Binomial distribution

A random variable $X$ follows the binomial distribution $\beta(n,p)$ if it counts the successes in $n$ independent trials, each with the same success probability $p$ (and failure probability $q = 1-p$). Its probability mass function is

$P(X=x) = \binom{n}{x}p^x q^{n-x}, \qquad x = 0,1,\dots,n.$

Conditions: $n$ is fixed and finite, each trial has two outcomes, the trials are independent, and $p$ is constant.

Mean $= np$, variance $= npq$, moment generating function $M_X(t) = (q+pe^t)^n$.

Recurrence relation for the central moments

Let $\mu_r = E\big[(X-np)^r\big]$ be the $r$-th moment about the mean, so $\mu_0 = 1$, $\mu_1 = 0$.

$\mu_r = \sum_{x=0}^{n}(x-np)^r\binom{n}{x}p^xq^{n-x}.$

Differentiate with respect to $p$, using $\dfrac{dq}{dp} = -1$:

$\frac{d\mu_r}{dp} = \sum_{x=0}^{n} -nr(x-np)^{r-1}\binom{n}{x}p^xq^{n-x} + \sum_{x=0}^{n}(x-np)^r\binom{n}{x}\frac{d}{dp}\big(p^xq^{n-x}\big).$

Now

$\frac{d}{dp}\big(p^xq^{n-x}\big) = p^xq^{n-x}\left(\frac{x}{p}-\frac{n-x}{q}\right) = p^xq^{n-x}\,\frac{x-np}{pq},$

because $xq - (n-x)p = x - np$. Therefore

$\frac{d\mu_r}{dp} = -nr\,\mu_{r-1} + \frac{1}{pq}\sum_{x=0}^{n}(x-np)^{r+1}\binom{n}{x}p^xq^{n-x} = -nr\,\mu_{r-1} + \frac{\mu_{r+1}}{pq}.$

Rearranging,

$\boxed{\mu_{r+1} = pq\left(nr\,\mu_{r-1} + \frac{d\mu_r}{dp}\right)}$

Checks. For $r=1$: $\mu_2 = pq\,(n\cdot 1 + 0) = npq$, the variance. For $r=2$: $\mu_3 = pq\left(0 + \dfrac{d}{dp}(np - np^2)\right) = npq(1-2p) = npq(q-p)$, the known third central moment.

Note that the relation holds for moments about the mean. For raw moments it fails, since $r=1$ would give $2npq \ne \mu_2' = npq + n^2p^2$.

Normal Probability Distribution

A continuous random variable (X) is said to follow a normal probability distribution if its probability density function is given by

[ f(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right], \qquad -\infty<x<\infty ]

where (\mu) and (\sigma) are the parameters of the distribution, with (-\infty<\mu<\infty) and (\sigma>0).

It is denoted by

[ X\sim N(\mu,\sigma^2) ]

The normal distribution is a bell-shaped, symmetric continuous distribution. Its mean, median and mode coincide. The total area under the curve is 1. If (\mu=0) and (\sigma=1), it is called the standard normal distribution, denoted by (Z\sim N(0,1)).

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Mean of Normal Distribution

The mean of a continuous random variable is

[ E(X)=\int_{-\infty}^{\infty}x f(x),dx ]

For the normal distribution,

[ E(X)=\int_{-\infty}^{\infty}x\frac{1}{\sigma\sqrt{2\pi}} \exp\left[-\frac{(x-\mu)^2}{2\sigma^2}\right]dx ]

Let

[ z=\frac{x-\mu}{\sigma} ]

Then

[ x=\mu+\sigma z,\qquad dx=\sigma,dz ]

Also, when (x\to -\infty), (z\to -\infty), and when (x\to \infty), (z\to \infty).

Therefore,

[ E(X)=\int_{-\infty}^{\infty}(\mu+\sigma z)\frac{1}{\sqrt{2\pi}}e^{-z^2/2},dz ]

[ E(X)=\mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2/2},dz + \sigma\int_{-\infty}^{\infty}z\frac{1}{\sqrt{2\pi}}e^{-z^2/2},dz ]

Now,

[ \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-z^2/2},dz=1 ]

and

[ \int_{-\infty}^{\infty}z e^{-z^2/2},dz=0 ]

because (z e^{-z^2/2}) is an odd function and the limits are symmetric about 0.

Hence,

[ E(X)=\mu(1)+\sigma(0) ]

[ \boxed{E(X)=\mu} ]

Thus, the mean of the normal distribution (N(\mu,\sigma^2)) is (\mu).

Difference between Primary Data and Secondary Data

Sources of Secondary Data

1. Published sources

   • Government publications: census reports, Economic Survey, Statistical Year Book, Nepal Rastra Bank reports, Department of Education reports.
   • International publications: reports of the UN, WHO, World Bank, IMF, UNESCO, ILO.
   • Semi-government and private publications: reports of Nepal Rastra Bank, Federation of Nepalese Chambers of Commerce and Industry, commercial banks and insurance companies.
   • Journals and periodicals: economic and statistical journals, newspapers, magazines.
   • Books and research reports: textbooks, theses, dissertations, research papers.

2. Unpublished sources

   • Records maintained by government offices, municipalities, schools and hospitals.
   • Records of business firms, banks and cooperatives.
   • Research works of students and scholars kept in libraries.
   • Personal records and diaries.

Note: Secondary data should be used only after checking its reliability, suitability and adequacy for the enquiry.

Show that the sum of squares of deviations is least when measured from the mean

Let the variable $x$ take the values $x_1, x_2, \dots, x_n$ with mean $\bar{x}$, where

$\bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$

Let $A$ be any arbitrary number from which the deviations are measured. Define the sum of squares of deviations from $A$ as

$S(A) = \sum_{i=1}^{n}(x_i - A)^2$

We must show that $S(A)$ is minimum when $A = \bar{x}$.

Step 1: Write the deviations in terms of the mean.

Add and subtract $\bar{x}$ inside the bracket:

$S(A) = \sum_{i=1}^{n}\left[(x_i - \bar{x}) + (\bar{x} - A)\right]^2$

Step 2: Expand the square.

$S(A) = \sum_{i=1}^{n}(x_i - \bar{x})^2 + 2(\bar{x} - A)\sum_{i=1}^{n}(x_i - \bar{x}) + \sum_{i=1}^{n}(\bar{x} - A)^2$

Step 3: Use the property of the mean.

The sum of deviations from the mean is always zero:

$\sum_{i=1}^{n}(x_i - \bar{x}) = \sum x_i - n\bar{x} = n\bar{x} - n\bar{x} = 0$

So the middle term vanishes, and the last term (a constant added $n$ times) gives

$S(A) = \sum_{i=1}^{n}(x_i - \bar{x})^2 + n(\bar{x} - A)^2$

Step 4: Interpret the result.

The first term $\sum (x_i - \bar{x})^2$ is a fixed non-negative quantity, while the second term $n(\bar{x} - A)^2 \ge 0$ depends on $A$. Hence

$S(A) \ge \sum_{i=1}^{n}(x_i - \bar{x})^2$

with equality only when

$n(\bar{x} - A)^2 = 0 \quad \Rightarrow \quad A = \bar{x}$

Conclusion.

Therefore $S(A)$ attains its least (minimum) value when the deviations are measured from the mean $\bar{x}$, and this least value is $\sum (x_i - \bar{x})^2$. Hence the sum of squares of deviations is least when taken from the mean. $\blacksquare$

Let the moments about the value (A=2) be

[ m_1=1,\qquad m_2=16,\qquad m_3=-40 ]

where

[ m_r=\frac{1}{N}\sum (x-2)^r ]

1. Mean

[ \bar x=A+m_1=2+1=3 ]

[ \boxed{\bar x=3} ]

2. Variance

[ \mu_2=m_2-m_1^2 ]

[ \mu_2=16-1^2=15 ]

[ \boxed{\text{Variance}=15} ]

3. Skewness

Third central moment:

[ \mu_3=m_3-3m_1m_2+2m_1^3 ]

[ \mu_3=-40-3(1)(16)+2(1)^3 ]

[ \mu_3=-40-48+2=-86 ]

Moment coefficient of skewness:

[ \gamma_1=\frac{\mu_3}{\mu_2^{3/2}} ]

[ \gamma_1=\frac{-86}{15\sqrt{15}} ]

[ \gamma_1=-1.48 \quad \text{(approximately)} ]

Also,

[ \beta_1=\frac{\mu_3^2}{\mu_2^3} =\frac{(-86)^2}{15^3} =\frac{7396}{3375}=2.191 ]

Hence,

[ \boxed{\text{Mean}=3,\quad \text{Variance}=15,\quad \text{Coefficient of skewness}=-1.48} ]

The distribution is negatively skewed.

Let the regression line of (Y) on (X) be [ a_1X + b_1Y = c_1 \quad \Rightarrow \quad Y = -\frac{a_1}{b_1}X + \frac{c_1}{b_1}. ] Thus the regression coefficient of (Y) on (X) is [ b_{yx} = -\frac{a_1}{b_1}. ]

Similarly, the regression line of (X) on (Y) is [ a_2X + b_2Y = c_2 \quad \Rightarrow \quad X = -\frac{b_2}{a_2}Y + \frac{c_2}{a_2}. ] Thus the regression coefficient of (X) on (Y) is [ b_{xy} = -\frac{b_2}{a_2}. ]

We know that the product of the two regression coefficients equals the square of the correlation coefficient: [ b_{yx} b_{xy} = r^2. ] Since (-1 \le r \le 1), we have (r^2 \le 1). Therefore, [ \left(-\frac{a_1}{b_1}\right)\left(-\frac{b_2}{a_2}\right) \le 1 ] [ \frac{a_1 b_2}{a_2 b_1} \le 1. ] In the usual regression form, (a_2 > 0) and (b_1 > 0), so (a_2 b_1 > 0). Multiplying both sides by (a_2 b_1) gives [ a_1 b_2 \le a_2 b_1. ] Hence proved.

Theorem: If (X) and (Y) are independent, then they are uncorrelated

Let (X) and (Y) be independent random variables with finite means and variances.
The covariance of (X) and (Y) is

[ \operatorname{Cov}(X,Y)=E(XY)-E(X)E(Y). ]

Since (X) and (Y) are independent,

[ E(XY)=E(X)E(Y). ]

Therefore,

[ \operatorname{Cov}(X,Y)=E(X)E(Y)-E(X)E(Y)=0. ]

The correlation coefficient is

[ \rho(X,Y)=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}. ]

Since (\operatorname{Cov}(X,Y)=0), we get

[ \rho(X,Y)=0. ]

Hence independent variates are uncorrelated.

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Converse is not necessarily true

Uncorrelated means (\operatorname{Cov}(X,Y)=0), but this does not always imply independence.

Example:

Let (X) take the values (-1,0,1) each with probability (\frac13), and define

[ Y=X^2. ]

Then (Y) takes the values (1,0,1). Clearly (Y) depends on (X), so (X) and (Y) are not independent.

Now,

[ E(X)=\frac{-1+0+1}{3}=0, ]

[ E(Y)=E(X^2)=\frac{1+0+1}{3}=\frac23, ]

and

[ E(XY)=E(X^3)=\frac{(-1)^3+0^3+1^3}{3}=0. ]

Thus,

[ \operatorname{Cov}(X,Y)=E(XY)-E(X)E(Y)=0-\left(0\cdot \frac23\right)=0. ]

So (X) and (Y) are uncorrelated.

But they are not independent, because

[ P(X=1,;Y=1)=P(X=1)=\frac13, ]

whereas

[ P(X=1)P(Y=1)=\frac13\cdot \frac23=\frac29. ]

Since

[ \frac13\ne \frac29, ]

(X) and (Y) are not independent.

Therefore, uncorrelated variates need not be independent. This proves that the converse of the theorem is not necessarily true.

Bayes' theorem. If $B_1, B_2, \dots, B_n$ are mutually exclusive and exhaustive events with $P(B_i) > 0$, and $A$ is any event with $P(A) > 0$, then

$P(B_i \mid A) = \frac{P(B_i)\,P(A \mid B_i)}{\sum_{j=1}^{n} P(B_j)\,P(A \mid B_j)}$

Numerical.

Let $A$, $B$, $C$ denote the events that the prisoner chooses road A, road B, road C respectively. Since the road is chosen at random,

$P(A) = P(B) = P(C) = \frac{1}{3}$

Let $E$ be the event that the prisoner escapes. Given:

$P(E \mid A) = \frac{1}{5}, \qquad P(E \mid B) = \frac{1}{8}, \qquad P(E \mid C) = \frac{1}{10}$

By the theorem of total probability,

$P(E) = P(A)P(E \mid A) + P(B)P(E \mid B) + P(C)P(E \mid C)$

$= \frac{1}{3}\cdot\frac{1}{5} + \frac{1}{3}\cdot\frac{1}{8} + \frac{1}{3}\cdot\frac{1}{10}$

$= \frac{1}{3}\left(\frac{1}{5} + \frac{1}{8} + \frac{1}{10}\right) = \frac{1}{3}\left(\frac{8 + 5 + 4}{40}\right) = \frac{1}{3}\cdot\frac{17}{40} = \frac{17}{120}$

By Bayes' theorem, the required probability is

$P(A \mid E) = \frac{P(A)\,P(E \mid A)}{P(E)} = \frac{\frac{1}{3}\cdot\frac{1}{5}}{\frac{17}{120}} = \frac{\frac{1}{15}}{\frac{17}{120}} = \frac{1}{15}\times\frac{120}{17} = \frac{8}{17}$

Answer: $\boxed{\dfrac{8}{17}}$

Random variable

A random variable is a real-valued function defined on the sample space of a random experiment. It assigns a real number to each possible outcome.
If the set of possible values is countable, it is called a discrete random variable.

Let (X) be the number of points obtained when a balanced die is rolled. Then
[ X=1,2,3,4,5,6 ] each with probability [ P(X=x)=\frac16. ]

Expected value of (X)

[ E(X)=\sum xP(X=x) ] [ E(X)=\frac{1+2+3+4+5+6}{6} =\frac{21}{6} =\frac72=3.5 ]

Variance of (X)

First find (E(X^2)): [ E(X^2)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} ] [ =\frac{1+4+9+16+25+36}{6} =\frac{91}{6} ]

Now, [ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 ] [ =\frac{91}{6}-\left(\frac72\right)^2 =\frac{91}{6}-\frac{49}{4} ] [ =\frac{182-147}{12} =\frac{35}{12} ] [ \boxed{\operatorname{Var}(X)=\frac{35}{12}\approx 2.9167} ]

Expected value of (g(X)=2X^3+1)

[ E[g(X)]=E(2X^3+1)=2E(X^3)+1 ]

Now, [ E(X^3)=\frac{1^3+2^3+3^3+4^3+5^3+6^3}{6} ] [ =\frac{1+8+27+64+125+216}{6} =\frac{441}{6} =\frac{147}{2} ]

Therefore, [ E[g(X)]=2\left(\frac{147}{2}\right)+1 =147+1 ] [ \boxed{E[g(X)]=148} ]

So, [ \boxed{E(X)=\frac72,\qquad \operatorname{Var}(X)=\frac{35}{12},\qquad E(2X^3+1)=148} ]

Poisson distribution

A discrete random variable $X$ is said to follow a Poisson distribution if it counts the number of occurrences of a rare event in a fixed interval of time, space or volume, and its probability mass function is

$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!},\qquad x=0,1,2,\dots$

where $\lambda>0$ is the average number of occurrences in the given interval. For a Poisson distribution,

$\text{Mean}=\text{Variance}=\lambda.$

It is the limiting form of the binomial distribution when the number of trials $n$ is very large and the probability of success $p$ is very small, with $\lambda=np$ finite.

Numerical

Let $X$ be the number of individuals who suffer an adverse reaction.

Given: $n=2000,\qquad p=0.001$

Since $n$ is large and $p$ is small, $X$ follows a Poisson distribution with

$\lambda=np=2000\times 0.001=2.$

We require $P(X=3)$:

$P(X=3)=\frac{e^{-2}2^3}{3!}$

$=\frac{e^{-2}\times 8}{6}$

$=\frac{4}{3}e^{-2}$

$=\frac{4}{3}\times 0.135335$

$=0.180447$

Hence, the probability that exactly 3 individuals will suffer an adverse reaction is approximately

$\boxed{0.1804}$

The given mass function must be the binomial PMF $P(x)={}^3C_x\left(\frac18\right)^x\left(\frac78\right)^{3-x},\qquad x=0,1,2,3,$ since otherwise the probabilities do not sum to 1.

The moment generating function is

$$M_X(t)=E(e^{tX}) =\sum_{x=0}^3 e^{tx}\,{}^3C_x\left(\frac18\right)^x\left(\frac78\right)^{3-x}$$ $$=\sum_{x=0}^3 {}^3C_x\left(\frac{e^t}{8}\right)^x\left(\frac78\right)^{3-x} =\left(\frac78+\frac{e^t}{8}\right)^3$$ $$\boxed{M_X(t)=\frac{(7+e^t)^3}{512}}$$

Now,

$$M_X'(t)=\frac{3e^t(7+e^t)^2}{512}$$

so

$$\mu_1'=M_X'(0)=\frac{3(8)^2}{512}=\frac{192}{512}=\boxed{\frac38}$$

Also,

$$M_X''(t)=\frac{3e^t(7+e^t)(7+3e^t)}{512}$$

Hence

$$\mu_2'=M_X''(0)=\frac{3(8)(10)}{512}=\frac{240}{512}=\boxed{\frac{15}{32}}$$

Therefore,

$$\boxed{\mu_1'=\frac38,\qquad \mu_2'=\frac{15}{32}}$$

For a probability density function,

$\int_{-\infty}^{\infty} f(x)\,dx = 1$

Here,

$f(x)=kx^2,\quad 0<x<1$

So,

$\int_0^1 kx^2\,dx=1$

$k\left[\frac{x^3}{3}\right]_0^1=1$

$k\cdot \frac{1}{3}=1$

$\boxed{k=3}$

Hence,

$f(x)=3x^2,\quad 0<x<1$

Now,

$P\left(\frac13<X<\frac12\right)=\int_{1/3}^{1/2}3x^2\,dx$

$= \left[x^3\right]_{1/3}^{1/2}$

$= \left(\frac12\right)^3-\left(\frac13\right)^3$

$= \frac18-\frac1{27}$

$= \frac{27-8}{216}$

$\boxed{P\left(\frac13<X<\frac12\right)=\frac{19}{216}}$

So,

$\boxed{k=3,\qquad P\left(\frac13<X<\frac12\right)=\frac{19}{216}}$

Uniform Distribution

A random variable $X$ is said to have a uniform (rectangular) distribution if it takes any value in a finite interval $[a, b]$ with constant probability density. It is written as $X \sim U(a, b)$.

Probability density function:

$$f(x) = \begin{cases} \dfrac{1}{b-a}, & a \le x \le b \\[6pt] 0, & \text{otherwise} \end{cases}$$

Properties:

• Total area under the curve: $\displaystyle\int_a^b \frac{1}{b-a}\,dx = 1$
• Mean: $E(X) = \dfrac{a+b}{2}$
• Variance: $V(X) = \dfrac{(b-a)^2}{12}$
• The distribution is symmetric about its mean.

Moment Generating Function (MGF)

The MGF of $X$ is defined as $M_X(t) = E(e^{tX})$.

$$M_X(t) = \int_a^b e^{tx} \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a}\int_a^b e^{tx}\,dx$$ $$= \frac{1}{b-a}\left[\frac{e^{tx}}{t}\right]_a^b = \frac{e^{tb} - e^{ta}}{t(b-a)}, \quad t \neq 0$$

For $t = 0$, expanding or taking the limit gives $M_X(0) = 1$.

$$\boxed{M_X(t) = \frac{e^{tb} - e^{ta}}{t(b-a)}, \quad t \neq 0}$$

Verification of mean and variance:

$$M_X'(t) = \frac{d}{dt}\left[\frac{e^{tb} - e^{ta}}{t(b-a)}\right]$$

Using the quotient rule and taking the limit as $t \to 0$:

$$E(X) = M_X'(0) = \frac{a+b}{2}$$

Similarly, from $M_X''(0)$:

$$E(X^2) = \frac{a^2 + ab + b^2}{3}$$ $$V(X) = E(X^2) - [E(X)]^2 = \frac{a^2+ab+b^2}{3} - \frac{(a+b)^2}{4} = \frac{(b-a)^2}{12}$$

Thus the MGF correctly reproduces the mean and variance of the uniform distribution.

i. Principles governing the choice of class intervals

(a) Number of class intervals:

• It should be neither too small nor too large; usually 5 to 20 classes are taken.
• It should be decided according to the total number of observations, often by Sturges’ rule:
  [ k = 1 + 3.322\log_{10}N ]
• It should bring out the main characteristics of the data clearly.

(b) Length of class intervals:

• Class intervals should preferably be of equal width.
• Class width is found by
  [ h=\frac{\text{Range}}{k} ]
• It should be a convenient and rounded value, and should cover the whole range of the data.

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ii. Examples of nominal and ratio scales

Nominal scale:

1. Gender: male, female
2. Blood group: A, B, AB, O

Ratio scale:

1. Weight in kg
2. Height in cm

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iii. Specificity of a diagnostic test

Specificity is the probability that a diagnostic test correctly identifies a person who does not have the disease as negative.
[ \text{Specificity}=\frac{TN}{TN+FP} ] where (TN) = true negative and (FP) = false positive.

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iv. Sample of 4 numbers with mean 5 and standard deviation zero

If standard deviation is zero, all observations must be equal.
So the sample is:

[ 5,\ 5,\ 5,\ 5 ]

Mean (=\frac{5+5+5+5}{4}=5), and standard deviation (=0).

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v. Binomial distribution

Given: [ np=1.6,\qquad npq=1.28 ] [ q=\frac{npq}{np}=\frac{1.28}{1.6}=0.8 ] [ p=1-q=0.2 ] [ n=\frac{np}{p}=\frac{1.6}{0.2}=8 ]

So (X\sim B(8,0.2)).

[ P(X\ge 1)=1-P(X=0) ] [ =1-\binom{8}{0}(0.2)^0(0.8)^8 ] [ =1-(0.8)^8 ] [ =1-0.16777216 ] [ =0.83222784 ] [ \boxed{P(X\ge 1)\approx 0.8322} ]

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vi. Standard normal probability

[ P(Z\le 1.52)=\Phi(1.52) ] From the standard normal table, [ \Phi(1.52)=0.9357 ] [ \boxed{P(Z\le 1.52)\approx 0.9357} ]

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vii. Correlation coefficient

Let [ b_{xy}=-0.84,\qquad b_{yx}=-0.20 ] Then [ r^2=b_{xy}\times b_{yx} ] [ r^2=(-0.84)(-0.20)=0.168 ] [ r=\pm \sqrt{0.168} ] Since both regression coefficients are negative, (r) is negative: [ r=-\sqrt{0.168}\approx -0.4099 ]

Interpretation: There is a moderate negative correlation between (x) and (y). As (x) increases, (y) tends to decrease, but the relationship is not very strong.

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viii. Moment generating function of Poisson distribution

If (X\sim \text{Poisson}(\lambda)), then the moment generating function is

[ M_X(t)=E(e^{tX})=\exp\left[\lambda(e^t-1)\right] ]

[ \boxed{M_X(t)=e^{\lambda(e^t-1)}} ]

Geometric Mean (GM) vs Harmonic Mean (HM)

Uses of Geometric Mean

1. Used to average ratios, proportions and percentages (e.g. growth rates, price relatives).
2. Used in index numbers, since index numbers are based on ratios.
3. Used when data changes geometrically, such as population growth or compound interest.
4. Gives equal weight to equal relative changes, so it is not distorted by very large values.

Uses of Harmonic Mean

1. Used to average rates and speeds, e.g. average speed when distances are equal.
2. Used to average time rates, prices, and ratios where the numerator is constant.
3. Used in business and economics for averaging price–quantity ratios.
4. Gives greater weight to smaller values, so it is useful when small items matter more.

Key point: AM is used for general averages, GM for ratios and growth, and HM for rates and speeds; among the three, $AM \ge GM \ge HM$ for positive data.

Dispersion is the extent to which the observations of a series scatter about a central value. Averages alone do not describe a distribution: the sets $A=\{48,49,50,51,52\}$ and $B=\{10,30,50,70,90\}$ both have mean 50, yet $A$ is tightly clustered and $B$ is widely spread. Dispersion measures this difference.

Measures of dispersion

Examples. For $A$: deviations $-2,-1,0,1,2$; M.D. $=6/5=1.2$; $\sigma=\sqrt{10/5}=\sqrt2\approx1.41$. For $B$: deviations $-40,-20,0,20,40$; $\sigma=\sqrt{4000/5}=\sqrt{800}\approx28.28$; C.V. $=\dfrac{28.28}{50}\times100\approx56.56\%$. Same mean, much larger dispersion in $B$.

Most important measure: the standard deviation. It uses every observation, is rigidly defined, is capable of further algebraic treatment (combined s.d. of groups, correlation, regression, sampling theory), and is least affected by sampling fluctuations. Together with C.V. it is the standard tool for comparing consistency of series.

Raw moments (about the origin). For values $x_i$ with frequencies $f_i$ and $N=\sum f_i$, the $r$-th raw moment is $\mu_r'=\frac1N\sum f_i x_i^{\,r},\qquad \mu_0'=1,\ \ \mu_1'=\bar x .$

Central moments (about the mean). $\mu_r=\frac1N\sum f_i(x_i-\bar x)^r,\qquad \mu_0=1,\ \ \mu_1=0,\ \ \mu_2=\sigma^2 .$

Relation. Expanding by the binomial theorem and using $\bar x=\mu_1'$: $\mu_r=\frac1N\sum f_i\sum_{k=0}^{r}\binom rk x_i^{\,k}(-\bar x)^{r-k}=\sum_{k=0}^{r}\binom rk(-\mu_1')^{r-k}\mu_k' .$

Second: $\mu_2=\mu_2'-2\mu_1'\mu_1'+(\mu_1')^2=\mu_2'-(\mu_1')^2 .$

Third: $\mu_3=\mu_3'-3\mu_1'\mu_2'+3(\mu_1')^2\mu_1'-(\mu_1')^3=\mu_3'-3\mu_1'\mu_2'+2(\mu_1')^3 .$

Fourth: $\mu_4=\mu_4'-4\mu_1'\mu_3'+6(\mu_1')^2\mu_2'-4(\mu_1')^3\mu_1'+(\mu_1')^4=\mu_4'-4\mu_1'\mu_3'+6(\mu_1')^2\mu_2'-3(\mu_1')^4 .$

These are used to find skewness $\beta_1=\dfrac{\mu_3^2}{\mu_2^3}$ and kurtosis $\beta_2=\dfrac{\mu_4}{\mu_2^2}$.

Definitions

Positive correlation: If two variables change in the same direction, i.e. when one variable increases, the other also increases (and vice versa), the correlation is called positive correlation. Its coefficient is (r>0).

Negative correlation: If two variables change in opposite directions, i.e. when one variable increases, the other decreases (and vice versa), the correlation is called negative correlation. Its coefficient is (r<0).

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Computation of Karl Pearson’s coefficient of correlation

Let (x) and (y) be the given variables, and (n=8).

So, [ \Sigma x=92,\quad \Sigma y=93,\quad \Sigma x^2=1200,\quad \Sigma y^2=1163,\quad \Sigma xy=1157 ]

Karl Pearson’s coefficient of correlation is [ r=\frac{n\Sigma xy-\Sigma x\Sigma y} {\sqrt{\left[n\Sigma x^2-(\Sigma x)^2\right]\left[n\Sigma y^2-(\Sigma y)^2\right]}} ]

Substituting the values: [ r=\frac{8(1157)-(92)(93)} {\sqrt{\left[8(1200)-(92)^2\right]\left[8(1163)-(93)^2\right]}} ]

[ r=\frac{9256-8556} {\sqrt{(9600-8464)(9304-8649)}} ]

[ r=\frac{700}{\sqrt{1136\times 655}} ]

[ r=\frac{700}{\sqrt{744080}} ]

[ r=\frac{700}{862.60}=0.8115 ]

[ \boxed{r \approx 0.812} ]

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Interpretation

The coefficient of correlation is (r \approx 0.812), which is positive and close to (+1). Hence, there is a high degree of positive correlation between (x) and (y). This means that as (x) increases, (y) also tends to increase.

Let the common mean of (x) and (y) be (\bar x=\bar y=\bar X).

Since both regression lines pass through the point of means ((\bar x,\bar y)), we have

[ \bar y=\alpha \bar x+b ]

and

[ \bar x=a\bar y+\beta . ]

Because (\bar x=\bar y=\bar X),

[ \bar X=\alpha \bar X+b ]

[ \Rightarrow b=\bar X(1-\alpha) ]

and

[ \bar X=a\bar X+\beta ]

[ \Rightarrow \beta=\bar X(1-a). ]

Dividing the two results,

[ \frac{b}{\beta}

\frac{\bar X(1-\alpha)}{\bar X(1-a)}

\frac{1-\alpha}{1-a}, ]

which proves the required relation.

Also, from the above,

[ \bar X=\frac{b}{1-\alpha}=\frac{\beta}{1-a}. ]

Hence the common mean is

[ \boxed{\bar x=\bar y=\frac{b}{1-\alpha}=\frac{\beta}{1-a}}. ]

Equivalently, solving the two regression equations simultaneously gives the same intersection point:

[ \boxed{\bar x=\bar y=\frac{ab+\beta}{1-a\alpha} =\frac{b+\alpha\beta}{1-a\alpha}}. ]

So the common mean is the point of intersection of the two regression lines.

Let (x) and (y) have standard deviations (\sigma_x,\sigma_y) and correlation coefficient (r). The two regression coefficients are

[ b_{yx}=r\frac{\sigma_y}{\sigma_x},\qquad b_{xy}=r\frac{\sigma_x}{\sigma_y}. ]

Their arithmetic mean is

[ A=\frac{b_{yx}+b_{xy}}{2} =\frac{r}{2}\left(\frac{\sigma_y}{\sigma_x}+\frac{\sigma_x}{\sigma_y}\right). ]

Set

[ u=\frac{\sigma_y}{\sigma_x}>0. ]

Then

[ A=\frac{r}{2}\left(u+\frac1u\right). ]

By AM–GM inequality,

[ u+\frac1u\ge 2. ]

Hence

[ |A|=\frac{|r|}{2}\left(u+\frac1u\right)\ge |r|. ]

Therefore,

[ \boxed{\left|\frac{b_{yx}+b_{xy}}{2}\right|\ge |r|}. ]

Equality holds iff (u=1) or (r=0), i.e. (\sigma_x=\sigma_y) or (r=0). If (r>0), then (A\ge r), with equality iff (\sigma_x=\sigma_y). If (r<0), then (A\le r), with equality iff (\sigma_x=\sigma_y). Thus the statement is true in absolute value; algebraically it is not true for negative (r).

Given: (A) and (B) are independent events, so
[ P(A\cap B)=P(A)P(B). ]

To prove: (\bar A) and (\bar B) are independent, i.e. [ P(\bar A\cap \bar B)=P(\bar A)P(\bar B). ]

Proof:

By De Morgan’s law, [ \bar A\cap \bar B=\overline{A\cup B}. ] Therefore, [ P(\bar A\cap \bar B)=P(\overline{A\cup B}) =1-P(A\cup B). ]

Using the addition theorem of probability, [ P(A\cup B)=P(A)+P(B)-P(A\cap B). ] Since (A) and (B) are independent, [ P(A\cap B)=P(A)P(B). ] Hence, [ P(A\cup B)=P(A)+P(B)-P(A)P(B). ]

So, [ P(\bar A\cap \bar B) =1-\bigl[P(A)+P(B)-P(A)P(B)\bigr] ] [ =1-P(A)-P(B)+P(A)P(B) ] [ =\bigl[1-P(A)\bigr]\bigl[1-P(B)\bigr] ] [ =P(\bar A)P(\bar B). ]

Thus, [ P(\bar A\cap \bar B)=P(\bar A)P(\bar B). ]

Therefore, by the definition of independent events, (\bar A) and (\bar B) are independent. (\square)

Bayes' theorem. If $B_1,\dots,B_n$ are mutually exclusive and exhaustive events with $P(B_i)>0$, and $A$ is any event with $P(A)>0$, then $P(B_i\mid A)=\frac{P(B_i)\,P(A\mid B_i)}{\sum_{j=1}^{n}P(B_j)\,P(A\mid B_j)} .$

Numerical. The question text is garbled; we read it as: Box 1 holds 10 Navy Polo shirts and 10 super T-shirts (20 in all), and Box 2 holds 25 mesh Polo shirts and 5 super T-shirts (30 in all).

Let $B_1,B_2$ be the events that box 1, box 2 is chosen, and $S$ the event that the shirt is a super T-shirt.

$P(B_1)=P(B_2)=\tfrac12,\quad P(S\mid B_1)=\tfrac{10}{20}=\tfrac12,\quad P(S\mid B_2)=\tfrac{5}{30}=\tfrac16 .$

$P(B_1\mid S)=\frac{\tfrac12\cdot\tfrac12}{\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac16}=\frac{1/4}{1/3}=\frac34 .$

The probability that the super T-shirt came from box 1 is $\dfrac34$.

Question. Obtain the moment generating function for the distribution

$dp = y_0\, e^{-x/\sigma}\,dx,\qquad \sigma>0,\ 0\le x<\infty,$

$y_0$ being a constant. [5 marks]

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Step 1: Find the constant $y_0$.

Since the total probability is 1,

$\int_0^{\infty} y_0 e^{-x/\sigma}\,dx = 1.$

$y_0 \left[-\sigma e^{-x/\sigma}\right]_0^{\infty} = y_0\,\sigma = 1 \quad\Rightarrow\quad y_0 = \frac{1}{\sigma}.$

So the p.d.f. is

$f(x) = \frac{1}{\sigma} e^{-x/\sigma},\qquad 0\le x<\infty.$

This is the exponential distribution with mean $\sigma$.

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Step 2: Definition of the m.g.f.

$M_X(t) = E\left(e^{tX}\right) = \int_0^{\infty} e^{tx}\, f(x)\,dx = \int_0^{\infty} e^{tx}\cdot \frac{1}{\sigma} e^{-x/\sigma}\,dx.$

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Step 3: Evaluate the integral.

$M_X(t) = \frac{1}{\sigma}\int_0^{\infty} e^{-x\left(\frac{1}{\sigma}-t\right)}dx.$

Put $a = \dfrac{1}{\sigma}-t$. Then

$M_X(t) = \frac{1}{\sigma}\int_0^{\infty} e^{-ax}\,dx = \frac{1}{\sigma}\cdot\frac{1}{a} = \frac{1}{\sigma\left(\frac{1}{\sigma}-t\right)} = \frac{1}{1-\sigma t},$

provided $a>0$, i.e. $t<\dfrac{1}{\sigma}$.

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Result.

$\boxed{M_X(t) = \frac{1}{1-\sigma t} = \left(1-\sigma t\right)^{-1},\qquad t<\frac{1}{\sigma}}$

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Step 4: Verification (mean and variance).

Expanding, $M_X(t) = 1 + \sigma t + \sigma^2 t^2 + \sigma^3 t^3 + \cdots$

$M'(t) = \sigma(1-\sigma t)^{-2} \Rightarrow M'(0) = \sigma = \mu_1' = E(X).$

$M''(t) = 2\sigma^2(1-\sigma t)^{-3} \Rightarrow M''(0) = 2\sigma^2 = \mu_2'.$

$\text{Variance} = \mu_2' - (\mu_1')^2 = 2\sigma^2 - \sigma^2 = \sigma^2.$

Hence the m.g.f. gives mean $\sigma$ and variance $\sigma^2$, as expected for this distribution.

(i) Probability density function (p.d.f.)

Let $X$ be a continuous random variable. A function $f(x)$ is called the probability density function of $X$ if it satisfies the following conditions:

1. $f(x) \geq 0$ for all $x \in (-\infty, \infty)$
2. $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$
3. For any two real numbers $a$ and $b$ with $a < b$, $P(a < X < b) = \int_a^b f(x)\,dx$

Thus $f(x)$ itself is not a probability; it is the density of probability at the point $x$. The probability is obtained only by integrating $f(x)$ over an interval. Also, for a continuous random variable, $P(X = c) = 0$ for any single value $c$.

(ii) Distribution function

The distribution function (also called the cumulative distribution function, c.d.f.) of a random variable $X$, denoted by $F(x)$, is defined as $F(x) = P(X \leq x), \quad -\infty < x < \infty$

For a discrete random variable it is obtained by summing the probabilities: $F(x) = \sum_{x_i \leq x} P(X = x_i)$

For a continuous random variable it is obtained by integrating the p.d.f.: $F(x) = \int_{-\infty}^{x} f(t)\,dt$

Properties of $F(x)$:

1. $0 \leq F(x) \leq 1$
2. $F(x)$ is a non-decreasing function of $x$
3. $F(-\infty) = 0$ and $F(\infty) = 1$
4. $P(a < X \leq b) = F(b) - F(a)$

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Distribution function of X

Given: $X$ takes the values $0, 1, 2$ with probabilities $P(X = 0) = \frac{1}{3}, \qquad P(X = 1) = \frac{1}{6}, \qquad P(X = 2) = \frac{1}{2}$

Check: $\dfrac{1}{3} + \dfrac{1}{6} + \dfrac{1}{2} = \dfrac{2+1+3}{6} = 1$. ✓

Now $F(x) = P(X \leq x)$.

• For $x < 0$: no value of $X$ is less than or equal to $x$, so $F(x) = 0$.
• For $0 \leq x < 1$: only $X = 0$ is included, so $F(x) = P(X = 0) = \dfrac{1}{3}$.
• For $1 \leq x < 2$: $X = 0$ and $X = 1$ are included, so $F(x) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$
• For $x \geq 2$: all three values are included, so $F(x) = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = 1$

Hence the distribution function is

$$F(x) = \begin{cases} 0, & x < 0 \\[4pt] \dfrac{1}{3}, & 0 \leq x < 1 \\[4pt] \dfrac{1}{2}, & 1 \leq x < 2 \\[4pt] 1, & x \geq 2 \end{cases}$$

The probability distribution of $X$ may also be shown in tabular form:

Binomial distribution. A random variable $X$ is said to have the binomial distribution with parameters $n$ and $p$ if it counts the number of successes in $n$ independent Bernoulli trials, each having the same probability of success $p$. Its probability mass function is

$P(X=x) = \binom{n}{x} p^x q^{\,n-x}, \qquad x = 0, 1, 2, \dots, n,$

where $q = 1-p$ and $\binom{n}{x} = \dfrac{n!}{x!\,(n-x)!}$.

The four conditions required are:

1. The number of trials $n$ is fixed in advance.
2. Each trial has only two possible outcomes — success or failure.
3. The trials are independent of one another.
4. The probability of success $p$ remains constant from trial to trial.

Real-life situation. A student appears in a 20-question multiple-choice test in which every question has 4 options and exactly one is correct. If the student guesses every answer, then each question is a trial with only two outcomes (correct or wrong), the probability of a correct answer is $p = \tfrac14$ on every question, the questions are answered independently, and the number of questions $n = 20$ is fixed. Hence the number of correct answers $X$ follows the binomial distribution with $n = 20$ and $p = \tfrac14$; the probability of getting exactly $x$ correct answers is

$P(X=x) = \binom{20}{x}\left(\frac14\right)^{x}\left(\frac34\right)^{20-x}, \qquad x = 0,1,\dots,20.$

(b) weight of a person.

Weight is a ratio scale measurement. It has a true (absolute) zero point — a weight of 0 means no weight at all — and the differences between values are equal and meaningful. This allows all arithmetic operations (addition, subtraction, multiplication, division) and comparison of ratios (e.g., 80 kg is twice 40 kg).

In contrast, rank of posting is an ordinal scale measurement. It only shows order or position (1st, 2nd, 3rd) without equal intervals between ranks or a true zero, so it cannot be used for arithmetic operations.

If the standard deviation is zero, all five numbers must be equal. For the mean to be 8, each number must be 8.

Sample: 8, 8, 8, 8, 8

Verification:

$\bar{x} = \frac{8+8+8+8+8}{5} = \frac{40}{5} = 8$

$\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sqrt{\frac{0+0+0+0+0}{5}} = 0$

Hence the mean is 8 and the S.D. is zero.

Kurtosis is a statistical measure that describes the peakedness or flatness of a frequency distribution curve relative to the normal curve. It indicates how much of the data is concentrated in the tails and around the centre of the distribution.

Based on kurtosis, distributions are classified as:

• Mesokurtic (β₂ = 3): Normal peakedness, like the normal curve.
• Leptokurtic (β₂ > 3): More peaked than normal, with heavier tails.
• Platykurtic (β₂ < 3): Flatter than normal, with lighter tails.

It is commonly measured by the coefficient β₂ = μ₄ / μ₂² (or by γ₂ = β₂ − 3).

(iv) Zero correlation does not imply independence — example:

Let $X$ be a random variable taking the values $-1,\ 0,\ 1$ each with probability $\tfrac13$, and define $Y = X^2$.

Correlation: $E(X) = 0$, so $\operatorname{Cov}(X,Y) = E(XY) - E(X)E(Y) = E(X^3) - 0 = \tfrac13(-1+0+1) = 0$. Hence $r = 0$ — the variables are uncorrelated.

But they are not independent: $Y$ is completely determined by $X$. For instance, $P(X = 0,\ Y = 0) = \tfrac13 \quad\text{but}\quad P(X=0)\cdot P(Y=0) = \tfrac13 \cdot \tfrac13 = \tfrac19 .$ Since $P(X=0,\ Y=0) \neq P(X=0)\,P(Y=0)$, the variables are dependent.

Conclusion: Zero correlation means only the absence of a linear relationship; a strong non-linear (here quadratic) relationship may still exist. So zero correlation does not necessarily imply independence.