Introduction to Business Statistics

Mode = 28

Using Karl Pearson’s empirical relation for a moderately asymmetric distribution:

$Mode = 3\,Median - 2\,Mean$

$Mode = 3(20) - 2(16) = 60 - 32 = 28$

Quartile deviation,
$QD = \frac{Q_3 - Q_1}{2} = \frac{70 - 40}{2} = 15$

Quartile deviation = 15

124.29

$\bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1+n_2} = \frac{20(120)+15(130)}{20+15} = \frac{2400+1950}{35} = \frac{4350}{35} = 124.29$

Given: $r = 0.8$, $\text{Cov}(X,Y) = 20$, $\text{Var}(X) = 16$

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

Using the formula:

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

$0.8 = \frac{20}{4 \times \sigma_Y}$

$0.8 = \frac{5}{\sigma_Y}$

$\sigma_Y = \frac{5}{0.8} = 6.25$

Standard deviation of Y = 6.25

Given:

• Quartile deviation, $QD = 2$
• $P_{90} = 24$, $P_{10} = 16$

Percentile coefficient of kurtosis is:

$K = \frac{QD}{P_{90}-P_{10}}$

$K = \frac{2}{24-16} = \frac{2}{8} = 0.25$

Kurtosis = 0.25

(0.6)

$S_k = \frac{\bar{x} - M_o}{\sigma} = \frac{65 - 62}{5} = \frac{3}{5} = 0.6$

Since $S_k > 0$, the distribution is positively (right) skewed.

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$= 0.5 + 0.4 - 0.2 = 0.7$

Answer: 0.7

For a Poisson distribution,

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

Given $\lambda=2$ and $x=2$:

$P(X=2)=\frac{e^{-2}2^2}{2!}=\frac{4e^{-2}}{2}=2e^{-2}$

$P(X=2)=0.2707 \text{ approximately}$

Answer: (P(x=2)=0.2707)

Standard error of mean (with finite population correction):

$SE_{\bar x}=\frac{\sigma}{\sqrt n}\sqrt{\frac{N-n}{N-1}}$

Given: (N=500,\ n=50,\ \sigma=5)

$SE_{\bar x}=\frac{5}{\sqrt{50}}\sqrt{\frac{500-50}{500-1}}$

$=\frac{5}{7.0711}\sqrt{\frac{450}{499}}$

$=0.7071 \times 0.9496$

$=0.6715$

Standard error of mean = 0.6715 (approximately).

The types of random (probability) sampling techniques are:

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

Let the missing frequencies be:

• $f_1$ = workers in class 20–30
• (f_2) = workers in class 40–50

Total workers:

$4+6+f_1+20+f_2+10=60$

$f_1+f_2=20 \quad \text{...(i)}$

Calculation of mean:

Given mean = 35:

$\bar{x}=\frac{\sum fm}{N}$

$35=\frac{1360+25f_1+45f_2}{60}$

$2100=1360+25f_1+45f_2$

$25f_1+45f_2=740$

Dividing by 5:

$5f_1+9f_2=148 \quad \text{...(ii)}$

From (i), (f_1=20-f_2). Substitute in (ii):

$5(20-f_2)+9f_2=148$

$100-5f_2+9f_2=148$

$4f_2=48$

$f_2=12$

Then,

$f_1=20-12=8$

Missing frequencies:

• Workers in 20–30 = 8
• Workers in 40–50 = 12

Given:
$n = 400$, $\bar{x} = 128$ mm, $s = 13$ mm, confidence level = 95%

Formula:
$95\%\ \text{confidence limits} = \bar{x} \pm z_{\alpha/2}\frac{s}{\sqrt{n}}$

For 95% confidence, $z_{0.025} = 1.96$.

Calculation:
$SE = \frac{s}{\sqrt{n}} = \frac{13}{\sqrt{400}} = \frac{13}{20} = 0.65$

$Margin\ of\ error = 1.96 \times 0.65 = 1.274$

$Lower\ limit = 128 - 1.274 = 126.726\ \text{mm}$

$Upper\ limit = 128 + 1.274 = 129.274\ \text{mm}$

Result:
The 95% confidence limits for the population mean blood pressure are:

$126.73\ \text{mm} \quad \text{to} \quad 129.27\ \text{mm}$

So, we are 95% confident that the true mean systolic blood pressure of males lies between 126.73 mm and 129.27 mm.

Given:

Total number of persons,
$N = 1520$

First quartile ($Q_1$):

$Q_1 \text{ position} = \frac{N}{4} = \frac{1520}{4} = 380$

$380$ lies in the class 1000–1999.

$Q_1 = L + \frac{\frac{N}{4} - cf}{f} \times h$

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

$Q_1 = 1000 + 660 = \text{Rs } 1660$

Third quartile ($Q_3$):

$Q_3 \text{ position} = \frac{3N}{4} = \frac{3 \times 1520}{4} = 1140$

$1140$ lies in the class 3000–3999.

$Q_3 = L + \frac{\frac{3N}{4} - cf}{f} \times h$

$Q_3 = 3000 + \frac{1140 - 1105}{100} \times 1000$

$Q_3 = 3000 + 350 = \text{Rs } 3350$

Coefficient of quartile deviation:

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

$= \frac{3350 - 1660}{3350 + 1660}$

$= \frac{1690}{5010} = 0.3373$

$\boxed{\text{Coefficient of quartile deviation} \approx 0.337 \text{ or } 33.73\%}$

Given data:

Formula:

$\text{Coefficient of Variation (C.V.)} = \frac{\sigma}{\bar{x}} \times 100$

Calculation:

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

For Brand B: $C.V._B = \frac{6}{12} \times 100 = 50.00\%$

Result:

Conclusion:

Since the coefficient of variation of Brand A (45.45%) is less than that of Brand B (50.00%), Brand A shows greater consistency in its performance regarding life. Hence, Brand A is more consistent.

Mode (grouped data)

The highest frequency is 45, so the modal class is 20–30.

$Mode = L + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$

Where:

• $L = 20$ (lower limit of modal class)
• $f_1 = 45$ (frequency of modal class)
• $f_0 = 20$ (frequency of preceding class)
• $f_2 = 15$ (frequency of succeeding class)
• $h = 10$ (class width)

$Mode = 20 + \frac{45 - 20}{2(45) - 20 - 15} \times 10$

$= 20 + \frac{25}{90 - 35} \times 10$

$= 20 + \frac{25}{55} \times 10$

$= 20 + 4.545$

$= 24.545$

Mode = 24.55 marks (approximately).

Top 10% of 500 students = $10\% \times 500 = 50$ students.

Cumulative frequency:

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

$P_{90}=L+\frac{\frac{90N}{100}-cf}{f}\times h$

Here, $P_{90}$ lies in class $60–80$: $L=60,\quad \frac{90N}{100}=450,\quad cf=390,\quad f=60,\quad h=20$

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

$=60+\frac{60}{60}\times 20=60+20=80$

Therefore, the lowest marks of the top 10% students is 80 marks.

Step 1: Mean and standard deviation (unequal class widths, so use mid-values directly)

$\bar x=\frac{\sum fm}{N}=\frac{23{,}900}{500}=47.8,\qquad \sigma=\sqrt{\frac{227{,}580}{500}}=\sqrt{455.16}=21.33$

Step 2: Quartiles and median ($N/4=125$, $N/2=250$, $3N/4=375$)

$Q_1=20+\frac{125-50}{100}\times20=35,\qquad M=40+\frac{250-150}{150}\times10=46.67,\qquad Q_3=50+\frac{375-300}{90}\times10=58.33$

Step 3: Coefficient of skewness

Bowley's coefficient: $SK_B=\frac{Q_3+Q_1-2M}{Q_3-Q_1}=\frac{58.33+35-93.33}{23.33}=0$

Pearson's coefficient (median form, because the classes have unequal widths and the mode is unreliable): $SK_P=\frac{3(\bar x-M)}{\sigma}=\frac{3(47.8-46.67)}{21.33}=+0.16$

Interpretation: Both coefficients are zero or very close to zero, so the distribution of marks is practically symmetrical. There is only a very slight positive skew (mean 47.8 is a little above median 46.67), meaning a few high scorers pull the average up slightly. Most of the 500 students score around 45–47, and the marks are spread almost evenly on both sides of the middle.

Total workers, (N=160).

Percentile formula:

[ P_i=L+\frac{\frac{iN}{100}-cf_b}{f}\times h ]

Percentiles

[ P_{10}:\quad \frac{10N}{100}=16 ]

Lies in class (100–199), so (L=99.5,\ cf_b=5,\ f=50,\ h=100).

[ P_{10}=99.5+\frac{16-5}{50}\times 100 =99.5+22=121.5 ]

[ P_{25}:\quad \frac{25N}{100}=40 ]

Lies in class (100–199).

[ P_{25}=99.5+\frac{40-5}{50}\times 100 =99.5+70=169.5 ]

[ P_{75}:\quad \frac{75N}{100}=120 ]

Lies in class (300–399), so (L=299.5,\ cf_b=110,\ f=10,\ h=100).

[ P_{75}=299.5+\frac{120-110}{10}\times 100 =299.5+100=399.5 ]

[ P_{90}:\quad \frac{90N}{100}=144 ]

Lies in class (400–499), so (L=399.5,\ cf_b=120,\ f=30,\ h=100).

[ P_{90}=399.5+\frac{144-120}{30}\times 100 =399.5+80=479.5 ]

Percentile coefficient of kurtosis

[ K=\frac{Q_3-Q_1}{2(P_{90}-P_{10})} =\frac{P_{75}-P_{25}}{2(P_{90}-P_{10})} ]

[ K=\frac{399.5-169.5}{2(479.5-121.5)} ]

[ K=\frac{230}{2\times 358} =\frac{230}{716} =0.3212 ]

[ \boxed{K=0.3212} ]

Comment

For a normal distribution, the percentile coefficient of kurtosis is (0.263). Here,

[ 0.3212>0.263 ]

So the distribution is platykurtic, i.e. it is flatter and less peaked than a normal distribution.

Let $X$ be the daily lunch expenditure (in Rs) of an employee.

$X \sim N(\mu = 120,\ \sigma = 20)$

Standard normal variable:

$z=\frac{X-\mu}{\sigma}=\frac{X-120}{20}$

(a) Probability between Rs 105 and Rs 140

$P(105<X<140)=P\left(\frac{105-120}{20}<z<\frac{140-120}{20}\right)$

$=P(-0.75<z<1.00)$

$=P(0<z<0.75)+P(0<z<1.00)$

$=0.2734+0.3413=0.6147$

So, the probability is:

$\boxed{0.6147}$

Expected number of employees:

$400 \times 0.6147 = 245.88 \approx 246$

(b) Probability between Rs 125 and Rs 150

$P(125<X<150)=P\left(\frac{125-120}{20}<z<\frac{150-120}{20}\right)$

$=P(0.25<z<1.50)$

$=P(0<z<1.50)-P(0<z<0.25)$

$=0.4332-0.0987=0.3345$

So, the probability is:

$\boxed{0.3345}$

Expected number of employees:

$400 \times 0.3345 = 133.8 \approx 134$

Answer:
(a) $0.6147$ or about $246$ employees
(b) $0.3345$ or about $134$ employees

Testing the hypothesis about population mean (large sample)

Given:

• Sample size, $n = 100$
• Sample mean, $\bar{x} = 65$ kg
• Sample standard deviation, $s = 20$ kg
• Hypothesised population mean, $\mu_0 = 60$ kg
• Level of significance, $\alpha = 0.05$

Hypotheses: $H_0: \mu = 60 \text{ kg} \quad \text{(population mean weight is 60 kg)}$ $H_1: \mu \neq 60 \text{ kg} \quad \text{(two-tailed test)}$

Test statistic: Since $n = 100$ is large ($n > 30$), we use the $Z$-test: $Z = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$

Substitution: $Z = \frac{65 - 60}{20/\sqrt{100}} = \frac{5}{20/10} = \frac{5}{2} = 2.5$

Critical value: At $\alpha = 0.05$ for a two-tailed test, the critical values are $Z_{0.025} = \pm 1.96$

Decision: Since the calculated value $|Z| = 2.5 > 1.96$, the test statistic falls in the critical (rejection) region. We reject $H_0$.

Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean weight of the population is not 60 kg. The sample mean of 65 kg differs significantly from the hypothesised mean.

Reading the table. The headings are misprinted. Taking the evident pattern, income $X$ has classes 0–500, 500–1000, 1000–1500, 1500–2000, 2000–2500 and expenditure $Y$ has classes 0–400, 400–800, 800–1200, 1200–1600, 1600–2000. $N=110$.

Use step deviations $u=\dfrac{X-1250}{500}$ and $v=\dfrac{Y-1000}{400}$.

Totals: $\sum f_xu=-52$, $\sum f_xu^2=176$, $\sum f_yv=-66$, $\sum f_yv^2=182$, $\sum fuv=114$.

(a) Regression lines

$\bar X=1250+500\left(\frac{-52}{110}\right)=1013.64,\qquad \bar Y=1000+400\left(\frac{-66}{110}\right)=760$

$\sigma_u=\sqrt{\tfrac{176}{110}-\left(\tfrac{-52}{110}\right)^2}=1.1733,\quad \sigma_v=\sqrt{\tfrac{182}{110}-\left(\tfrac{-66}{110}\right)^2}=1.1378$

$\sigma_X=500\times1.1733=586.63,\qquad \sigma_Y=400\times1.1378=455.11$

$\text{Cov}(u,v)=\frac{114}{110}-\left(\frac{-52}{110}\right)\left(\frac{-66}{110}\right)=0.7527$

$b_{yx}=\frac{\text{Cov}(u,v)}{\sigma_u^2}\cdot\frac{400}{500}=0.4375,\qquad b_{xy}=\frac{\text{Cov}(u,v)}{\sigma_v^2}\cdot\frac{500}{400}=0.7268$

Regression of $Y$ on $X$: $Y-760=0.4375(X-1013.64)\;\Rightarrow\; Y=316.57+0.4375X$

Regression of $X$ on $Y$: $X-1013.64=0.7268(Y-760)\;\Rightarrow\; X=461.25+0.7268Y$

(b) Correlation coefficient

$r=\frac{\text{Cov}(u,v)}{\sigma_u\sigma_v}=\frac{0.7527}{1.1733\times1.1378}=0.564$

Check: $r=\sqrt{b_{yx}b_{xy}}=\sqrt{0.4375\times0.7268}=0.564$.

(c) Probable error

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

$6\,PE=0.263$. Since $r=0.564$ is more than $6\,PE$, the correlation is significant. There is a moderately strong positive relationship: families with higher income spend more.

(d) Estimated expenditure at income Rs 4,000

$Y=316.57+0.4375(4000)=316.57+1750\approx \text{Rs }2{,}066$

(This lies well beyond the observed income range, so the estimate should be treated with caution.)

Total ways of forming a 5-member sub-committee from 10 persons:

$\binom{10}{5}=252$

Favourable ways: choose 3 girls from 6 and 2 boys from 4:

$\binom{6}{3}\times \binom{4}{2}=20\times 6=120$

Required probability:

$P=\frac{120}{252}=\frac{10}{21}$

Answer: $\frac{10}{21}$

For 99% confidence, $z_{\alpha/2}=2.576$.

Given:
$\sigma = 12$, $E = 3$

$n=\left(\frac{z_{\alpha/2}\sigma}{E}\right)^2$

$n=\left(\frac{2.576 \times 12}{3}\right)^2=(10.304)^2=106.17$

Rounding up, the required sample size is:

$\boxed{n=107}$

For a binomial distribution,

$P(r)= {n \choose r}p^r q^{n-r}, \quad q=1-p$

Given: $n=10,\ p=0.6,\ q=0.4,\ r=3$

$P(r=3)= {10 \choose 3}(0.6)^3(0.4)^7$

$=120 \times 0.216 \times 0.0016384$

$=0.04247$

Therefore, $P(r=3) \approx 0.0425$.

The data as printed are inconsistent. Only four values of $X$ (3, 7, 12, 18) are given, but there are five frequencies (2, 6, 10, 8, 4), so one value of $X$ is missing and the question cannot be solved exactly as written.

Method. For a discrete series, $Q_1$ is the size of the $\frac{N+1}{4}$th item and $Q_3$ the size of the $\frac{3(N+1)}{4}$th item, and $QD=\frac{Q_3-Q_1}{2}.$

Here $N=2+6+10+8+4=30$, so $Q_1$ is the 7.75th item and $Q_3$ the 23.25th item.

If the missing value is a fifth, larger value at the end of the series (the usual misprint), then:

• the 7.75th item lies in the group with $cf=8$, so $Q_1=7$;
• the 23.25th item lies in the group with $cf=26$, so $Q_3=18$.

$QD=\frac{18-7}{2}=5.5$

This result does not depend on what the fifth value is, as long as it is greater than 18. If instead the missing value lies elsewhere in the series, the frequencies must be re-aligned and the answer changes, so the original question paper should be checked.

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

Hypotheses: $H_0:\mu=250\qquad H_1:\mu>250\ \text{(right-tailed)}$

Test statistic (large sample, $n>30$, so a $Z$-test): $Z=\frac{\bar x-\mu_0}{s/\sqrt n}=\frac{300-250}{11/\sqrt{50}}=\frac{50}{1.5556}=32.14$

Critical value: $Z_{0.05}=1.645$ (one-tailed).

Decision: Since $32.14>1.645$, reject $H_0$.

Conclusion: At the 5% level of significance, the sample shows that the current average monthly wage rate is higher than Rs 250.

Population mean: [ \mu=\frac{3+5+8+10+14}{5}=\frac{40}{5}=8 ]

All possible samples of size 2 without replacement ((N=5,\ n=2)):

Number of samples: [ \binom{5}{2}=10 ]

Mean of sample means: [ E(\bar x)=\frac{4+5.5+6.5+8.5+6.5+7.5+9.5+9+11+12}{10} ] [ E(\bar x)=\frac{80}{10}=8 ] Hence, [ E(\bar x)=\mu=8 ]

So the mean of the sample means is equal to the population mean.

Population standard deviation: [ \sigma=\sqrt{\frac{\sum (X-\mu)^2}{N}} ] [ \sum (X-\mu)^2=(3-8)^2+(5-8)^2+(8-8)^2+(10-8)^2+(14-8)^2 ] [ =25+9+0+4+36=74 ] [ \sigma=\sqrt{\frac{74}{5}}=\sqrt{14.8}=3.85 ]

Standard deviation of sampling distribution: [ \sigma_{\bar x}=\frac{\sigma}{\sqrt n}\sqrt{\frac{N-n}{N-1}} ] [ \sigma_{\bar x}=\frac{\sqrt{14.8}}{\sqrt2}\sqrt{\frac{5-2}{5-1}} ] [ \sigma_{\bar x}=\sqrt{\frac{14.8}{2}\times\frac{3}{4}} =\sqrt{5.55}=2.36 ]

Now, [ \sigma_{\bar x}=2.36 < 3.85=\sigma ]

Therefore, the standard deviation of the sampling distribution is less than the population standard deviation.

Given: $N=150$ families. The middle 70% of families lie between $P_{15}$ and $P_{85}$, so the required range is $P_{85}-P_{15}$.

$P_{15}$: position $=\dfrac{15\times150}{100}=22.5$, which lies in the class 20–40. $P_{15}=20+\frac{22.5-15}{25}\times20=26$

$P_{85}$: position $=\dfrac{85\times150}{100}=127.5$, which lies in the class 60–80. $P_{85}=60+\frac{127.5-100}{30}\times20=78.33$

Range of the middle 70%: $P_{85}-P_{15}=78.33-26=52.33\ \text{hundred}\approx \text{Rs }5{,}233$

The middle 70% of families pay monthly rent between Rs 2,600 and Rs 7,833.

Sampling Techniques

Meaning: Sampling is the process of selecting a representative part (sample) from a large group (population) for the purpose of study and drawing conclusions about the whole population. The technique or method used to select this sample is called a sampling technique. A good sample must be representative, unbiased, and adequate in size.

Methods of Sampling Techniques

A. Probability (Random) Sampling — every unit of the population has a known, equal chance of being selected.

1. Simple Random Sampling: Each unit is selected purely by chance. Method: lottery method or random number table. Example: drawing 50 names from a box containing 1,000 names.
2. Systematic Sampling: Units are selected at a fixed interval (k = N/n) after a random start. Example: selecting every 10th student from a register of 500 students.
3. Stratified Sampling: The population is divided into homogeneous groups (strata) such as male/female or faculty-wise, and samples are drawn randomly from each stratum. It gives more accurate results.
4. Cluster Sampling: The population is divided into clusters (e.g., wards, schools), some clusters are selected randomly, and all units within them are studied. Useful when the population is widely spread.
5. Multi-stage Sampling: Sampling is done in several stages — first districts, then schools, then classes, then students.

B. Non-probability (Non-random) Sampling — selection depends on the researcher's judgment or convenience; the chance of selection is not equal.

1. Convenience Sampling: Units are chosen because they are easily available. Example: surveying friends and classmates.
2. Judgment (Purposive) Sampling: The researcher selects units based on his own knowledge and purpose. Example: selecting experts for an opinion survey.
3. Quota Sampling: A fixed number (quota) is assigned to each group and units are selected as per that quota. Example: 40 male and 60 female respondents.
4. Snowball Sampling: Existing respondents refer further respondents; used for rare or hidden populations.

Conclusion: Probability methods are more reliable and free from bias, while non-probability methods are cheaper, quicker and useful when a sampling frame is not available. The choice of method depends on the nature of the study, cost, time and availability of the population list.

Let
(X) = expenditure on advertisement in ’000 Rs
(Y) = volume of sales in ’000 Rs

[ \bar X=8,\qquad \bar Y=\frac{54}{7}=7.7143 ]

So, [ \sum x^2=18,\qquad \sum y^2=\frac{2324}{49},\qquad \sum xy=27 ]

1. Regression equation of sales on advertisement

[ b_{yx}=\frac{\sum xy}{\sum x^2}=\frac{27}{18}=\frac32 ]

[ a=\bar Y-b_{yx}\bar X=\frac{54}{7}-\frac32(8)=-\frac{30}{7} ]

Hence, [ \boxed{Y=-\frac{30}{7}+\frac32X} ] or [ \boxed{14Y=21X-60} ]

When advertisement expenditure is Rs 15,000, (X=15):

[ Y=-\frac{30}{7}+\frac32(15) =\frac{255}{14} =18.2143 ]

So estimated sales (=18.2143) thousand Rs
[ \boxed{\text{Estimated sales} \approx \text{Rs }18{,}214} ]

2. Regression equation of advertisement expenditure on sales

[ b_{xy}=\frac{\sum xy}{\sum y^2} =\frac{27}{2324/49} =\frac{189}{332} ]

[ a=\bar X-b_{xy}\bar Y =8-\frac{189}{332}\left(\frac{54}{7}\right) =\frac{599}{166} ]

Hence, [ \boxed{X=\frac{599}{166}+\frac{189}{332}Y} ] or [ \boxed{332X=189Y+1198} ]

When sales is Rs 20,000, (Y=20):

[ X=\frac{599}{166}+\frac{189}{332}(20) =\frac{2489}{166} =14.9939759 ]

So estimated advertisement expenditure (=14.9939759) thousand Rs
[ \boxed{\text{Estimated advertisement expenditure} \approx \text{Rs }14{,}994} ]

3. Correlation coefficient

[ r=\frac{\sum xy}{\sqrt{\sum x^2\sum y^2}} ]

[ r=\frac{27}{\sqrt{18\left(\frac{2324}{49}\right)}} =\frac{189}{\sqrt{41832}} =\sqrt{\frac{567}{664}} ]

[ \boxed{r \approx 0.9241} ]

Thus, there is a strong positive correlation between advertisement expenditure and sales.

Class width $h=5$, assumed mean $A=12$ (mid-value of 10–14), $d'=\dfrac{m-12}{5}$, $N=100$ for each type.

Means $\bar x_A=12+\frac{16}{100}\times5=12.8,\qquad \bar x_B=12+\frac{22}{100}\times5=13.1$

Standard deviations $\sigma_A=5\sqrt{\frac{112}{100}-\left(\frac{16}{100}\right)^2}=5\sqrt{1.0944}=5.23$ $\sigma_B=5\sqrt{\frac{160}{100}-\left(\frac{22}{100}\right)^2}=5\sqrt{1.5516}=6.23$

Coefficients of variation $CV_A=\frac{5.23}{12.8}\times100=40.9\%,\qquad CV_B=\frac{6.23}{13.1}\times100=47.5\%$

Since $CV_A<CV_B$, Type A companies have more uniform profit.

Investment decision: Invest in Type A. Its profit varies less relative to its average (40.9% against 47.5%), so the return is more consistent and the risk lower. Type B's average profit (Rs 13.1 hundred) is only slightly higher than Type A's (Rs 12.8 hundred), which does not compensate for the greater variability.

Total days, (N = 110+60+15+10+5 = 200)

Total accidents, (\sum fx = (0\times110)+(1\times60)+(2\times15)+(3\times10)+(4\times5)=140)

Mean of Poisson distribution:

[ \lambda = \bar x = \frac{\sum fx}{N} = \frac{140}{200} = 0.7 ]

Expected frequency for (x) accidents:

[ f_e = N \cdot \frac{e^{-\lambda}\lambda^x}{x!} ]

Here, (e^{-0.7}=0.4966)

So, the expected frequencies are approximately:

[ \boxed{99.32,\ 69.52,\ 24.33,\ 5.68,\ 0.99,\ 0.16} ]

for (0,1,2,3,4,) and (5) or more accidents respectively.

Given: MA Economics: $n_1=60$, $\bar x_1=68$. MBS (management): $n_2=50$, $\bar x_2=69$. $\sigma=2.24$ (known), $\alpha=0.05$.

Hypotheses: $H_0:\mu_1=\mu_2\qquad H_1:\mu_2>\mu_1\ \text{(management students are taller; one-tailed)}$

Test statistic: $z=\frac{\bar x_2-\bar x_1}{\sigma\sqrt{\frac1{n_1}+\frac1{n_2}}}=\frac{69-68}{2.24\sqrt{\frac1{60}+\frac1{50}}}=\frac{1}{2.24\times0.1915}=2.33$

Decision: The critical value at 5% (one-tailed) is $1.645$. Since $2.33>1.645$, reject $H_0$.

Conclusion: At the 5% level, we conclude that management (MBS) students are taller than economics (MA) students.

Construction of the frequency distribution (30 observations)

The smallest value, 66, lies below the first stated class 68–80. To include every observation, the first class is taken as 60–80. A value equal to a class limit goes into the higher class (e.g. 100 into 100–120, 120 into 120–140).

Central tendency $\bar x=\frac{3680}{30}=122.67$ Median class 120–140 ($N/2=15$): $M=120+\dfrac{15-13}{6}\times20=126.67$. Modal class 80–100: $Z=80+\dfrac{10-2}{2(10)-2-1}\times20=89.4$. (The series is bimodal, with a second peak at 140–160, so the mode is not very reliable.)

Dispersion $\sigma=\sqrt{\frac{\sum f(m-\bar x)^2}{N}}=\sqrt{1132.9}=33.66$ $Q_1=80+\frac{7.5-2}{10}\times20=91,\quad Q_3=140+\frac{22.5-19}{8}\times20=148.75$ $CV=\frac{33.66}{122.67}\times100=27.4\%$

Skewness $SK_B=\frac{Q_3+Q_1-2M}{Q_3-Q_1}=\frac{148.75+91-253.33}{57.75}=-0.24$ $SK_P=\frac{3(\bar x-M)}{\sigma}=\frac{3(122.67-126.67)}{33.66}=-0.36$ $\beta_1=\frac{\mu_3^2}{\mu_2^3}=0.044$

Kurtosis $\beta_2=\frac{\mu_4}{\mu_2^2}=\frac{2{,}580{,}444}{(1132.9)^2}=2.01$

Conclusions

• The average value is about 123, with the median (126.7) slightly higher than the mean, and the data are moderately dispersed (CV about 27%).
• Skewness is small and negative ($\beta_1\approx0.04$), so the distribution is nearly symmetrical with a slightly longer tail towards the lower values.
• $\beta_2=2.01<3$, so the distribution is platykurtic, that is, flatter than the normal curve. This agrees with the two separate peaks in the table.

Step 1: Total expenditure $400+120+250+100+50+180=\text{Rs }1{,}100$

Step 2: Angle of each sector, $\text{angle}=\dfrac{\text{item}}{1100}\times360^\circ$

Step 3: Drawing. Draw a circle with a suitable radius (say 4 cm) and mark a horizontal radius. Using a protractor, measure the angles above one after another clockwise, starting with Food (130.9°). Shade or label each sector with its item name and amount, and give the diagram a title, "Monthly expenditure of a family".

Reading the diagram: Food takes the largest sector (over one-third of the budget), followed by clothes, while energy takes the smallest.

Note on the data: The printed frequencies add up to $10+35+70+20+10+5=150$, not 100 as the question states. The answer below uses the figures exactly as printed ($N=150$). If the third frequency was meant to be 20 (so that $N=100$), the results are given at the end.

Convert to continuous classes (width $h=5$):

i. Lowest income of the richest 30% is $P_{70}$. Position $=\dfrac{70\times150}{100}=105$, which lies in the class 9.5–14.5. $P_{70}=9.5+\frac{105-45}{70}\times5=9.5+4.286=13.786\ (\text{'000})\approx \text{Rs }13{,}786$

ii. Highest income of the poorest 25% is $P_{25}$. Position $=\dfrac{25\times150}{100}=37.5$, which lies in the class 4.5–9.5. $P_{25}=4.5+\frac{37.5-10}{35}\times5=4.5+3.929=8.429\ (\text{'000})\approx \text{Rs }8{,}429$

If the third frequency is 20 ($N=100$): $P_{70}=14.5+\frac{70-65}{20}\times5=15.75$, that is Rs 15,750, and $P_{25}=4.5+\frac{25-10}{35}\times5=6.64$, that is Rs 6,643.