}{\sum p_0q_0}$.
3. Unit test: the index should not depend on the units in which prices or quantities are quoted.
4. Circular test: an extension of the time reversal test for shifting the base: $P_{01}\times P_{12}\times P_{20}=1$.
5. Proportionality test: if all prices change in the same proportion $k$, the index should be $k$ times the base.

Fisher's ideal index satisfies the time reversal and factor reversal tests.

A cumulative frequency curve (ogive) is a graph obtained by plotting cumulative frequencies against the class limits (upper limits for "less than" and lower limits for "more than") and joining the points with a smooth curve.

Uses

• To find the median, quartiles, deciles and percentiles graphically (the median is the value corresponding to $N/2$).
• To find the number or percentage of observations above or below a given value.
• To compare two or more frequency distributions on the same graph.
• To determine the partition values quickly without calculation.
• To check the shape of the distribution (the "less than" and "more than" ogives intersect at the median).

(a) Secondary data

Secondary data are data that have already been collected and processed by someone else for some other purpose and are now used by the investigator. Sources include published reports of government bodies (Central Bureau of Statistics, Nepal Rastra Bank), journals, newspapers, research papers, websites and unpublished records of institutions.

Factors to consider before using secondary data

• Reliability of the source: the agency or collector should be competent and honest.
• Suitability: the objective, scope and definitions of units used in the original study should match the present need.
• Adequacy: the data should cover the required area and period sufficiently.
• Accuracy: the method of collection, sample size and degree of accuracy achieved should be checked.
• Time period: the data should be up to date and refer to a comparable period.
• Bias: any possible bias or errors in collection and processing should be examined.

(b) Histogram and frequency curve

The class widths are unequal (10, 20, 20, ...), so frequencies must be adjusted to a common width of 20. Adjusted frequency $=f\times\dfrac{20}{\text{class width}}$.

Histogram: take marks on the X-axis and adjusted frequency on the Y-axis. Draw adjacent rectangles on each class with heights 20, 25, 40, 80, 60, 45, 15. The tallest bar, at 60-80, shows the modal class.

Frequency curve: mark points at the mid-values above the bar tops, (15, 20), (30, 25), (50, 40), (70, 80), (90, 60), (110, 45), (130, 15), and join them with a smooth free-hand curve. It rises to a peak at about 70 and then falls, with a long tail to the right.

(a) Central tendency

Central tendency is the tendency of the observations of a distribution to cluster around a central value that represents the whole data. Measures: mean, median, mode, geometric mean, harmonic mean.

Properties of a good measure

• It should be rigidly defined.
• It should be easy to understand and calculate.
• It should be based on all observations.
• It should be capable of further algebraic treatment.
• It should have sampling stability (be least affected by fluctuations of sampling).
• It should not be unduly affected by extreme values.

(b) Converting "less than" cumulative frequencies to class frequencies:

$\bar X=\frac{\sum fm}{N}=\frac{15+105+250+490+270+550}{50}=\frac{1680}{50}=33.6$

$\text{Median}=L+\frac{N/2-c.f.}{f}\times h=30+\frac{25-20}{14}\times10=33.57$

$\text{Mode}=L+\frac{f_1-f_0}{2f_1-f_0-f_2}\times h=30+\frac{14-10}{28-10-6}\times10=33.33$

The mean, median and mode are close to each other (about 33.6, 33.6, 33.3), so the average mark is about 33.6 and the median (33.57) is the most suitable single measure.

Minimum pass marks: 60% pass, so 40% fail, i.e. $0.4\times50=20$ students fail. The 20th student corresponds to cumulative frequency 20, which is reached at 30 marks. So the passing students obtained at least 30 marks (minimum pass mark is 30).

(a) Kurtosis

Kurtosis is the degree of peakedness or flatness of the top of a frequency curve compared with the normal curve. Curves are:

• Leptokurtic: more peaked than normal ($\beta_2>3$).
• Mesokurtic: normal curve ($\beta_2=3$).
• Platykurtic: flatter than normal ($\beta_2<3$).

Methods of measuring kurtosis

1. Karl Pearson's coefficient: $\beta_2=\dfrac{\mu_4}{\mu_2^2}$ and $\gamma_2=\beta_2-3$.
2. Percentile coefficient of kurtosis: $\kappa=\dfrac{Q.D.}{P_{90}-P_{10}}$ (0.263 for a normal curve).
3. Graphical comparison with the normal curve.

(b) The competitor with the greater coefficient of variation is more variable.

Competitor A: $\sum X=6+12+12+10+7+9+15+8=79$, $\bar X=9.875$

$\sum X^2=36+144+144+100+49+81+225+64=843$

$\sigma_A=\sqrt{\frac{843}{8}-9.875^2}=\sqrt{105.375-97.516}=\sqrt{7.859}=2.80$

$C.V._A=\frac{2.80}{9.875}\times100=28.39\%$

Competitor B: $\sum X=12+15+18+7+5+4+6+4=71$, $\bar X=8.875$

$\sum X^2=144+225+324+49+25+16+36+16=835$

$\sigma_B=\sqrt{\frac{835}{8}-8.875^2}=\sqrt{104.375-78.766}=\sqrt{25.609}=5.06$

$C.V._B=\frac{5.06}{8.875}\times100=57.02\%$

Since $C.V._B>C.V._A$, Competitor B is more variable (less consistent), while A is more consistent although A also has the higher average.

Take $A=85$ and $d=\dfrac{m-85}{10}$, $N=88$.

Moments about $A$ (in units of class width):

$m_1'=\frac{-18}{88}=-0.2045,\quad m_2'=\frac{282}{88}=3.2045,\quad m_3'=\frac{-138}{88}=-1.5682,\quad m_4'=\frac{1770}{88}=20.1136$

Central moments (class width $h=10$):

$\mu_2=(m_2'-m_1'^2)h^2=(3.2045-0.0418)\times100=316.27$

$\mu_3=(m_3'-3m_2'm_1'+2m_1'^3)h^3=(-1.5682+1.9662-0.0171)\times1000=381.1$

$\mu_4=(m_4'-4m_3'm_1'+6m_2'm_1'^2-3m_1'^4)h^4\approx(20.1136-1.2830+0.8043-0.0052)\times10^4\approx1.963\times10^5$

Mean $=85+10(-0.2045)=82.95$.

Coefficient of skewness

$\beta_1=\frac{\mu_3^2}{\mu_2^3}=\frac{(381.1)^2}{(316.27)^3}=0.0046,\qquad \gamma_1=\sqrt{\beta_1}=\frac{\mu_3}{\mu_2^{3/2}}=\frac{381.1}{5624.7}=0.068$

(and $\beta_2=\mu_4/\mu_2^2\approx1.96$).

Interpretation: $\mu_3>0$, so $\gamma_1=+0.068$ is positive but very close to zero. The distribution is practically symmetrical, with only a very slight positive skewness (a slightly longer tail to the right). $\beta_2<3$ shows it is somewhat flat-topped (platykurtic).

(a) Scatter diagram

A scatter diagram is a graph in which paired observations $(X,Y)$ are plotted as points on a plane, with $X$ on the horizontal axis and $Y$ on the vertical axis.

Role in understanding correlation

• If the points rise from lower left to upper right, there is positive correlation; if they fall, negative correlation.
• Points close to a straight line indicate high correlation; widely scattered points indicate low correlation; no pattern indicates no correlation.
• Points lying exactly on a line mean perfect correlation ($r=\pm1$).
• It shows whether the relationship is linear or non-linear and reveals outliers.
• It is simple and quick, but gives only a rough idea; it does not give the numerical degree.

(b) Let $X$ = promotional expenses, $Y$ = sales, $n=9$.

$\sum X=180$, $\sum Y=360$, $\sum XY=7393$, $\sum X^2=3720$

$\bar X=20$, $\bar Y=40$

$b_{yx}=\frac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\frac{9(7393)-180(360)}{9(3720)-180^2}=\frac{66537-64800}{33480-32400}=\frac{1737}{1080}=1.608$

$a=\bar Y-b\bar X=40-1.608(20)=7.83$

Regression equation of sales on promotional expenses:

$Y=7.83+1.608X$

For $X=22$:

$Y=7.83+1.608(22)=43.22$

Estimated sales $\approx$ Rs 43,220 (that is, 43.22 thousand).

(a) Problems in constructing index numbers

• Purpose: the objective of the index must be clearly defined, since this decides what data and method to use.
• Selection of commodities: a representative, manageable set of items must be chosen; they should be of standard quality.
• Selection of the base period: it should be a normal period, not too distant, free from abnormal events.
• Collection of prices: prices must be obtained from reliable, representative markets and sources (wholesale or retail).
• Choice of average: arithmetic mean or geometric mean (geometric mean is theoretically better).
• Choice of weights: relative importance of items must be fixed fairly (base year or current year quantities).
• Choice of formula: Laspeyres, Paasche, Fisher, etc.; each has its own bias.
• Choice of units and changes in quality: units must be uniform, and changes in quality or new items make comparison difficult.

(b) Price per unit $p=\text{value}/\text{quantity}$.