Binomial distribution

Consider $n$ independent trials, each with only two outcomes, success (probability $p$) and failure (probability $q=1-p$), $p$ being the same in every trial. The number of successes $X$ has the binomial distribution $X\sim B(n,p)$:

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

Its important properties are:

• Parameters: $n$ (a positive integer) and $p$ ($0<p<1$).
• Mean $=np$, variance $=npq$ (so the variance is always less than the mean).
• $\mu_2=npq$, $\mu_3=npq(q-p)$, and the distribution is symmetric if $p=q=\tfrac12$.
• It has mode(s) near $(n+1)p$.
• It tends to the Poisson distribution when $n\to\infty$, $p\to0$ with $np$ fixed, and to the normal distribution when $n$ is large.
• Example: the number of defective items in a sample of 10 from a lot with 5% defectives.

Recurrence relation for central moments

Let $\mu_r=\sum_{x=0}^{n}(x-np)^r\binom{n}{x}p^xq^{n-x}$. Differentiate with respect to $p$ (with $q=1-p$, so $dq/dp=-1$):

$\frac{d\mu_r}{dp}=\sum_x\binom{n}{x}\left[(x-np)^r\frac{d}{dp}\left(p^xq^{n-x}\right)-nr(x-np)^{r-1}p^xq^{n-x}\right].$

Now

$\frac{d}{dp}\left(p^xq^{n-x}\right)=xp^{x-1}q^{n-x}-(n-x)p^xq^{n-x-1}=p^{x-1}q^{n-x-1}\left[xq-(n-x)p\right]=p^{x-1}q^{n-x-1}(x-np),$

since $xq-(n-x)p=x-np$. Hence

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

Multiplying by $pq$ and rearranging:

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

Check: $\mu_0=1,\ \mu_1=0$. Put $r=1$: $\mu_2=pq(n\cdot1\cdot\mu_0+0)=npq$. Put $r=2$: $\mu_3=pq\left(2n\mu_1+\dfrac{d(npq)}{dp}\right)=pq\cdot n(q-p)=npq(q-p)$.

Normal Probability Distribution

Definition. A continuous random variable $X$ is said to follow a normal probability distribution with parameters $\mu$ (mean) and $\sigma^2$ (variance) if its probability density function is given by

$f(x)=\frac{1}{\sigma\sqrt{2\pi}}\,e^{-\frac{(x-\mu)^2}{2\sigma^2}},\qquad -\infty<x<\infty,\ -\infty<\mu<\infty,\ \sigma>0$

It is denoted by $X\sim N(\mu,\sigma^2)$. The curve is bell-shaped, symmetric about $x=\mu$, and the total area under it is 1. The standard normal variate is $z=\dfrac{x-\mu}{\sigma}\sim N(0,1)$.

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Mean of the Normal Distribution

The mean is the expected value of $X$:

$E(X)=\int_{-\infty}^{\infty}x\,f(x)\,dx=\int_{-\infty}^{\infty}x\cdot\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$

Substitution. Put $z=\dfrac{x-\mu}{\sigma}$, so that $x=\mu+\sigma z$ and $dx=\sigma\,dz$. When $x\to\pm\infty$, $z\to\pm\infty$. Hence

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

$=\frac{\mu}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-z^2/2}dz+\frac{\sigma}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z\,e^{-z^2/2}dz$

First integral. This is the total area of the standard normal curve:

$\int_{-\infty}^{\infty}e^{-z^2/2}dz=\sqrt{2\pi}\quad\Rightarrow\quad \frac{\mu}{\sqrt{2\pi}}\cdot\sqrt{2\pi}=\mu$

Second integral. The integrand $z\,e^{-z^2/2}$ is an odd function of $z$, so its integral over a symmetric interval $(-\infty,\infty)$ is zero:

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

Therefore,

$E(X)=\mu+0=\mu$

$\boxed{\text{Mean of the normal distribution } = E(X) = \mu}$

Thus the parameter $\mu$ appearing in the p.d.f. is itself the mean of the distribution. (Similarly, by the same substitution, the variance is $E(X-\mu)^2=\sigma^2$, so $\sigma$ is the standard deviation.)

Difference Between Primary Data and Secondary Data

Sources of Secondary Data

1. Published sources

• Government publications: Central Bureau of Statistics (CBS) reports, national census, Nepal Living Standards Survey, Economic Survey (Ministry of Finance), Nepal Rastra Bank reports, statistical yearbooks.
• International publications: UN Statistical Yearbook, World Bank and IMF reports, WHO, ILO, FAO publications.
• Semi-government and private publications: Reports of Nepal Rastra Bank, Federation of Nepalese Chambers of Commerce and Industry, commercial banks, insurance companies, research institutes.
• Books, journals and newspapers: Textbooks, research journals, magazines, daily and weekly newspapers.
• Electronic sources: Websites of government and international agencies, online databases, e-journals.

2. Unpublished sources

• Records and registers maintained by government offices, municipalities, schools, hospitals, banks and business firms.
• Research reports, theses and dissertations of students and scholars.
• Personal records, diaries, letters and files of individuals and organisations.

Note: Secondary data should be used only after checking its reliability, relevance, adequacy and accuracy (unit of measurement, time period and method of original collection).

Show that the sum of squares of the deviations of the values of the variable (x) is least when the deviations are measured from the mean.

Let the variable (x) take the values [ x_1, x_2, x_3, \dots, x_n ] with arithmetic mean (\bar{x}), so that [ \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 the deviations from (A) as [ S(A) = \sum_{i=1}^{n} (x_i - A)^2. ]

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

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Step 1: Express (S(A)) in terms of (\bar{x})

Write [ x_i - A = (x_i - \bar{x}) + (\bar{x} - A). ]

Squaring both sides and summing over (i), [ S(A) = \sum_{i=1}^{n} \left[(x_i - \bar{x}) + (\bar{x} - A)\right]^2. ]

Expanding, [ S(A) = \sum_{i=1}^{n} (x_i - \bar{x})^2 + 2(\bar{x} - A)\sum_{i=1}^{n}(x_i - \bar{x}) + n(\bar{x} - A)^2. ]

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Step 2: Use the property of the mean

A basic property of the arithmetic mean is [ \sum_{i=1}^{n}(x_i - \bar{x}) = \sum_{i=1}^{n} x_i - n\bar{x} = n\bar{x} - n\bar{x} = 0. ]

So the middle term vanishes, and [ S(A) = \sum_{i=1}^{n} (x_i - \bar{x})^2 + n(\bar{x} - A)^2. ]

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Step 3: Minimise (S(A))

In the above expression, the first term (\sum (x_i - \bar{x})^2) is a constant (it does not depend on (A)), and the second term (n(\bar{x} - A)^2) is a perfect square, so it is always non-negative: [ n(\bar{x} - A)^2 \ge 0. ]

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}. ]

Therefore the least value of (S(A)) is [ S_{\min} = \sum_{i=1}^{n} (x_i - \bar{x})^2, ] and this least value is attained only when the deviations are measured from the mean, i.e. when (A = \bar{x}).

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Alternative proof (by calculus)

Treat (A) as a continuous variable and differentiate (S(A)) with respect to (A): [ \frac{dS}{dA} = -2\sum_{i=1}^{n}(x_i - A) = -2\left(\sum x_i - nA\right). ]

For a stationary point, [ \frac{dS}{dA} = 0 \quad \Rightarrow \quad \sum x_i - nA = 0 \quad \Rightarrow \quad A = \frac{1}{n}\sum x_i = \bar{x}. ]

The second derivative is [ \frac{d^2S}{dA^2} = 2n > 0, ] which is positive for all (A). Hence (S(A)) has a minimum at (A = \bar{x}).

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Conclusion: The sum of squares of the deviations (\sum (x_i - A)^2) is least when the deviations are taken from the arithmetic mean (\bar{x}). This least value is the variance times (n), i.e. (\sum (x_i - \bar{x})^2).

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

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

1. Mean

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

[ \boxed{\bar x = 3} ]

2. Variance

[ \sigma^2 = m_2 - m_1^2 ]

[ \sigma^2 = 16 - (1)^2 = 15 ]

[ \boxed{\sigma^2 = 15} ]

Standard deviation:

[ \sigma = \sqrt{15} = 3.873 ]

3. 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 ]

4. Skewness

Coefficient of skewness:

[ \gamma_1 = \frac{\mu_3}{\sigma^3} ]

[ \gamma_1 = \frac{-86}{(\sqrt{15})^3} ]

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

[ \gamma_1 = -1.480 ]

[ \boxed{\gamma_1 = -1.48} ]

Since (\gamma_1) is negative, the distribution is negatively skewed.

Final Answer

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

The distribution is negatively skewed.

Let the regression line of (Y) on (X) be

[ a_1X+b_1Y=c_1. ]

Solving for (Y),

[ Y=-\frac{a_1}{b_1}X+\frac{c_1}{b_1}. ]

So the regression coefficient of (Y) on (X) is

[ b_{yx}=-\frac{a_1}{b_1}. ]

Similarly, let the regression line of (X) on (Y) be

[ a_2X+b_2Y=c_2. ]

Solving for (X),

[ X=-\frac{b_2}{a_2}Y+\frac{c_2}{a_2}. ]

So the regression coefficient of (X) on (Y) is

[ b_{xy}=-\frac{b_2}{a_2}. ]

Let (r) be the correlation coefficient between (X) and (Y). A standard result is

[ b_{yx},b_{xy}=r^2. ]

Substituting the values of (b_{yx}) and (b_{xy}),

[ \left(-\frac{a_1}{b_1}\right)\left(-\frac{b_2}{a_2}\right)=r^2 ]

[ \frac{a_1b_2}{a_2b_1}=r^2. ]

Since the square of the correlation coefficient cannot exceed (1),

[ r^2\le 1. ]

Hence,

[ \frac{a_1b_2}{a_2b_1}\le 1. ]

Under the usual convention that (a_2b_1>0), multiplying both sides by (a_2b_1) gives

[ a_1b_2\le a_2b_1. ]

Equality holds when (r^2=1), i.e. when the two regression lines coincide, which happens in the case of perfect linear correlation.

Theorem

If two random variables (X) and (Y) are independent, then they are uncorrelated, i.e. [ \operatorname{Cov}(X,Y)=0 ] and hence the correlation coefficient [ \rho_{XY}=\frac{\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}=0. ]

Proof

By definition, [ \operatorname{Cov}(X,Y)=E\left[(X-E(X))(Y-E(Y))\right]. ] Expanding, [ \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. ] Thus, [ \rho_{XY}=\frac{0}{\sigma_X\sigma_Y}=0, ] provided (\sigma_X>0) and (\sigma_Y>0). Hence independent variables are uncorrelated.

Converse is not necessarily true

The converse would be: “If (X) and (Y) are uncorrelated, then they are independent.” This is false. Uncorrelated variables may still be dependent.

Example

Let (X) take the values (-1,0,1) each with probability (\frac13). Define [ Y=X^2. ] Then (Y) is completely determined by (X), so (X) and (Y) are dependent.

Now, [ E(X)=(-1)\frac13+0\cdot\frac13+1\cdot\frac13=0. ] Also, [ E(Y)=E(X^2)=(-1)^2\frac13+0^2\frac13+1^2\frac13=\frac23. ] And, [ E(XY)=E(X^3)=(-1)^3\frac13+0^3\frac13+1^3\frac13=0. ] Therefore, [ \operatorname{Cov}(X,Y)=E(XY)-E(X)E(Y) =0-0\cdot\frac23=0. ] So (X) and (Y) are uncorrelated. But they are not independent, since, for example, [ P(X=1,Y=1)=\frac13, ] whereas [ P(X=1)P(Y=1)=\frac13\cdot\frac23=\frac29. ] Since these are not equal, (X) and (Y) are not independent.

Hence, independence implies uncorrelatedness, but uncorrelatedness does not necessarily imply independence.

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 of choosing road A, road B, road C, and let $E$ denote the event of escaping.

Given: $P(A) = P(B) = P(C) = \frac{1}{3}$ $P(E \mid A) = \frac{1}{5}, \quad P(E \mid B) = \frac{1}{8}, \quad P(E \mid C) = \frac{1}{10}$

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

Result: The probability that the prisoner chose road A, given that he escaped, is $\boxed{\dfrac{8}{17}}$.

Random variable: A random variable is a real-valued function defined on the sample space of a random experiment. It assigns a numerical value to each possible outcome of the experiment. It may be discrete or continuous.

Let (X) be the number of points obtained when a balanced die is rolled. Then

[ X=1,2,3,4,5,6 ]

and each value has probability (\frac16).

Expected value of (X)

[ E(X)=\sum xP(X=x) ]

[ E(X)=\frac{1+2+3+4+5+6}{6} ]

[ E(X)=\frac{21}{6}=\frac{7}{2}=3.5 ]

Variance of (X)

[ \operatorname{Var}(X)=E(X^2)-[E(X)]^2 ]

Now,

[ E(X^2)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6} ]

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

So,

[ \operatorname{Var}(X)=\frac{91}{6}-\left(\frac{7}{2}\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} ]

[ E(X^3)=\frac{1+8+27+64+125+216}{6} ]

[ E(X^3)=\frac{441}{6}=\frac{147}{2} ]

Therefore,

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

[ =147+1=148 ]

[ \boxed{E[g(X)]=148} ]

Hence,

[ \boxed{E(X)=\frac{7}{2},\quad \operatorname{Var}(X)=\frac{35}{12},\quad E(2X^3+1)=148} ]

Poisson distribution

A Poisson distribution is a discrete probability distribution that gives the probability of a given number of events occurring in a fixed interval of time or space, when these events occur independently and with a known constant average rate. If $X$ is a Poisson random variable with parameter $\lambda>0$, then

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

where $\lambda$ is the mean number of occurrences in the given interval. For a binomial distribution with large $n$ and very small $p$, the Poisson distribution is used as an approximation with $\lambda=np$.

Numerical

Given:

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

So,

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

We need $P(X=3)$.

$$P(X=3)=\frac{e^{-2}2^3}{3!}$$ $$=\frac{e^{-2}\times 8}{6} =\frac{4}{3}e^{-2}$$

Using $e^{-2}=0.135335$,

$$P(X=3)=\frac{4}{3}\times 0.135335 =0.18045$$

Answer: The probability that exactly 3 individuals out of 2000 will suffer an adverse reaction is approximately

$$\boxed{0.1804}$$

The given pmf is the binomial pmf with
$n=3,\quad p=\frac18,\quad q=\frac78.$ So the intended pmf is $P(x)={}^3C_x\left(\frac18\right)^x\left(\frac78\right)^{3-x},\qquad x=0,1,2,3.$

Moment generating function

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

First raw moment

$$M_X'(t)=3\left(\frac78+\frac{e^t}{8}\right)^2\cdot \frac{e^t}{8}.$$

At (t=0),

$$\mu_1'=M_X'(0) =3(1)^2\cdot \frac18 =\frac38.$$

Second raw moment

$$M_X''(t)=3\left[2\left(\frac78+\frac{e^t}{8}\right)\left(\frac{e^t}{8}\right)^2 +\left(\frac78+\frac{e^t}{8}\right)^2\frac{e^t}{8}\right].$$

At (t=0),

$$\mu_2'=M_X''(0) =3\left[2(1)\left(\frac18\right)^2+(1)^2\left(\frac18\right)\right] =3\left(\frac{2}{64}+\frac18\right) =3\left(\frac{1}{32}+\frac{4}{32}\right) =\frac{15}{32}.$$

Hence,

$$\boxed{M_X(t)=\left(\frac78+\frac{e^t}{8}\right)^3,\qquad \mu_1'=\frac38,\qquad \mu_2'=\frac{15}{32}.}$$

Given probability density function: $f(x)=kx^2,\quad 0<x<1$ and (f(x)=0) elsewhere.

1. Determine (k)

For a valid probability density function, $\int_{-\infty}^{\infty} f(x)\,dx=1$

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

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

$k\left(\frac{1}{3}-0\right)=1$

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

$\boxed{k=3}$

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

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2. Find (P\left(\frac13<X<\frac12\right))

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

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

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

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

$=\frac18-\frac1{27}$

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

$=\frac{19}{216}$

$\boxed{P\left(\frac13<X<\frac12\right)=\frac{19}{216}\approx 0.088}$

Uniform Distribution

A random variable $X$ is said to have a uniform distribution (also called rectangular distribution) if it takes any value in a finite interval $[a, b]$ with constant probability density throughout the interval. It is the simplest continuous distribution, used when all outcomes in an interval are equally likely.

Probability Density Function (p.d.f.)

If $X \sim U(a, b)$, then

$$f(x) = \frac{1}{b-a}, \quad a \le x \le b$$

and $f(x) = 0$ elsewhere.

Verification: $\displaystyle \int_a^b \frac{1}{b-a}\,dx = \frac{b-a}{b-a} = 1$, so it is a valid p.d.f.

Moment Generating Function (M.G.F.)

The m.g.f. of $X$ is defined as

$$M_X(t) = E(e^{tX}) = \int_a^b e^{tx} f(x)\,dx$$

Substituting $f(x) = \dfrac{1}{b-a}$:

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

For $t = 0$: Since $M_X(0) = E(e^{0}) = 1$, we define

$$M_X(0) = 1$$

Alternative form (multiplying numerator and denominator by $e^{-ta}$):

$$M_X(t) = \frac{e^{ta}\left(e^{t(b-a)} - 1\right)}{t(b-a)}$$

Note: The m.g.f. exists for all real $t$ because the interval $[a, b]$ is finite. From this m.g.f. we can obtain the mean and variance by differentiating at $t = 0$:

$$E(X) = M_X'(0) = \frac{a+b}{2}, \qquad \text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = \frac{(b-a)^2}{12}$$

i. Principles governing the choice of class intervals

(a) Number of class intervals

• It should be neither too small nor too large; usually between 5 and 20.
• Sturges’ formula may be used:
  [ k = 1 + 3.322\log_{10}N ] where (N) is the total number of observations.
• It should reveal the shape and pattern of the distribution clearly.
• Empty or very thin classes should be avoided.

(b) Length of class intervals

• Class width is found by
  [ h = \frac{L-S}{k} ] where (L) = largest value, (S) = smallest value, (k) = number of classes.
• It should be a convenient round number.
• The same width should be used for all classes.
• Classes should be mutually exclusive and exhaustive, with no gaps or overlaps.

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ii. Examples of nominal and ratio scales

Nominal scale

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

Ratio scale

1. Weight in kg
2. Income in Rs

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iii. Specificity of a diagnostic test

Specificity is the ability of a diagnostic test to correctly identify persons who do not have the disease. It is the proportion of true negatives among all non-diseased persons:

[ \text{Specificity} = \frac{TN}{TN+FP} ]

where (TN) = true negatives and (FP) = false positives.

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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{20}{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 - {8 \choose 0}(0.2)^0(0.8)^8 ]

[ = 1 - (0.8)^8 ]

[ = 1 - 0.16777216 ]

[ = 0.83222784 \approx 0.8322 ]

[ \boxed{P(X \ge 1) \approx 0.8322} ]

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vi. Standard normal variate

From the standard normal table:

[ P(Z \le 1.52) = 0.9357 ]

[ \boxed{P(Z \le 1.52) = 0.9357} ]

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vii. Correlation coefficient

Given:

[ b_{xy} = -0.84,\qquad b_{yx} = -0.20 ]

[ r = \pm \sqrt{b_{xy} \times b_{yx}} ]

[ r = \pm \sqrt{(-0.84)(-0.20)} ]

[ r = \pm \sqrt{0.168} ]

[ r = \pm 0.4099 ]

Since both regression coefficients are negative, (r) must be negative:

[ \boxed{r = -0.41 \text{ approximately}} ]

Interpretation: There is a moderate negative correlation between (x) and (y). As (x) increases, (y) tends to decrease.

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viii. Moment generating function of Poisson distribution

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

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

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

Geometric Mean vs Harmonic Mean

Geometric Mean (GM)

The geometric mean of $n$ positive values $x_1, x_2, \dots, x_n$ is the $n$th root of their product:

$GM = \left(x_1 \cdot x_2 \cdots x_n\right)^{1/n} = \text{antilog}\left(\frac{1}{n}\sum \log x_i\right)$

Harmonic Mean (HM)

The harmonic mean is the reciprocal of the arithmetic mean of the reciprocals:

$HM = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n}} = \frac{n}{\sum \frac{1}{x_i}}$

Points of Difference

Uses of Geometric Mean

1. Average rate of growth — used to find the average percentage increase in population, production, sales or profits over a period.
2. Index numbers — used in constructing price and quantity index numbers, since it gives equal weight to relative changes.
3. Averaging ratios and ratios of change — suitable when data are in the form of ratios or percentages.
4. Compound interest and depreciation — used to compute the average rate of compound interest or depreciation.
5. Highly skewed data — preferred when the distribution is highly skewed, as it is less affected by extreme values than the AM.

Uses of Harmonic Mean

1. Average speed — used when the same distance is covered at different speeds; the average speed is the HM of the speeds.
2. Average price — used when equal amounts of money are spent on different commodities at different prices.
3. Averaging rates and ratios — used for data expressed as "per unit" quantities, such as kilometres per hour or rupees per kilogram.
4. Time and work problems — used to find the average time taken by workers or machines working at different rates.
5. Averaging time rates — preferred when the variable is a rate of change with respect to time.

Dispersion (also called variation or spread) refers to the extent to which the individual values of a series deviate from one another or from a central value. Two series may have the same average but differ greatly in the way their values are scattered. For example, the daily wages (in Rs) of two groups of five workers are:

• Group A: 100, 100, 100, 100, 100 — mean Rs 100, no dispersion
• Group B: 20, 60, 100, 140, 180 — mean Rs 100, but values are widely scattered

Both groups have the same mean, yet Group B shows much greater dispersion. Dispersion is thus a measure of the reliability of an average and of the homogeneity (consistency) of the data.

Measures of dispersion

1. Range — the difference between the largest and smallest value: $R = L - S$. For Group B, $R = 180 - 20 = 160$. It is simple but uses only two extreme values, so it is unreliable.

2. Quartile deviation (semi-interquartile range) — $QD = \dfrac{Q_3 - Q_1}{2}$, where $Q_1$ and $Q_3$ are the first and third quartiles. It covers the middle 50% of the data and ignores extreme values, but it too is based on only two positional values.

3. Mean deviation — the average of the absolute deviations of the values from an average (usually the mean or median): $MD = \frac{\sum |x - \bar{x}|}{n}$ For Group B, $\bar{x} = 100$ and $\sum |x - \bar{x}| = 80 + 40 + 0 + 40 + 80 = 240$, so $MD = 240/5 = 48$. It uses every observation but ignores signs, which makes it unsuitable for further algebraic treatment.

4. Standard deviation — the positive square root of the mean of the squared deviations from the arithmetic mean: $\sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$ For Group B, $\sum (x - \bar{x})^2 = 6400 + 1600 + 0 + 1600 + 6400 = 16000$, so $\sigma = \sqrt{16000/5} = \sqrt{3200} = 56.57$. It is based on all observations and is mathematically tractable.

5. Coefficient of variation — a relative measure, free of units: $CV = \frac{\sigma}{\bar{x}} \times 100$ For Group B, $CV = \dfrac{56.57}{100} \times 100 = 56.57\%$. It is used to compare the variability of two or more series with different units or different means.

Most important measure

The standard deviation is the most important measure of dispersion. It takes every observation into account, is rigidly defined, is capable of further algebraic treatment (for example, in computing the combined standard deviation and in correlation and regression), and forms the basis of the coefficient of variation, which is the best relative measure for comparing consistency. Its only drawback — giving greater weight to extreme values — is more than offset by its mathematical usefulness.

Raw moments. For a frequency distribution with values $x_i$, frequencies $f_i$ and $N=\sum f_i$, the $r$th raw moment (moment about the origin) is

$m'_r = \frac{1}{N}\sum f_i x_i^{\,r}, \qquad m'_1=\bar{x}.$

Central moments. The $r$th central moment (moment about the mean $\bar{x}$) is

$m_r = \frac{1}{N}\sum f_i (x_i-\bar{x})^{r}, \qquad m_1=0,\ m_2=\sigma^2.$

Central moments in terms of raw moments. Expanding by the binomial theorem with $\bar{x}=m'_1$:

$m_r=\sum_{k=0}^{r}(-1)^k\binom{r}{k}(m'_1)^k\,m'_{r-k},\qquad m'_0=1.$

For $r=2,3,4$:

$m_2 = m'_2 - 2(m'_1)^2 + (m'_1)^2 = m'_2 - (m'_1)^2$

$m_3 = m'_3 - 3m'_1m'_2 + 3(m'_1)^3 - (m'_1)^3 = m'_3 - 3m'_1m'_2 + 2(m'_1)^3$

$m_4 = m'_4 - 4m'_1m'_3 + 6(m'_1)^2m'_2 - 4(m'_1)^4 + (m'_1)^4 = m'_4 - 4m'_1m'_3 + 6(m'_1)^2m'_2 - 3(m'_1)^4$

Definitions

Positive correlation: Two variables are said to have positive correlation when they move in the same direction. That is, an increase in one variable is accompanied by an increase in the other, and a decrease in one is accompanied by a decrease in the other. Its coefficient lies between $0$ and $+1$.

Negative correlation: Two variables are said to have negative correlation when they move in opposite directions. That is, an increase in one variable is accompanied by a decrease in the other. Its coefficient lies between $-1$ and $0$.

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Computation of Karl Pearson’s coefficient of correlation

Karl Pearson’s coefficient of correlation is given by:

$$r=\frac{n\sum xy-\sum x\sum y} {\sqrt{\left[n\sum x^2-(\sum x)^2\right]\left[n\sum y^2-(\sum y)^2\right]}}$$

Here,

$$n=8,\quad \sum x=92,\quad \sum y=93,\quad \sum x^2=1200,\quad \sum y^2=1163,\quad \sum xy=1157$$

Substituting the values:

$$r=\frac{8(1157)-(92)(93)} {\sqrt{\left[8(1200)-(92)^2\right]\left[8(1163)-(93)^2\right]}}$$ $$r=\frac{9256-8556} {\sqrt{(9600-8464)(9304-8649)}}$$ $$r=\frac{700}{\sqrt{1136\times 655}}$$ $$r=\frac{700}{\sqrt{744080}}$$ $$r=\frac{700}{862.60}=0.8115$$ $$\boxed{r=0.8115}$$

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Interpretation

The coefficient of correlation is $r=0.8115$, 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. The relationship is not perfect, but it is strong.

Let the common mean of $x$ and $y$ be $m$, so that

[ \bar x=\bar y=m. ]

Every regression line passes through the point $(\bar x,\bar y)$, i.e. through $(m,m)$.

1. Using the first regression equation

[ y=\alpha x+b ]

Put $x=m,\ y=m$:

[ m=\alpha m+b ]

[ b=m-\alpha m=m(1-\alpha) ]

So,

[ b=m(1-\alpha) \quad (1) ]

2. Using the second regression equation

[ x=ay+\beta ]

Put $x=m,\ y=m$:

[ m=am+\beta ]

[ \beta=m-am=m(1-a) ]

So,

[ \beta=m(1-a) \quad (2) ]

3. Required result

Dividing (1) by (2),

[ \frac{b}{\beta}=\frac{m(1-\alpha)}{m(1-a)} ]

Since $m\neq 0$,

[ \boxed{\frac{b}{\beta}=\frac{1-\alpha}{1-a}} ]

Hence proved.

4. Common mean

From (1),

[ m=\frac{b}{1-\alpha} ]

and from (2),

[ m=\frac{\beta}{1-a} ]

Therefore, the common mean is

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

provided $\alpha\neq 1$ and $a\neq 1$.

Let $b_{yx}=r\dfrac{\sigma_y}{\sigma_x}$ and $b_{xy}=r\dfrac{\sigma_x}{\sigma_y}$ be the two regression coefficients, where $\sigma_x,\sigma_y>0$. Their arithmetic mean is

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

By the AM-GM inequality, for positive numbers,

$\frac{\sigma_y}{\sigma_x}+\frac{\sigma_x}{\sigma_y}\ \ge\ 2\sqrt{\frac{\sigma_y}{\sigma_x}\cdot\frac{\sigma_x}{\sigma_y}}=2,$

with equality only when $\sigma_x=\sigma_y$.

Case $r>0$ (the usual case): multiplying by $\dfrac r2>0$,

$A\ \ge\ r,$

so the AM of the regression coefficients is greater than $r$, and equal to $r$ only when $\sigma_x=\sigma_y$.

Case $r<0$: both regression coefficients are negative and the inequality reverses, but in magnitude the same result holds:

$|A|=\frac{|r|}{2}\left(\frac{\sigma_y}{\sigma_x}+\frac{\sigma_x}{\sigma_y}\right)\ge |r|.$

Conclusion: $\dfrac{b_{yx}+b_{xy}}{2}\ge r$ for $r\ge0$, and in general $\left|\dfrac{b_{yx}+b_{xy}}{2}\right|\ge|r|$; the AM exceeds the correlation coefficient strictly unless $\sigma_x=\sigma_y$.

Given: (A) and (B) are independent events.

So, [ P(A\cap B)=P(A)P(B) ]

We have to prove that (\bar A) and (\bar B) are independent, i.e. [ P(\bar A\cap \bar B)=P(\bar A)P(\bar B) ]

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) ]

Now, [ P(\bar A\cap \bar B)=1-[P(A)+P(B)-P(A)P(B)] ] [ =1-P(A)-P(B)+P(A)P(B) ] [ =[1-P(A)][1-P(B)] ] [ =P(\bar A)P(\bar B) ]

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

Therefore, (\bar A) and (\bar B) are independent events. Hence proved.

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. Each box holds 30 shirts. Let $B_1,B_2$ be the events that box 1 or box 2 is selected and $S$ the event that a super T-shirt is drawn.

$P(B_1)=P(B_2)=\frac12,\qquad P(S\mid B_1)=\frac{10}{30}=\frac13,\qquad P(S\mid B_2)=\frac{5}{30}=\frac16.$

$P(B_1\mid S)=\frac{\frac12\cdot\frac13}{\frac12\cdot\frac13+\frac12\cdot\frac16}=\frac{1/6}{1/4}=\frac23.$

The probability that the super T-shirt came from box 1 is $\dfrac23$.

Question (as given): Obtain the moment generating function for the distribution

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

where $y_0$ is a constant.

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Step 1: Find the constant $y_0$.

Since $dp$ is a probability density,

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

$y_0\int_0^{\infty} e^{-x/\sigma}\,dx = y_0\Big[-\sigma e^{-x/\sigma}\Big]_0^{\infty} = y_0(0+\sigma) = y_0\sigma.$

Hence

$y_0\sigma = 1 \quad\Rightarrow\quad y_0 = \frac{1}{\sigma}.$

So the p.d.f. is

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

i.e. the exponential distribution with mean $\sigma$.

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

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

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Step 3: Evaluate the integral.

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

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

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

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

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Result:

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

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Step 4: Verification (mean and variance).

$M_X'(t) = \sigma(1-\sigma t)^{-2}, \qquad M_X'(0) = \sigma = \mu_1'.$

$M_X''(t) = 2\sigma^2(1-\sigma t)^{-3}, \qquad M_X''(0) = 2\sigma^2 = \mu_2'.$

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

These agree with the known mean $\sigma$ and variance $\sigma^2$ of the exponential distribution, confirming the m.g.f. is correct.

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

A probability density function of a continuous random variable $X$ is a function $f(x)$ such that:

1. $f(x) \ge 0$ for all $x$,
2. $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$,
3. For any interval $(a, b)$, $\displaystyle P(a < X < b) = \int_a^b f(x)\,dx$.

Thus $f(x)$ itself is not a probability; probabilities are obtained by integrating $f(x)$ over an interval. For a discrete random variable the corresponding function $P(X = x)$ is called the probability mass function.

(ii) Distribution function

The distribution function (or cumulative distribution function, c.d.f.) of a random variable $X$ is defined as

$F(x) = P(X \le x), \qquad -\infty < x < \infty.$

It gives the probability that $X$ takes a value not exceeding $x$. Its properties are: $F(-\infty) = 0$, $F(\infty) = 1$, $F(x)$ is non-decreasing, and $P(a < X \le b) = F(b) - F(a)$.

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Distribution function of X

Given:

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

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

• For $x < 0$: no value of $X$ is $\le x$, so $F(x) = 0$.
• For $0 \le x < 1$: only $X = 0$ is included, $F(x) = P(X = 0) = \frac{1}{3}.$
• For $1 \le x < 2$: $X = 0, 1$ are included, $F(x) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}.$
• For $x \ge 2$: all values are included, $F(x) = \frac{1}{3} + \frac{1}{6} + \frac{1}{2} = 1.$

Hence the distribution function is

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

This is a step function, increasing by jumps of $\frac{1}{3}, \frac{1}{6}, \frac{1}{2}$ at $x = 0, 1, 2$ respectively.