SSC CGL (JSO) 2018 Official Paper (Held On: 14 Sept 2019 ) (Previous Year Paper)

Fomula used

Simple aggregative method = P₀₁ = ∑p₁/∑p₀  × 100

∑p₁ = Price of all commodities in the current year

∑p0  =  Price of all commodities in base year

Calculation:

∑p0 = 600 + 700 + 400 + 300 = 2000

∑p1 = 700 + 600 + 480 + 270 = 2050

Net Price changes in Percentage

= P₀₁ = 2050/2000 × 100

= 102.5

∴ The price of commodities in 2017 is increased by 2.5% from 2016

Important Points

Simple Aggregative Method = In it the total price of commodities in a given (current) year is divided by the total price of commodities in a base year and expressed as a percentage. 

Formula = 

P01 = ∑p1/∑p0  × 100

∑p1 = Price of all commodities in the current year

∑p0  =  Price of all commodities in base year

Given

Average working hours per month of staff aged over 50 years = 160

Average working hours per month of staff aged below 50 years = 200

Mean average = 200

Concept:

Average = (Sum of Values/Number of Values)

Calculation:

Let number of staff aged over 50 years = x

Number of staff aged below 50 = y

According to question

⇒ 160x + 210y = 200(x + y)

⇒ 40x = 10y

⇒ x : y =  1 : 4

∴ The ratio of the number of the staff aged over 50 to that the staff aged under 50 is 1 : 4

Alternate Method

By alligation method

∴ The ratio of the number of the staff aged over 50 to that the staff aged under 50 is 1 : 4

Formula used

D = iN/10

D = Decile

i = 1, 2, 3, 4, 5, 6,---

N = Total number of frequency

Calculation

Here We have to find 4th decile 

∴ i = 4

⇒ D = 4 × 256/10

⇒ D = 25.6 × 4 = 102.4

∴ 102.4 lies in cumulative frequency 167 which is in x = 4

∴ 4th decile = 4

Formula used

Mean Deviation about median= $∑\rm \frac{|x_{i} - M|}{n}$

Where, 

x_i = individual term 

M = Median

n = total number of terms

Calculation:

52, 56, 66, 70, 75, 80, 82

Median = 70

⇒  M = 70

⇒ The value of (x_i - M) are

⇒ -18, -14, -4, 0, 5, 10, 12

⇒ Mean Deviation = $\rm \frac{18 + 14 + 4 + 0 + 5 + 10 + 12}{7}$

∴ The mean deviation about median is 9

Important Points

The mean is the average of a data set.

The median is the middle of the set of numbers.

Mean deviation about mean 

\(∑\rm \frac{|x_{i} - x̅|}{n}\) where x̅ = mean 

x_i = individual term 

n = total number of terms

Explanation

For a random variable, central moment(μ) of all order exist

Square of (2j + 1)^th moment (μ²_{2j + 1})

Let j = 1 then (2j + 1)^th moment is given by

(μ²₃) = [(∑(x – x̅)³/N]²

⇒ (∑x³/N)²    ------(i)

If j = 1 then μ₂ = ∑x²/N

⇒  μ_{2j + 2} if j = 1 then

⇒ μ₄ = ∑x⁴/N

⇒ μ₂ × μ_{2j + 2} = ∑x²/N × ∑x⁴/N    ------(ii)

Compare both equation

(∑x³)²/N² = (∑x² × ∑x⁴)/N²  

⇒ (∑x³)² ≤ (∑x² × ∑x⁴)

∴ (μ_{2j + 1})² ≤ μ_2j × μ_{2j + 2}

Explanation:

Memoryless property  = A variable X is memoryless with respect to t if for all s with t ≠ 0

⇒ P(x > s + t I x > t) = P(x > s)

⇒ P(x > t, x > T)/P(x > T) = P(x > s)

⇒ P(x > s + t) = P(x > s) P(x > t)

This property is satisfied by exponential distribution

P(x > t) = e^-λt

⇒ P(x > s + t) = e^{-λ(s + t)}

∴ P(x > s + t) = P(x > s) P (x > t)

⇒ e^-λs × e^-st

⇒ e^{-λ(s + t)}

∴ This is only memoryless random distribution

The only continuous probability distributions that are memoryless are the exponential distributions. If a continuous X has the memoryless property

(over the set of reals) X is necessarily an exponential.

Important Points

Continuous uniform distribution = A random variable X is said to follow a continuous uniform (rectangular) distribution over an interval (a, b) if its probability density function is given by 

\(f(x) = \left{ \begin{matrix} \dfrac{1}{b-a} \ \text{for} \ a < x < b \\ 0 , \ \ \text{otherwise} \end{matrix} \right.\)

Normal distribution = A continuous random variable X is said to follow normal distribution with parameters μ(-  ∞ <  μ <  ∞) and σ² (> 0) if it takes on any real value and its probability density function is given by

$f(x)=\dfrac{1}{\sqrt[\sigma]{2\pi}} e^{\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

Gamma distribution = if n > 0, the integral $\displaystyle\int_0^\infty$ is called gamma function and t is denoted by 1(n̅)

Explanation:

The population correction factor is given by the formula :

$\sqrt {\frac{{N - n}}{{N - 1}}}$

Important Points

Sample size = In statistics, the sample size is the measure of the number of individual samples used in an experiment for example,  if we are testing 50 samples of people who watch movie in a city, then the sample size is 50.

Sample mean = The sample mean from a group of observations is an estimate of the population mean (μ).

Given a sample of size n, consider n independent random variables X₁, X₂, ..., X_n, each corresponding to one randomly selected observation. Each of these variables has the distribution of the population, with mean(μ)  and standard deviation (σ).

The sample mean = x̄ = ( Σ xi ) / n

Explanation:

Let X be a discrete random variable with the geometric distribution with parameter p for some 0 < p < 1.

⇒ X(Ω) = {0,1,2,…} = N

⇒ Pr (X = k) = p(1−p)^k ∴ Then the excess kurtosis γ₂ of X is given by: $6 + \dfrac{p^2}{1-p}$

Important Points Kurtosis = Kurtosis is a statistical measure that defines how heavily the tails of distribution differ from the tails of a normal distribution   Excess kurtosis is a metric that compares the kurtosis of distribution against the kurtosis of a normal distribution. The kurtosis of a normal distribution equals 3.   Therefore, Excess Kurtosis = Kurtosis – 3   The geometric distribution is a discrete distribution for n= 1, 2, 3,---- having probability distribution function   p(n) = p(1 - p)ⁿ   p(n) = pqⁿ   Where0 <p <1 and q = 1 - p

Explanation:

Let {Xi, i ≥ 1} be independent and identically distributed random variables with P(Xi = 1) = p = 1 - P(Xi =0), Sn = $\sum^n_{i=1}{X_i}$i. The distribution of Sn is :

Binomial distribution with parameter n and p

Important Points

1 - Suppose a discrete random variable X has the following pmf P(X = k) = qk P, 0 ≤ k < ∞ The X is said to have Geometric distribution with parameter p

2 - A discrete random variable X is said to follow Bernoulli distribution with parameter p if its probability mass function is given by

,P[X = x] = {p^x(1 - p)^{1 -x}  x = 0,1 and 0, elsewhere

Explanation:

The population mean and standard deviation are two common parameters. In statistics, Greek symbols usually represent population parameters, such as μ (meu) for the mean and σ (sigma) for the standard deviation.

Important Points

 \(̅{X}\) = The sample mean formula is:

\(̅{X}\) = ( Σ xi ) / n

Population standard deviation = Standard deviation measures the spread of a data distribution. It measures the typical distance between each data point and the mean.

σ =√([Σ(x - u)2]/N)]

S = Sample standard deviation = σ =√([Σ(x - x̅ )2]/(n - 1)]

Given

∑x = 0

∑x² = 28

∑xy = 56

Formula

Trend equation is given by

Y_c = a + bx

Calculation

⇒ a= ∑y/N = 616/7

⇒ a = 88

⇒ b = ∑xy/∑x²

⇒ 56/28 = 2

Put a and b in trend equation

Y_c = 88 + 2 × x

⇒ 81 = Y₁₉₇₂ = 88 + 2 × ( - 3.5)

⇒ Y₁₉₇₆ = 88 + 2 × 4 ( difference in 1976 – 1972 = 4)

∴ Y₁₉₇₆ = 96

Given

μ₂ = 4

μ₄ = 18

Formula used

Karl pearson's coefficient of skewness β_z = μ₄/μ²₂

Calculaton

Karl Pearson defined the following four coefficients, based upon the first four moments"about mean:

⇒ βz = 18/4²

⇒ βz = 18/16

∴ The value of Pearson's coefficient of skewness βz is 1.125

Important Points

Moment = The rth moment of a variable x about any point x = A, usually denotted by μ_r' and it is given by

μr' = (1/N)[∑f_i(x_i - A)^r , ∑f_i = N

⇒  (1/N)[∑f_id_i^r  d_i = (x_i - A)

Moment about mean = The rth moments of a variable about the mean x̅  denoted by μ =(1/N)[∑fi(xi - x̅ )r = (1/N)[∑f_iz_i^r  where z_i = xi - x̅

Explanation:

Seasonal index for time series can be calculated using the following method

1. Method of averages

2. Link Relative Method

3. Moving Average Method

But cannot be calculated using Mathematical equations.

Hence Option 4 is correct.

Important Points

Moving average method = A moving average is a technique to get an overall idea of the trends in a data set; it is an average of any subset of numbers.

The method of moving average is useful to measure the future cost value and forcasting

Link relative method = This method was developed by Prof. Karl Pearson so, it is also known as karl person of seasonal variant.

Concept:

Coefficient of variation = $\rm \frac{\sigma _{x}}{\bar{x}}$

Where, 

$\rm \sigma _{x}$ = Standard Deviation

$\rm \bar{x}$ = Mean of the observations

Calculation:

Coefficient of variation = $\rm \frac{17.4}{148}$

∴ Coefficient of variation is  0.12

Explanation:

As there is no spread of variables around the mean, the variance for the degenerate distribution is zero (Var(X) = 0).

Important Points

The symbol of Sample variance = S

S = ∑(x - x̅ )²/(n - 1)

x̅ = mean of all observation

x = value of the observation

n = number of observations

variance is the expectation of the squared deviation of a random variable from its mean

Explanation:

Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.

Hence Option 3 is correct.

Important Points

Statistics = a branch of mathematics dealing with the collection, analysis, interpretation, and presentation of masses of numerical data.

In statistics we read many topics like central of tendency, Index number, Time series,Standard deviation, variance,

Formula used

$\rm Mode =L + \left [ \frac{f1 - f0}{2f1 - f0 - f2} \right ]\times i$

i = class interval or class size

f1 = frequency of modal class

fo = frequency of pre modal class

f2 = frequency of success modal class

L = lower limit of modal class

Calculation:

Here maximum frequency is 28 ,the class corresponding to this 40-50

So modal class is 40-50

Now lower limit of modal class (l) = 40

Frequency (f1) of modal class = 28

frequency (f0) of preceding the modal class  = 14

frequency (f2) of the class succeeding the modal class = 20

Class size = 10    

$\rm Mode = 40 + \left [ \frac{28 - 14}{2 \times 28 - 14 - 20} \right ]\times 10$

= 40 + 6.36

∴ Mode = 46.36

Important Points

Another formula to calculate mode = 3 Median - 2 Mean

Explanation:

There are various types of sampling methods such as:

• • Simple random sampling
• Systematic sampling
• Stratified sampling
• Probability-proportional-to-size sampling
• Cluster sampling
• Quota sampling
• Minimax sampling
• Accidental sampling
• Voluntary Sampling
• Line-intercept sampling
• Panel sampling
• Snowball sampling
• Theoretical sampling
• Purposive sampling

Hence Option 3 is correct.

Important Points

Sampling = sampling consists of obtaining information from a larger group or a universe

Systematic sampling= k =  population size/sample size

Systematic sampling it is defined as probability sampling method where the researcher chooses elements from a target population by selecting a random starting point and select sample members.

Stratified sampling = stratified sampling is a method of sampling from a population which can be partitioned into subpopulations

Purposive sampling = the main goal of purposive sampling is to focus on particular characteristics of a population that are of interest

Concept:

The formula of the probability of an event is:

P(A) = $\rm \frac{Number of favourable outcomes}{Total number of favourable outcomes}$

Calculation:

Total number of ways A, B, C, D and E can sit together = 5! = 120

Number of ways in which A and B sit together 

lets stick A and B together and treat them as 1 unit let us name this unit as U now

we have UCDE

number of ways of arranging this is 4!

but number of ways of arranging U is 2!

so total number of ways are 4!*2! = 4 $\times$ 3  $\times$ 2 $\times$ 1 $\times$ 2 = 48

The probability that A and B sit next to each other is

= $\rm \frac{48}{120}$

= $\dfrac{2}{5}$

Important Points

Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur

Probability always lies from 0 to 1

The sum of probability of all possible outcomes is always 1

Formula

Poisson distribution = P(X = x) = [ e^-m m^x/x! for x = 0, 1, 2, 3 -----]

                                                          [    0                       ; otherwise]

Calculation

Poisson distribution has 2 modes at x = 1 and x = 2 it means they have same frequency at x = 1 and 2and have same probability

P(x = 1) = λ¹. e^-λ/1!

⇒ P(x = 2) = λ². e^-λ/2!

According to question P(x = 1) = P(x = 2)

λ¹. e^-λ/1! = λ¹. e^-λ/2!

⇒ λ¹. e^-λ = λ². e^-λ/2

⇒ λ = 2

⇒ P(x = 1) = 2¹. e^-2/1!

⇒ 2e^-2

⇒ P(x = 2) = 2². e^-2/2!

⇒ 2.e^-2

⇒  P(x) = P(x = 1) + P(x = 2)

⇒ 2e^-2 + 2.e^-2

∴ P(x) = 4e^-2

Explaination

Index number is a number used to measure the level of certain phenomena as compared to the level of the same phenomenon

We also defined as the index number is used to for comaring between the price, quantity etc

We can take an example to understand index number  at a certain place or a period of time and that of another place or period of time

Basically Index number are used for the base of planned economy

∴ Option second is correct answer

Given

In discrete random variable X follows uniform distribution

Assumed values of X = 5. 9, 11, 15, 18, 20

Calculation

I X – 14 I < 5

⇒ - 5 <  X – 14 < 5

⇒ 9 < X <, 19

∴ Favourable case = {11, 15, 18} = 3

Total sample space = {8, 9, 11, 15, 18, 20} = 6

⇒ P(IX – 14I < 5) = P(9 < X < 19)

⇒ Favourable case/total sample space = 3/6

∴ The value of P(IX – 14I < 5 is ½

Explanation:

The Marshall Edgeworth Method for the index number, credited to Marshall (1887) and Edgeworth (1925), is a weighted relative of the current period to base period sets of prices. This index uses the arithmetic average of the current and based period quantities for weighting.

Factor Reversal Test: 

Fisher has described this test in the following words: Just as each formula should permit the interchange of the two items without giving inconsistent results, so it ought to permit interchanging the prices and quantities without giving inconsistent result, i.e., the two results multiplied together should give the true value ratio.”In simple words, the test means that the change in the price multiplied by the change in the quantity should be equal to the total change in the value. The total value of a given commodity in a given year is the product of the quantity and the price per unit (total value = p x q) . The ratio of the total value in one year to the total value in the preceding year is p1q1 / p0q0. Symbolically the test is represented as

P01 x Q01 = Σp1q1 / Σp0q0 V01

Where P01 denotes price index and Q01 , quantity index number.

This test is satisfied by Fisher’s method only.

Circular test

According to this, if indices are constructed for year one based on year zero, for year two based on year one and for year zero based on year two, the product of all the indices should be equal to 1.

Symbolically

P₀₁ ×  P₁₂ ×  P₂₀ = 1

This test is satisfied by

2-Simple aggregative method and

1-Kelly’s method.

Explanation

Ogive = When X – coordinates are the upper limits of the class interval and y coordinates are corresponding cumulative frequency on a curve by joining points is called ogivi curve

Histogram = A histogram is a graphical display of data using bars of different heights. In a histogram, each bar groups numbers into ranges.

Pie curve = A pie chart is a type of graph that represents the data in the circular graph

Frequency polygon = A frequency polygon is almost identical to a histogram, which is used to compare sets of data or to display a cumulative frequency distribution. It uses a line graph to represent quantitative data

∴  option 2 is correct

Calculation:

To find the first quartile of X, we need to find the value x such that the area under the probability density function f(x) to the left of x is 0.25.

We can begin by finding the cumulative distribution function (CDF) of X:

F(x) = ∫[0, x] f(t) dt

Since f(x) is a uniform distribution between 0 and 5, we have:

f(x) = 1/5 for 0 ≤ x ≤ 5

Therefore, the CDF of X is:

F(x) = ∫[0, x] (1/5) dt = x/5

To find the first quartile of X, we need to find the value x such that:

F(x) = x/5 = 0.25

Solving for x, we get:

x = 0.25 × 5 = 1.25

Therefore, the first quartile of X is 1.25.

Given

P(A) = 0.60

P(B/A) = 0.10 

P(B/A)’ = 0.80

Concept:

P(A | B) ⋅ P(B) = P(B | A) ⋅ P(A) 

Calculation:

Let A be the event : the employees are graduate

Let B be the event : the employees are in sales

⇒ P(B) = P(A).P(B/A) + P(A’). P(B/A)’

⇒ 0.60 × 0.10 + 0.40 ×  0.80 

∴ The probability that an employee selected at random is in sales, is 0.38

Important Points

Probability always lies between 0 to 1

Explaination

Seasonal index for four quarters given by calculating average of quarterly average = Grand average and calculating quarterly averages = average age of quarter

Seasonal index = (Average of quarter/Grand average) × 100

Seasonal index for first term is given by I₁ = S₁/S × 10

S₁ = Average of first term

S = average of all term SS_J/k

J = 1, 2, 3, 4 ----

K = 12 for monthly data

K = 4 for quarterly data

∴ Seasonal index = (Average of quarter/Grand average) × 100

Concept:

General Formula for multiple correlation coefficient

\(\rm R^2_1{.}{2}{3} = \frac{r{12}^{2}+r{13}^{2}-2r_{12}r_{13}r_{23}}{1-r_{23}^{2}}\)

Given

 r12 = +0.80, r13 = -0.40 and r23 = -0.56

Calculation

 $\rm R_{1.23}^{2} = \frac{r_{12}^{2}+r_{13}^{2}-2r_{12}r_{13}r_{23}}{1-r_{23}^{2}}$

$\rm R^2_1{.}{_2}{_3}= \frac{0.80^{2}+(-0.40)^{2}-2(0.80)(-0.40)(-0.56)}{1-(-0.56)^{2}}$

$\rm R^2_1{.}{_2}{_3}$= 0.64  0.16 - 2 × 0.8 × (- 0.40)(-0.56)/[1 - (0.56)²]/[1 - (0.56)²

⇒ (0.80 - 0.3584)/(1 - 0.3136)

∴ $\rm R^2_1{.}{_2}{_3}$ = 0.6434

Explanation:

The general formula for multiple correlation coefficient

\(\rm R^2_1{.}{2}{3} = \frac{r{12}^{2}+r{13}^{2}-2r_{12}r_{13}r_{23}}{1-r_{23}^{2}}\)

A multiple correlation coefficient is a non-negative coefficient.

It is the value ranges between 0 and 1. It cannot assume a minus value.

If the multiple correlation coefficient of X1 on X2 and X3 is zero, then:

If R1.23 = 0 then r12 = 0 and r13 = 0

R1.23 is the same as R1.32

Explanation:

The hypotheses of interest in an ANOVA are as follows:

• H₀: μ₁ = μ₂ = μ₃ ... = μ_k
• H₁: Means are not all equal.

Where k = the number of independent comparison groups.

In this example, the hypotheses are:

• H₀: μ₁ = μ₂ = μ₃ = μ₄
• H₁: The means are not all equal.

The null hypothesis in ANOVA is always that there is no difference in means. The research or alternative hypothesis is always that the means are not all equal and is usually written in words rather than in mathematical symbols.

Formula

The general formula of multiple correlation coefficient

\(\rm R^2_1{.}{2}{3} = \frac{r{12}^{2}+r{13}^{2}-2r_{12}r_{13}r_{23}}{1-r_{23}^{2}}\)

Explanation:

Multiple correlation coefficient is a non-negative coefficient.

It is the value ranges between 0 and 1. It cannot assume a minus value.

If R_1.23 = 0 then r₁₂ = 0 and r₁₃ = 0

R_1.23 is the same as R_1.32

Explanation:

The second differential in time series will help to eliminate the quadratic trend. The quadratic trend is considered an appropriate model for secular trend components of time series when data appear not to fall in a straight line

Time series =An arrangement of statistical data according to time of occurrence or in chronological order is called time series

Time series are used in forecasting 

The time series is used in analysing the past behaviour of the variables.

Formula:

ⁿC_r = $\rm \frac{n!}{r! (n-r)!}$

Calculation:

In a  game of bridge 13 cards are dealt t each of the 4 players from a single deck of cards.

The total number of ways of getting 13 cards in  one hand are ^⁵²C_₁₃

To get 9 cards of the same suit, one can choose the particular suit in 4 ways since there are 4 different suits and then 9 cards to be dealt from that particular suit and the remaining 4 cards should have been dealt from the remaining 3 suites

So, 9 cards out of 13 cards of the same suit can be selected in ¹³C₉ ways and the remaining 4 cards out of 39 cards can be selected in ³⁹C₄ ways and a particular suite can be chosen in 4 different ways

Hence the total number of selections are 4 × ¹³C₉ × ³⁹C₄

Hence the probability of getting 9 cards of the same suit in one hand is

⇒  4 × ¹³C₉ × ³⁹C₄  / ⁵²C₁₃

∴ The required probability is 4 × ¹³C₉ × ³⁹C₄  / ⁵²C₁₃

Personal approach is NOT an approach for assigning the probability of the event.

Explanation:

Relative frequency approach​:

Relative frequency can be calculated by taking the count of an individual kind of outcome and divide by the total counts for all kinds of outcomes.  For a ratio of 1:1, there are two total outcomes, so the relative frequency of heads is ½ or 0.5 and the relative frequency of tails is the same.

Classical approach:

• Rolling a fair die. It’s equally likely you would get a 1, 2, 3, 4, 5, or 6.
• Selecting bingo balls. Each numbered ball has an equal chance of being chosen.
• Guessing on a test. If you guessed on a multiple-choice test with four possible answers A, B, C and D, each choice has the same odds of being picked (assuming you pick randomly and don’t follow a pattern).

The probability of a simple event happening is the number of times the event can happen, divided by the number of possible events.

The “math” way of writing the formula is P(A) = f / N.

P(A) means “probability of event A” (event A is whatever event you are looking for, like winning the lottery).

“f” is the frequency or number of possible times the event could happen.

N is the number of times the event could happen.

Statistical approach:

The basic approach statistical methods adopt to deal with uncertainty is via the axioms of probability: Probabilities are (real) numbers in the range 0 to 1. A probability of P(A) = 0 indicates total uncertainty in A, P(A) = 1 total certainty and values in between some degree of (un)certainty.

Concept:

A, B, and C are mutually exclusive events,

P(A) + P(B) + P(C) = 1

Calculation:

P(B) = $\rm \frac{3}{2}P(A)$ and P(C) = $\rm \frac{1}{2}P(B)$

⇒ Let P(A) = p

⇒ P(B) = $\rm \frac{3p}{2}$ and P(C) = $\rm \frac{3p}{4}$

Since A, B, C are mutually exclusive and exhaustive events associated with a random experiment.

⇒ A ∪ B ∪ C = S

⇒ P(A ∪ B ∪ C) = P(S)

⇒ P(A ∪ B ∪ C) = 1

⇒ P(A) + P(B) + P(C) = 1

⇒ p + $\rm \frac{3p}{2}$ + $\rm \frac{3p}{4}$ = 1

⇒ p = $\rm \dfrac{4}{13}$

∴ P(A) = $\rm \dfrac{4}{13}$

Concept:

Fisher-Price Index = (Laspeyres Price Index × Passche's Price Index)^0.5

Calculation:

The laspeyres price index of a commodity is 208

Passche's index of the same commodity is 52

Fisher-Price Index = (208 × 52)^0.5 = 104

the value of the Fisher index number will be 104

Laspeyre's index number = If we use base period  quantities(q_o) as the weight in the general weighted aggregative index formula 

L = (∑p_nq_o/∑p_oq_o) × 100

p_n =  current year price

p_o = Price at base year

q_o = Quantity at base price

Paasche's index = If we use current year quantities(q_n) as weights in the general aggregative formula we get paasche's formula

P = (∑(p_nq_n/∑p_oq_o) × 100

Explanation:

Variance is independent of change of origin as the change in origin is uniformly added to all the values and hence the mean also and hence, when (x − $\rm \overline{x}$)² is calculated, it remains unaffected. But, change of scale alters all values unevenly and hence, variation changes. 

The correlation coefficient is dependent on the choice of both origin and scale of observation. 

Hence, both (I) and (II) is correct.

Given:

Coefficient of variation (CV) is 20

Variance is 16

Concept

The coefficient of variation (CV)  is defined as the ratio of the standard deviation σ to the mean μ

CV = $\rm \rm \frac{\sigma }{\mu }$ × 100

Calculation:

Standard deviation (σ) = 4

⇒ CV = $\rm \frac{\sigma }{\mu }$ × 100

⇒ 20 = $\rm \frac{4}{\mu }$ × 100

⇒ μ = 20

∴ The arithmetic mean is: 20

Explanation

Poisson distributionis a limiting case of binomial distribution if it follows conditions

n, the number trials is indefinitely large  that means n tends to infinite

p, the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(1 – p)^{n – x}

∴ (n/x)(p/(1 – p)]^x(1 – p)^{n - x}

p , the constant probability of success for each trial is indefinitely small p tends to 0

np = λ , is finite so λ/n = p, q = 1 – p

⇒ (1 – λ/n), λ is positive integer

The probability of x successes in a series of n independent trials is

⇒ b(x, n, p) = (n/x)p^xq^{n – x}, x = 0, 1, 2, 3…….n

⇒ b(x, n, p) = (n/x)p^x(1 – p)^{n – x}

∴ (n/x)(p/(1 – p)]^x(1 – p)^{n - x}

⇒ **[**n(n - 1)(n - 2)------(n - x + 1)/x!] × (λ/n)^x/(1 - λ /n)^x[1 - λ/n]ⁿ

⇒ [(1 - 1/n)(1 - 2/n)-----( 1 - (x - 1)/n/x!(1 - λ/n)^x] × λ^x[1 - λ/n]ⁿ

⇒ Lim x → ∞ b(x, n, p) = e^-λ × λ^x/x! ; x = 0, 1, 2, 3, 4 -------,n

Poisson distribution = A random variables X is said to follow poisson distribution if it assumes only non - negative values and its proportionality mass function i s given

by P)X = x) = e-λ × λx/x! where x = 0, 1, 2, 3 ------n and  λ > 0

⇒ p(x, λ) = ∑P(X - x)

⇒ e^-λ∑λ^x/x!

⇒ e^-λ× e^-λ = 1

∴ The corresponding distribution function is F(x) = P(X = x) = ∑P(r) =  e-λ ∑λ²/r!; x = 0, 1, 2 .......

Given

Anova two way classification to test two types of cltths in fachion trends is given by

Formula

SSA + SSB + SSE = SST

SSA = Sum of square of column A

SSB = Sum of square of Column B        

SSE = Sum of square of error

SST = Total sum of square

Df = degree of freedom

Calculation

According to question

⇒ 280 + α + 20 = 640

⇒ α = 640 – 280 – 20

⇒ α = 340

∴ SSB = 340

Simillarly,

df(A) +  df(B) + df(C) = df(total)

⇒ 2 + 3 + β = 11

⇒ β = 11 – 2 – 3

⇒ β = 6

Variety β Mean sum of square ratio of rows (MSR)

MSR = SSB/df

⇒ MSR = 340/3

Error residual MSS = MSE = 3.33

F – ratio of variety β = MSR/MSE

⇒ 340/3/3.33

⇒ 340/(3 × 3.33)

⇒ 34.03

∴ (α, β, γ) are (340, 6, 34.03)

Concept:

Mean of the classed Data:

$\rm Mean =\dfrac{ \sum\limits_{i=1}^{i=n}f_i\overline x_i}{\sum\limits_{i=1}^{i=n}f_i}$, where n is the total number of class and f and x̅ are the frequency and mean of the class

Calculation:

Mean = $\dfrac{5\times12+15\times18+25\times27+35\times20+45\times17+55\times60}{12+18+27+20+17+60}$

Mean = $5770\over154$ = 37.467 ≈ 38

Explanation

Approximate Median of the poisson distribution with parameter λ

⇒ [λ + 1/3 – 0.02/λ]

⇒ λ  + 1/3 – 1/50λ

Poisson distribution  = It was discovered by French mathematician cum physicist

Simeon denis  poisson in 1837

It is also a discrete distribution

He derieved it as a limiting case of binomial distribution

Poisson distribution is given by = P(X = x) = [ e^-mm^x/x! for x = 0, 1, 2 ------

                                                                      [ 0,                   otherwise

Here m is parameter of the distribution and m > 0

Explanation 

If x₁, x_{2--------------,}x_n is a single random sample without replacement of size n is from a finite population N with mean μ and variance σ² the covariance(x_i, x_j) is given by

∴ Cov(x_i, x_j) = σ²/(n – 1)

Explanation

Multiplicative model assumes  that the observed value is obtained by multiplying the Trend(T) by the rates of three other components

Y = T × S × C × I

where as

Y = Original data

T = Trend

S = Seasonal variation

C = Cyclical variation

I = Irregular movement

Explanation

Let X and Y be two variables

Var(X) = s₁² = 1990

Var(Y) = s₂² = 796

n₁ = 11

n₂ = 9

Calculation

σ₁² = n₁ × s₁²/(n₁ – 1)

⇒ 11 × 1990/10

⇒ σ₁² = 11 × 199

σ₂² = n₂ × s₂²/(n₂ – 1)

⇒ 9 × 796/8

F(10, 8) = σ₁²/ σ₂²

⇒ (11 × 199)/9 × 796/8

⇒ 4378/1791

∴ F(10, 8) is 2.5

F_0.05 (10, 8) = 3.35 ( given value from table)

∴ (F_calculated = 2.5) < (F_0.05(1.98) = 3.35)

Given

Arithmetic mean =  (μ) = 10

Coefficient of variation = CV = 40% or 0.4

Formula

CV = σ/μ

σ = Standard deviation

μ = Mean

Calculation

0.4 = σ/10

⇒ σ/ = Standard deviation = 0.4 × 10 = 4

Now, Variance= V = (standard deviation)² = σ²

⇒ Variance = 4² = 16

We know that V(ax + b) = a²(V(x)

⇒ V(10 - 2x) = (-2)² V(X)

⇒  V(10 - 2x) = 4V(x)

∴ V(10 - 2x) = 4 × 16 = 64

Important Points

Symbol of Standard deviation = σ

Variance = square of standard deviation = σ²

Coefficient of variation = CV = CV = σ/μ

μ = Population mean

Given

MSA = Mean sum of square due to factor A

MSE = Mean sum of square due to error

Calculation

If null hypothesis for one way classification is not true means

⇒ MSA ≠ MSE

⇒ MSA > MSE

⇒ MSA < MSE then

If MSA = MSE then F value  = 1 otherwise

∴ F = MSA/MSE > 1

Null means nothing or invalid, Let assume that mean of the population is μ_o and mean of sample is x̅. Since we have assumed that the population has a mean of μ_o it is null hypothesis and it is written as

H_o : μ = μ_o

Where as H_o is null hypothesis and the alternative hypothesis is H₁ : μ ≠ μ_o

The rejection of the null hypothesis will show that the mean of the population is not μ_o

Given

Formula

Laspeyres price index = (∑p₁q_o/∑p₀q₀) × 100

P₁ = current year

P₀ = base year

q₀ =  base quantity

Calculation

Laspeyres price index for year 2006

⇒ (3 × 2.5 + 4 × 3.0 + 2 × 2.5)/(3 × 2 + 4 × 2.5 + 2 × 3.0) × 100

⇒ (7.5 + 12 + 5)/(6 + 10 + 6) × 100

⇒ (24.5/22) × 100

∴ Laspeyres price index for year 2006 is 111.36

Explanation

If Z_1,  Z₂, Z₃, ----------Z_n are n independent normal variates then

∴ ∑Z_i²  I = 1, 2, 3, -------n will be chi – square distribution with degree of freedom n

The chi-square distribution has one parameter: a positive integer n that specifies the number of degrees of freedom (the number of random variables being summed, Z_i s).

In probability theory and statistics, the chi-square distribution (also chi-squared or χ²-distribution) with k degrees of freedom is the distribution of a sum of the squares of k independent standard normal random variables

Given

Coefficient of correlation r between X and Y = r(X, Y) = r_xu = r

Standard deviation for X and Y = σ_x and σ_y

Calculation

The tangent of angle between two lines are given by

⇒ (1 - r)²( σx × σ_y)/r (σ_x² + σ _y²)

Equation of line of regression Y on X is

⇒ Y - y̅ = r × σ_y(X - x̅)/σ_x

The equation of line of regression X on Y is 

⇒ ⇒ X - x̅ = r × σ_x(Y - y̅ )/σ_y

Here slope of the line are

⇒ r × σ_y/σ_x and r × σ_x/σ_y

If θ is the angle between the two lines of regression

⇒ Tan θ =(r × σy/σ_x - r × σx/σ_x)/(1 + r × σy/σ_x × σx/σ_y)

⇒ (r² - 1)(σ_x σ_y)/(r × σ_x² × σ _y²)

⇒ Tan θ = (1 - r²)(σx σy)/(r × σx2 × σ y2)

⇒ θ = tan^-1{ (1 - r2)(σx σy)/(r × σx2 × σ y2)

Whenever two lines intersect, there are two angles between them, one is acute and another obtuse angle

If tan θ > 0 or o < θ < π/2 i.e θ is acute angle

.If  tan θ < 0 or π/2 < θ < 0 i.e θ is obtuse angle

∴ 0 < r² < 1, the acute angle (θ₁) and Obtuse angle (θ₂) between two lines of regression are given by

θ₁ = tan-1{ (1 - r2)(σx σy)/(r × σx2 × σ y2)

θ₂ = tan-1{ (r² - 1)(σx σy)/(r × σx2 × σ y2)

Explanation

Income of employees in a state is assumed to be normal distributed with

X̅ = 15000

σ² = 900

Calculation

The income of employees follows normally distribution then

Mean = median = mode

∴ Median = 15000

Calculation

Normal distributiln = A normal distribution comes with a perfectly symmetrical shape that means distribution curve can divided in the middle to produce two equal halves.

The normal distribution function is denotted by

$f\left( x \right) = \frac{1}{{\sigma \sqrt {2\pi } }}{e^{ - \frac{1}{2}}}{\left( {\frac{{x - \mu }}{\sigma }} \right)^2}$

f(x) = Probability density function

σ = Standard deviation

μ = Mean

In normal distribution the symmetry about the center and in it the mean = medan = mode = μ

∴ The mean = medan = mode = μ in normal distribution

Explanation

The mode of geometric distribution with parameter p is equal to 1 if 0 < p < 1

p, If 0 < p ≤ 1 then Mode = 0

Given

Mean = M

Median = M_d

Mode = M₀

Calculation

According to Bowley's measure of skewness is based on quartiles and media.

In a symmetrical distribution, the distance between first quartile and median is equal to distance between median and third quartile

⇒ Median - Q₁ = Q₃ - Median

⇒ Median = (Q₃ + Q₁)/2

According to Bowley's measure of skewness 

⇒ Q₃ + Q₁ - 2M_d

The bowley's relative measure of skewness is Bowley's coefficient of skewness(Sk_B)

∴ Sk_B = (Q3 + Q1 - 2M_d)/(Q3 - Q₁)

Given

38, 39, 40, 52, 59, 67, 73, 77, 149, 248

Calculation

The number of observation =  = 10

⇒ Median observation number = (N + 1)/2

⇒ Median = (10 + 1)/2

∴ Number of median observation =  5.5

Median = M_d = [(N/2)TH ITEM + (N/2 + 1) th item]/2

⇒ 5^th observation + ½(6^th observaqtion – 5^th observation)

⇒ 59 + (67 – 59(/2

⇒ 59 + 8/2

⇒ 63

∴ Median of these observation is 63

Explanation

With reference to analysis of variance change of origin or scale will not affect the value of F.

The analysis of variance is a method of splitting the total variation of data into the constituent part which measures different sources of variation. The total variation split up into 2 components 

1 - Variation within the subgroups of samples

2 - Variation between the subgroups of the samples

The total variance is the sum of variation within the subgroups of samples and Variation between the subgroups of the samples.

F – Test = Variance ratio test

F = S²₁/S₂²

S₁² = [1/(N₁ – 1)]∑(X – X̅)²

S₂² = {1/(N₂ – 1)}∑(Y – Y̅)²

N₁ = Size of first sample

N₂ = Size of the second sample

X̅ and Y̅ are sample means of the random variable(X and Y)

The sources from where we get information are called data sources and these may comprise documents, humans, institutions as well as mass media like newspapers, radio, television, etc.

All recorded sources of information irrespective of their contents and forms come under documentary sources which can be categorized based on their content into primary, secondary, and tertiary sources of data.

Primary Sources refer to the sources which:

• contain original/first-hand material or fresh information.
• contain documents of an event that are experienced by the author.
• includes Oral history, Autobiography, Information from correspondents, Collective questionnaires, etc.

Secondary Source: These are sources that are derived from the primary source after some analysis. For example, biographies, dissertation, indexes, Reports of committees and commissions, Newspapers and magazines, etc.

Hence, we conclude that Information from correspondents is a source of primary data.

Given

Mean = x̅ = 25

Median = M_d = 26

Standard devation = σ = 5

Formula

Karl pearson coefficient of skewness = SK_p = (Mean – Mode)/Standard deviation

Calculation

SK_P = (25 – 26)/5

⇒ -1/5 = -0.2

∴ The karl pearson coefficient of skewness is -0.2

Explanation

The product of partial regression coefficients b_12.3b_23.1b_31.2 is equal to r_12.3r_23.1r_31.2

Partial correlation coefficient

Let x₁, x₂, x₃ be three variables

r₁₂ = Simple correlation coefficient between x₁ and x₂

r₂₃ = Simple correlation coefficient between x₂ and x₃

r₃₁ = Simple correlation coefficient between x₃ and x₁

r_12.3 = Partial correlation coefficient between x₁ and x₂ keeping x₃ as constant

r_23.1 = Partial correlation coefficient between x₂ and x₃ keeping x₁ as constant

r_13.2 = Partial correlation coefficient between x₁ and x₃ keeping x₂ as constant

⇒ r_12.3 = (r₁₂ – r₁₃. r₂₃)/[√(1-r²₁₃) × √(1-r²₂₃)]

⇒ r_13.2 = (r₁₃ – r₁₂. r₂₃)/(√1-r²₂₃) × √(1-r²₁₂)]

⇒ r_23.1 = (r₂₃ – r₁₂. R₁₃)/(√1-r²₁₃) × √(1-r²₁₂)]

⇒ b_12.3 = r_12.3 × (σ_1.23/σ_2.13)

⇒ b_13.2 = r_13.2 × (σ_2.31/σ_3.21)

⇒ b_23.1 = r_23.1 × (σ_2.31/σ_3.21)

∴ b_12.3b_13.2 × b_23.1 = r_12.3r_13.2 .r_23.1

The partial correlation is used ina situation where three or more variables are involved. The three variables may be age, height and weight. Correlation between height and weight can be computed by keeping age constant. The partial correlation is used to keep constant the effect of age.

This statistical technique is caleed as partial correlation betweenx and y and it is denottd by r_xy

Given

X_i/f_i I = 1, 2, ----n is a frtequency distribution

Standard deviation = σ = 15

Mean = x̅ = 30

Formula

Coefficient of variation = CV = σ/x̅ ) × 100

Calculation

⇒ CV = (15/30) × 100

∴ The coefficient of variance is 50%

Imp

**Standard deviation = σ = ∑(x – x̅)²]/**n

x = observations

x̅ = mean

n = number of observation

Coefficient of variation = The coefficient of variation shows the extent of variability of data in a sample in relation to the mean of the population.

The coefficient of variation allows investors to determine how much volatility, or risk, is assumed in comparison to the amount of return expected from investmernt.

Explanation

At a reservation counter-passenger are arriving for booking the tickets in poison fashion with mean = 60 passengers/hour

The poison distribution means exponential distribution, then the kurtosis of inter-arrival time of the passengers = 6

Exponential distribution = In probability theory and statistics, the exponential distribution is the probability distribution of the time between events in a Poisson i.e.,

a process in which events occur continuously and independently at a constant average rate 

f(x, λ) = [λe ^- λx    x >= 0

             [0            x < 0

Where, f(x, λ) = probability density function

λ = Rate parameter

x = Random variable

Explanation

The randomised design is based on the principle of replication

In the design of experiments, completely randomized designs are for studying the effects of one primary factor without the need to take another variable

into account.

In a completely randomized design, treatment levels or combinations are assigned to experimental units at random. This is typically done by listing the treatment

levels or treatment combinations and assigning a random number to each. By sorting on the random number, we produce a random order for the application of

the treatments to experimental units.

Given

∑p_oq_o == 160

∑p_oq₁ = 250

∑p₁q_o = 200

∑p₁q₁ = 288

Formula

Fisher’s ideal index = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁] × 100

Here,

P₁ = Price in current year

q₁ = quantity in current year

p_o = price in base year

q₀ = quantity in base year

Calculation

Fisher’s ideal index = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁] × 100

⇒ √[(200/160) × (288/250)] × 100

⇒ (6/5) × 100

∴ Fisher’s index number is 120

The Fisher’s index number obtained as geometric mean  of index obtained by laspeyres’ and paasche’s fpormula

F = √(L × P)

L = Laspeyre’s index = ∑p_nq_o/∑p_oq_o

P = Paasche’s index = ∑p_nq_n/∑p_oq_n

Explanation

The randomized design provides maximum number of degrees of freedom to error sum of squares