SSC CGL Tier-II (JSO ) 2019 Official Paper ( Held On : 17 Nov 2020 ) (Previous Year Paper)

Explanation 

The formula of Marshall and Edgeworth for constructing an index number is called Marshall Edgeworth’s method.\In this method they have suggested to take the arithmetic mean of the quantities of the base year and the current year as the weight of the items.

P₀₁ = {∑[p₁(q₀ + q₁)/2]/∑[p_o(q_o + q₁)/2]} × 100

The Marshall – Edgeworth’s index number is given by

∴ P₀₁(ME) = {(∑p₁q_o + ∑p₁q₁)}/(∑p_oq_o + ∑p_oq₁)} × 100

The Edgeworth – Marshall method gives a result which is approximately close to the results obtained by the Fisher’s ideal formula, and lies between thye result of Laspeyre and Paasche’s indices

∴ The arithematic average of Paasche’s and Laspeyre’s index number is Edgeworth – Marshall index number

A table is structured for organizing and displaying information with data ordered in row and column.

The object of table is to represent tabular information in a more meaningful way. The object of table is to  represent tabular information in row and olumn to understand the meaning of it in a simple and easy method

∴  The tabulation of th data is the systematic arrangement of the statistical data in columns and rows

Given

P(A I B) = 30% = 0.3

P(A) = 80% = 0.8

P(B I A’) = 70% = 0.7

P(A’) = 100% - 80% = 2-0% = 0.2

Formula

P(A I B) = P(A I B). P(A)/P(B) = [P(A I B). P(A)]/[P(A I B). P(A) + P(B I A’) P(A’)]

Calculation

P(A I B) = (0.3 × 0.8)/(0.3 × 0.8 + 0.7 + 0.2)

⇒ 0.24/(0.24 + 0.14)

⇒ 0.24/0.38

∴ The probability that its error was debugged correctly is 12/19

Given

μ'₁, μ₂, μ'₃, μ'₄ are raw moments

μ’₁ = 2

μ’₂ = 136

μ’₃ = 320

μ’₄ = 40,000

Calculation

We have to find the first our central moments

μ₁ = μ’₁ = 2

μ₂ = μ’₂ – (μ’₁)²

⇒ 136 – (2)²

_⇒ 136 – 4

⇒ 132

μ₃ = μ’₃ – 3 × μ’₂μ’₁ + 2 × (μ’₁)³

⇒ 320 – 3 × 136 × 2 + 2 × (2)³

⇒ 320 – 816 + 16

⇒ - 480

μ₄ = μ’₄ – 4 × μ’₁ × μ’₃ + 6 × μ’₂ × (μ’₁)² – 3 × (μ’₁)⁴

⇒ 40000 – 4 × 2 × 320 + 6 × 136 × 2² – 3 × 2⁴

⇒ 40000 – 2560 + 3264 – 48

⇒ 40656

The coefficient of skewness  denoted by β₁

⇒ β₁ = μ²₃/μ³₂

⇒ (-480)²/(132)³

∴ The coefficient of skewness is (-480)²/(132)³

Explanation

The coefficient of regression is given as = β_xIy and β_γIx

We know that the coefficient of correlation r = ± √(b_xy × b_yx) where b_xy and b_yx are coefficient of regression

⇒ Similarly from the above options there is only 1 option that satisfied this equation 

∴ The coefficient of correlation is $\pm\sqrt{{\beta_{X|y}\beta_{Y|x}}}$

Key Points Harmonic Mean: 

• The central tendency of a group of numbers is determined using the Harmonic Mean, a type of average.
• Smaller values in the set are given more weight by the harmonic mean.
• Since it considers the inverse relationship between two variables, the harmonic mean is very helpful when working with rates or ratios.
• Harmonic Mean = n / [1/x₁ + 1/x₂ + 1/x₃ + ... + 1/x_n]

Important Points 

Given:

n = 4

a = 8, b = 6, c = 12 and d = 4

n = number of observation

a, b, c, d are observations

Formula:

Harmonic mean

= $\frac{n}{\frac{1}{a}+\frac{1}{b}+\dots +\frac{1}{n^{th}\ term}}$

Calculation:

Harmonic mean

= $\frac{4}{\frac{1}{8}+\frac{1}{6}+\frac{1}{12}+\frac{1}{4}}$

= $\frac{4}{\frac{3+4+2+6}{24}}$

= (24 × 4)/15 = 32/5

= 6.4

∴ The harmonic mean of these observations is 6.40

 

The harmonic mean is the reciprocal of arithmetic mean

Explanation

Kelly’s skewness denoted by S_k

S_k = P₉₀ + P₁₀ – 2P₅₀

P₉₀, P₁₀, P₅₀ are percentile partition

Calculation

S_k = 63.831 + 58.983 + 2 × 61.345

⇒ 122.814 – 122.69

⇒ 0.124

Kelly’s coefficient of skewness denoted by S_p

S_P = (P₉₀ + P₁₀ – 2P₅₀)/P₉₀ – P₁₀

⇒ 0.124/4.848

∴ Kelly’s coefficient of skewness is 0.026

Formula

SSA + SSB + SSE = TOTAL

Df A + Df B + Df E = Total

SSA = Sum of square of variety A

SSB = Sum of square of variety B

SSE = Sum of square of error \Df = Degree of freedom

Calculation

According to question

⇒ 280 + α + 20 = 640

⇒ α  = 640 - 300

∴ α  = 340

⇒ 2 + 3 + β = 11

⇒ β = 11 - 5

∴ β = 6

∴ The value of α and β are 340, 6

Expalantion

For the 5^th Decile = \% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! \% MathType!MTEF!2!1!+- \% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qadaGfWbqabSWdaeaapeGaaGimaaWdaeaapeGaaGOmaaqdpaqaa8qa \% cqGHRiI8aaacbmGccaWFMbWaaeWaa8aabaWdbiaa-HhaaiaawIcaca \% GLPaaacaWFKbGaa8hEaiabg2da9iaaiwdacaGGVaGaaGymaiaaicda \% aaa!43C1! \mathop \smallint \limits_0^2 f\left( x \right)dx = 5/10\% MathType!End!2!1!

Calculation

\% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX! \% MathType!MTEF!2!1!+- \% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qadaGfWbqabSWdaeaapeGaaGimaaWdaeaapeGaaeiraiaaiwdaa0Wd \% aeaapeGaey4kIipaaOGaaG4maiaac+cacaaI0aGaamiEamaabmaapa \% qaa8qacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiaadsgacaWG \% 4baaaa!448B! \mathop \smallint \limits_0^{{\text{D}}5} 3/4x\left( {x - 2} \right)dx = = ¾(2x²/2 – x³/3)₀^d5

⇒ 5/10 = ¾(2D₅²/2 – D₅³/3)

Putting the values of D₅, Put the value of D₅ in the equation, the value which satisfy the equation will become the answer of this question

Option 1 – (1)

⇒ ¾(2 × (1)²/2_– (1)³/3)

⇒ ¾(1 – 1/3

⇒ ¾(2/3) = ½ or 5/10

∴ Option 1 satisfy the equation hence 5^th decile point of X is 1

Given

F(x) = e^-x

Calculation

 For $Q_3=\displaystyle\int^{Q_3}_0e^{-x}dx=\dfrac{75}{100}$

⇒ ¾ = (-e^-x)^Q3₀

⇒ ¾ = (-1 – e^-Q3)

⇒ ¼ = e^-Q3

⇒Q₃ = In 4

For $Q_1=\displaystyle\int^{Q_1}_0e^{-x}dx=\dfrac{25}{100}$

⇒ (-e^-xdx)^Q1₀ = 1/4

⇒ (- 1 – e^-Q1) = ¼

⇒ Log Q₁ = ¾

Q₁ = In 4/3

⇒ Interquartile range = Q₃ – Q₁

⇒ In 4 – In 4/3

⇒ In  4 –(In 4 – In 3)

∴ The interquartile range is In 3

The Probability density function = $\displaystyle\int_0^\infty f(x)\ dx=1$

Explanation

If the event B occurs does not change the probability that event A occurs, and event A and B are the independent event then

⇒ P(A I B) = P(A)

According to bayes’s theorem It states the relation between the probabilities of A and B, P(A) and P(B) and the conditional probabilities of A gives B and B gives A.

⇒ P(A I B) and P(B I A)

⇒ P(A I B) = P(B I A).P(A)/P(B)

⇒ P(B I A) = P(A ∩ B)/P(A)

⇒ P(A ∩ B) = P(A).P(B)

⇒ P(A I B) = P(A ∩ B).P(A)/P(A). P(B)

⇒ P(A I B) = P(A). P(B).P(A)/P(A).P(B)

⇒ P(A I B) = P(A)

∴ The value of P(A I B) is P(A)

Explanation

The relative frequency theory of probability holds that if an experiment is repeated an extremely large number of times and a particular outcome occurs a percentage

of the time, then that particular percentage is close to the probability of that outcome.

Let A be an event of interest and assume that we have performed the same experiment n times so that n is the number of times A could have occurred

∴ The experimenta is repeated an extremely large number of times of probability and this approach is known as Relative frequency approach

 

Classical approach = : based on equally likely events. If an experiment has n possible outcomes, this method would assign a probability of 1/n to each outcome.

It is necessary to determine the number of possible outcomes

Subjective approach = “In the subjective approach we define probability as the degree of belief that we hold in the occurrence of an event”

Objective probability refers to the chances or the odds that an event will occur based on the analysis of concrete measures rather than hunches or guesswork. ...

An independent event is an event whose outcome is not influenced by prior events

Explanation

The events are called pairwise independent if any two events in the collection are independent of each other, while saying that the event are mutually independent

(or collectively  independent) it means that each event is independent of any combination of other events in the collection.

∴ Events, A₁, A₂, and A₃ are pairwise independent events

Explanation

The probability distribution function of Normal distribution us was given as

⇒ F(x) =[1/σ√2π ]e^{-(x – μ)2/2σ2}

If we put Z = (x – μ)/σ in Normal distribution then it is called a standard normal distribution

The probability distribution function of the standard normal distribution is as follows

⇒ f(x) = [1/σ√2π]e^{- z^2/2}

The value of E(X³) in a standard normal distribution is 0

 

The standard deviation of normal distribution  = σ = 1

The coefficient of skewness of Normal distribution is β₁ = 0

The coefficient of kurtosis in normal distribution is β₃ = 3

Given

α = level of significant = 5% = 0.05

R = Row = 3

C = Column = Block = 4

Calculation

n₁ = R - 1

⇒ n₁ = 3 - 1 

⇒ 2

n₂ = (R - 1)(C - 1)

⇒ (3 - 1)(4 - 1)

⇒ n₂ = 2 × 3

⇒ n₂ = 6

F ≤ F_{n1,n2, 1 - α} 

⇒ F≤ F(_{R - 1),(R - 1)(C - 1),1 - α}

⇒ F≤ F_2,6,0.95

∴ the critical region to test the equality of mean purification of three methods at a 5% level of significance by randomised block design is F≤ F2,6,0.95

Formula 

In Binomial distribution f(x) = ⁿc_xp^xq^{n - x}

q = 1 - p

p = Probability of success of event

q = probability of failure of event

Calculation

n = 6

x = 5

⇒ The power of the test = ⁶c₅(0.6)⁵(0.4)^{6 - 5} + ⁶c₆(0.6)⁶(0.4)^{6 - 6}

⇒ (6!/5!(6 - 5)!⁵(0.4)¹ + ⁶c₆(0.6)⁶(0.4)⁰

⇒ 6 × (0.6)⁵ × 0.4 + (0.6)⁶ × 1

⇒ (0.6)⁵[6 × 0.4 + 0.6]

⇒ (0.6)⁵[2.4 + 0.6]

∴ The power of test is 0.233

Explanation

The exponential distribution is often concerned with the amount of time until some specific event occurs. It is the probability distribution of the time between events in

a poisson point process. The exponential distribution are commonly used in calculations of product reliability or the length of time a product tasts.

                                                                                   

Geometric distribution = The geometric distribution represents the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:

f(x) = (1 − p)^x − 1p

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.

Gamma distribution = if n > 0, the integral ∫ 0 to ∞ is called gamma function and t is denoted by f(n̅)

Given

Formula

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

P₁ = current year

P₀ = base year

q₀ =  base quantity

Calculation

Laspeyres price index for year 2020

⇒ (20.5/17.5) × 100

∴ Laspeyres price index for year 2020 is 117.14

Explanation

The mean is the most popular and well-known measure of central tendency. It can be used with both discrete and continuous data. Although its use is most often 

with continuous data. The mean is equal to the sum of all the values in the data set divided b the number of values in the data set

Mean = ∑x_i/N

N = number of observation

X = observation

Median = Median locates a central part which divides a distribution into two equal halves

Median = (N + 1/2) th item, N = number of observation

Mode = Mode refers that value in a distribution which occur most frequently 

Mode = M_o = L + [(F₁ - F_O)/(2F₁ - F_O -  F₂)] \TIMES C

C = Size of the interval

F₁ = Frequency of the modal class

F₂ = Frequency of the class succeeding the modal class

F₀ = Frequency of the class preceding the modal class

L = Lower limit of the modal class

Confusion Points

• Range is a measure of dispersion.
• Central tendency has three measures - Mean, Median and Mode.

Given

μ₄ = 199

μ₃ = 50

μ₂ = 8

Formula

Kurtosis a₄= μ₄/(μ₂)²

Excess kurtosis = a₄ – 3

Calculation

a₄ = 199/(8)²

⇒ 199/64

⇒ 3.109

⇒ Excess kurtosis is denoted by g₂

⇒ g₂ = a₄ – 3

⇒ 3.109 – 3

∴ Excess kurtosis is 0.109

Concept used

H_o: p < p_o

Z = (ṗ - p_o)/[√p_o(1 – p_o)/n]

Given

P = 0.136

P_o = 0.211

n = 3,536

Calculation

Z = (0.136 – 0.211)/√[0.211(1 – 0.211)/3536]

⇒ (- 0.075)/√(0.166479/3536)

⇒ (0.075)/√0.0000047

⇒ = -10.93

∴ The value of appropriate test statistic is – 10.93

Explanation

Bowley’s coefficient of skewness is based on the quartile. The formula for calculating coefficient of skewness is given by

⇒ S_k = [(Q₃ – Q₂) – (Q₂ – Q₁)]/(Q₃ – Q₁)

⇒ (Q₃ – 2Q­₂ + Q₁)/(Q₃ – Q_­1)

The value of S_k would be zero, if it is a symmetrical distribution

If the value is less than zero, then it is negatively skewed distribution

It will take value between + 1 and - 1

Explanation

Examples of relative measure of dispersion are

⇒ Coefficient of standard deviation

⇒ Variation

⇒ Quartile deviation

∴ variance is not an example of relative measure of dispersion

 

Standard deviation = σ = ∑(x_i – x̅)/n

Variance = σ² = average squared devation of value of mean

Variance is calculate after squaring in the standard deviation so it does not have the same unit.

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.

Key Points

• Secondary data refers to data that is collected by someone other than the primary user.
• It means that the information is already available, and someone analyses it.
• The secondary sources include magazines, newspapers, books, journals, censuses information collected by the government dept, organizational records, and data that was originally collected for other research purposes.
• For example, Suppose we want to know the number of persons in different income groups in a particular village. For this, the figures will be available with the Municipality / Panchayat Office. This is an example of data collected from secondary sources.

Hence, the data taken from the published report on dropouts from school will be considered as secondary data.

Hint

• Primary Sources refer to the sources which contain original/first-hand material or fresh information.

Given

μ₁ = 60.125

μ₂ = 61.345

μ₃ = 62.688

Calculation

Skewness coefficient is denoted by β₁

⇒ β₁ = (μ₃)²/(μ₂)²

⇒ (62.688)²/(61.345)³ 

⇒β₁ = 0.0170

Absolute skewness measurement = γ₁

⇒ γ₁ = √(β₁)

⇒ √(0.0170)

∴ Absolute skewness measurement is 0.123

Given

F_xy(x, y) = 10x²y, 0 ≤ y ≤ x ≤ 1

Concept used

Marginal probability is the probability of an event irrespective of the outcome of another variable.

Calculation

Marginal density f_y(y) = $\% MathType!MTEF!2!1!+- \% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qacqGHRiI8ieWacaWFMbWaaeWaa8aabaWdbiaa-HhacaGGSaGaa8hO \% aiaa-LhaaiaawIcacaGLPaaacaWFKbGaa8hEaaaa!401C! \smallint f\left( {x,;y} \right)dx$

⇒ \% MathType!MTEF!2!1!+- \% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qadaqfWaqabSWdaeaaieWapeGaa8hEaiabg2da9iaa-Lhaa8aabaWd \% biaaigdaa0WdaeaapeGaey4kIipaaaaa!3C61! \mathop \smallint \nolimits_{x = y}^1 (10x²ydx

After integration we get

⇒ 10y(x³/3)¹_{x = y}

⇒ (10y/3)(1 - y³)

∴ The marginal density fγ (y) is (10y/3)(1 - y³).

Given

Mean = 20

Mode  = 25

Formula

3 × Median = Mode + 2 × Mean

Calculation

Median = (Mode + 2 Mean)/3

⇒ (25 + 2 × 20)/3

⇒ (25 + 40)/3

⇒ 65/3

∴ Median = 21 2/3

Explanation

If A and B are two event and A̅  and B̅ are their complementary events

AB̅ and AB are two mutually exclusive and exhaustive events then

∴ P(A) = P(AB) + P(AB̅)

Explanation

Heartstroke is a condition caused by your body overheating, usually as a result of prolonged exposure to or physical exertion in high temperature.

This most serious form of heat injury, heartstroke can occur if your body temperature rises to 104° F or higher. 

∴ This condition is most common in the summer months. It is an example of seasonal variation

Concept used

H_o: p < p_o

Z = (ṗ - p_o)/[√p_o(1 – p_o)/n]

Given

P = 0.27

P_o = 0.24

n = 400

Calculation

Z = (ṗ - po)/[√po(1 – po)/n]

⇒ (0.27 - 0.24)/√[(0.24)(0.76)/400]

∴ Now required probability, P(Z > (0.27 - 0.24)/√[(0.24)(0.76)/400]

Explanation

For factor reversal test

P₀₁ × Q₀₁ = ∑P₁Q₁/∑P₀Q₀

⇒ P₀₁  = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁]

⇒ Q₀₁ = √[(p₀q₁/∑p₀q₀) × (∑p₁q₁/∑p₁q₀)

⇒ P₀₁ × Q₀₁ = √[(∑(p₁q_o/∑p_oq_o) × (∑p₁q₁/∑p_oq₁) × (p₀q₁/∑p₀q₀) × (∑p₁q₁/∑p₁q₀)]

⇒ p₀₁ × Q₀₁ = ∑ p₁q₁∑p₀q₀

⇒ Factor reversal test is also satisfied

∴ Fisher’s ideal index number satisfied both time reversal and factor reversal test

Explanation

The correlation Coefficient is a statistical measure of the strength of the relationship between the relative movements of two variables. The value range between -1 to 1

A calculated number greater than 1 or less than -1 means thta there was an error in the correlation measurement

Correlation 1 means a perfect positive correlation

Correlation -1 means a perfect negative correlation

0 correlation shows no linear relationship between the movement of the two variables

∴ The value of simple correlation coefficient in the interval of [-1, 1]

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

F(x) = 6/5(x² + x); 0 ≤ x ≤ 1

Calculation

F(x) = \mathop \smallint \nolimits_0^x f\left( x \right)dx

⇒ \% MathType!MTEF!2!1!+- \% feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn \% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr \% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 \% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x \% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 \% qadaqfWaqabSWdaeaapeGaaGimaaWdaeaaieWapeGaa8hEaaqdpaqa \% a8qacqGHRiI8aaaaaa!3A60! \mathop \smallint \nolimits_0^x 6/5(x² + x)dx

After integration we get,

⇒ 6/5(x³/3 + x²/2)

⇒ f(0.5) = (6/5)((0.5)³/3 - (0.5)²/2)

⇒ (6/5)(0.25/6 + 0.75/6)

⇒ (6/5)(1/6)

⇒ 1/5

∴ The value of F(0. 5) is:1/5

Given

n = 6

p = 1/4

Formula

S_k = (1 - 2p)/√(np(1 - p)

Calculation

Skewness of X = (1 - 2 × 1/4)/√(6 × 1/4(1 - 1/4)

⇒ (1 - 1/2(/√(3/2 × 3/4

⇒ 1/2/√(9/8)

∴ The value of skewness X is√2/3

 

Binomial distribution = A random variable X has a binomial distribution if its probability distribution is given by

b(x; n, p) = P(X = x) = (n x)p^xq^{n - x} x = 0, 1, 2, ----n

Normal distribution has the probability density function given by

f(x) = [1/(σ√2π)]e^{1/2(x - μ/σ)2}

Here, μ and σ are prameters of this distribution. The parameter μ is mean and σ is square root of variance of this distribution

-∞ < μ < ∞, σ > 0

The normal distribution is also called as the Gaussian distribution

Explanation

The second degree parabola = It describes the trend( non linear) in a time series where a change in the amount of change is constant per unit time.

The quadratic (parabolic)trend can be describe by equation

⇒ y = a + bx + cx²

The method of least square gives the normal equation as 

⇒ ∑y = na + b∑x + c∑x²

⇒ ∑xy = a∑x + b∑x² + c∑x³

⇒ ∑x²y = a∑x² + b∑x³ + c∑x⁴

However if ∑x = 0 then, the normal equation reduces to

⇒ ∑y = an + c∑x²

⇒ ∑xy = b∑x²

⇒ ∑x²y = a∑x² + c∑x⁴

The values of a, b, and c can be found as

⇒ c = (n∑x²y - ∑x²)(∑y)/[n∑x² - (∑x²)²]

⇒ a = (∑y - c∑x²)/n

⇒ b = ∑xy/∑x²

Explanation

Ballantine venn diagram = Regression model can be visually reprersented with Ballantine Venn diagram. In these diagram, each variable in the regression model is repesented by a different circle.

∴  Mutual and unique variances among multiple factors can be embodied in a diagram that comprises overlapping circles. The diagram is known as BALLANTINE DIAGRAM

 

 

Homogeneity diagram = A graph is connected-homogeneous if every isomorphism between two connected induced subgraphs can be extended to an automorphism of the whole graph

Path diagram = In statistics, path analysis is used to describe the directed dependencies among a set of variables. This includes models equivalent to any form of multiple regression analysis, factor analysis, canonical correlation analysis, discriminant analysis, as well as more general families of models in the multivariate analysis of variance and covariance analyses.

3-way scatterplot = Unlike a classic XY scatter chart, a 3D scatter plot displays data points on three axes (x, y, and z) in order to show the relationship between three variables. Therefore, it is often called an XYZ plot. Regrettably, there is no way to create a 3D scatter plot in Excel, even in the new version of Excel 2019

Given

Formula

Coefficient of Range = S = (x_m - x_o)/(x_m + x_o)

Where, x_m = Maximum value

x_o = Minimum value

Calculation

Coefficient of range = (22 - 10)/(22 + 10)

⇒ Coefficient of range = 12/32

∴ Coefficient of range is 0.375

Calculation

The coefficient of rank correlation = 1 – 6∑d²/n(n² – 1)

D = r₁ – r₂

r₁, r₂ ------ are ranks

According to spearman’s rank correlation coefficient ρ = 1 – 6∑d²/n(n² – 1)

⇒ 1 – 6 × 214/10(100 – 1)

⇒ 1 – 6 × 214/990

⇒ 1 – 1.297

∴ The value of rank correlation is approx. – 0.3

Explanation

Null hypothesis = The null hypothesis is a typical statistical theory that suggests that no statistical relationship and significance exists in a set of given single observed variables, between two sets of observed data and measured phenomena.

∴ In this, the correct option for the null hypothesis is μ1 = μ2 = μ3

Explanation

The word commensurable means having a commonsurable that means test of commensurability is that probability after changing the unit of measurement of the commodity the index number does not change.

∴The property that ‘on changing the unit of measurement of a commodity the index number does not change’ is called the test of commensurability.

Test of equity = Test if two population means are equal. The two-sample t-test is used to determine if two population means are equal. A common application

is to test if a new process or treatment is superior to a current process or treatment.

Test of homogeneity = In the test of homogeneity, we select random samples from each subgroup or population separately and collect data on a single categorical variable. The null hypothesis says that the distribution of the categorical variable is the same for each subgroup or population. Both tests use the same chi-square test statistic.

The difference is a matter of design. In the test of independence, observational units are collected at random from a population and two categorical variables are observed for each unit. In the test of homogeneity, the data are collected by randomly sampling from each sub-group separately

test of proportionality = Let us consider the parameter p of population proportion. For instance, we might want to know the proportion of males within a total population of adults when we conduct a survey. A test of proportion will assess whether or not a sample from a population represents the true proportion from the entire population

Explannation

For the positive skewed distribution the frequency curve has longer tail towards the right

Karl pearson’s measure of skewness is based upon the divergence of mean from mode in a skewed distribution

⇒ Sk = (Mean – Mode)/Standard deviation

The sign Sk gives the direction and its magnitude gives the extent of skewness

If Sk > 0 the distribution is positive skewed

If Sk < 0 the distribution is negative skewed

⇒ To obtained the  S_k < 0 we have one condition

∴ Mean will be less than the mode

Relation of mean, mode, and median for positive skewness is Mean > Median > Mode

Relation of mean, mode, and median for negative skewness is Mode > Median > Mean

Explanation

Primary data collection is the process of gathering data through surveys, interviews, or experiments. A typical example of primary data is a household survey. In this form of data collection, researchers can personally ensure that primary data meets the standards of quality, availability, statistical power, and sampling required for a particular research question.

∴ Primary data collection requires less amount of resources such as time, cost and manpower.

Secondary data collection = Secondary data refers to data that is collected by someone other than the primary user. Common sources of secondary data for social science include censuses, information collected by government departments, organizational records and data that was originally collected for other research purposes

Explanation

Cross - sectional data are the result of a data collections, carried out at a single point in time on a statistical unit. With cross - sectional data, we are not interested in the change of data over time, but in the current, valid opinion of the respondents about a question in a survey.

∴ The share price of 5 companies on a particular day is a cross - sectional data.

Given

Total employ in the office = 25

Mean salary of the employees = Rs 30,000/ month

An senior officer retired whose salary = 60,000/month

After adding new clerk the mean salary of the employees reduced by 1,000 per month.

Formula

Mean = Sum of the all observations/number of observations

Calculation

The mean salary of 25 employees = 30,000

⇒ The sum of the salary of all employees = 30000 × 25

⇒ 750,000

After retiring the senior officer of monthly salary 60,000

⇒ New sum of salary = 750,000 – 60,000

⇒ 690,000

After adding new member the average salary salary was reduced by 1000

⇒ New average salary = 29,000

⇒ New sum of salary of 25 employees = 29000 × 25

⇒ 725,000

The salary of new employees per month = 725,000 – 690,000

⇒ The salary of new employees be Rs 35,000 per month

Formula

For the exponential distribution r^th raw moment is given by

μ_r­’ = d^r/dt^r(λ/(λ – t)

Calculation

For the 4^th raw moment μ_r’ is given by

⇒ (d⁴/dt⁴⁾⁽λ/(λ – t)

After differentiate it 4 times we get the value

⇒ 24λ /(λ  - t)⁵

When t = 0

∴ The value of μ_r’ = 24/λ⁴

Explanation

Additive model assumes  that the observed value is obtained by adding 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

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

According to  Spearman's rank correlation coefficient when ranks are repeated is as follows - 

⇒ r_s = 1 - 6[(∑d² + (m₁³ - m₁)/12 + (m₂³ - m₂)/12 + -------)]/n(n² - 1)

m₁, m₂ ----- are the number of repetitions of the ranks

(m₁³ - m₁)/12 ----- are corresponding factors

∴ The factor is added for each repeating value in the both series is given by (m³ - m)/12 = m(m² - 1)/12 = m(m - 1)(m + 1)/12

Originally the spearsman rank correlation is denotted by rs = 1-6∑d2/n(n² - 1)

Explanation:

 

We know that for Fpr random variable X sum of Probaboility = 1

⇒ ∑P_i = 1

⇒ k + 2k + 3k + 5k + 5k + 4k + 3k + 2k + k = 1

⇒ 26k = 1

⇒ k = 1/26

∴ xf(x) = 103k = 103 × 1/26

∴ E(x) is 103/26

Explanation

If Z~ Φ (0, 1) it means Z - standard normal distribution with mean (μ) = 0 and standard deviation (σ) = 1 

The Standard Normal Distribution:

• The standard normal distribution is centered at zero and the degree to which a given measurement deviates from the mean is given by standard deviation.

Chi-square Distribution:

The sum of Standard Normal Distribution is the chi-square distribution with degrees of freedom equal to the number of standard normal variate are used in the summation.

If Suppose If X follows Normal Distribution, then

X∼N(μ,σ2)

It can be converted into standard normal distribution using the following transformation 

X - μ ∼ N(0, σ²)

$\frac{X-\mu}{\sigma^2}\sim(0, 1)$

If we square the standard normal we get the Chi-square Distribution with 1 degree of freedom.

and $\displaystyle\sum^n_{i = 1}{\left(\frac{X-\mu}{\sigma^2}\right)^2} \sim\chi^2_n$ follows the chi-square distribution at n degrees of freedom.

Explanation

Let each observation be y_xi

Where x indicates the category that observation is in and

i is the label of the particular observation.

n_x = The number of observation in category x

⇒ y̅_x= ∑iy_xi/n_x

⇒ y̅ = ∑_xn_xy̅_x/∑_xn_x

y̅_x = Mean of category x

y̅ = mean of the whole population

The correlation ratio E² = ∑_xn_x(y̅_x – y̅)²/∑_xi(y_xi – y̅)²

⇒ E² = σ²_y̅ /σ²_y

Where as, σ²_y̅  = ∑_xn_x(y̅_x – y̅)²/∑_xn_x

σ²_y = ∑_x,i(y_xi – y̅)²/n

The weighted variance of the category means divided by the variance of all samples.It is the relation between values of x and values of y̅_x is linear this will give the same result as the square of pearson’s correlation coefficient; otherwise the correlation ratio will be larger in magnitude. It can therefore be used for judging non linear relationship

∴ E takes the values between 0 and 1

Important Points

The E = 0  represents the special case of no dispersion among the mean of the different categories

E = 1 refers to no dispersion within the respective categories

E² = 0 to 1, we have to learn this result

The value of correlation coefficient = - 1 to 1

Given

μ₁ = 0

μ₃ = 0.7

μ₄ = 18.75

β₂ = 3 ( distribution is mesokurtic)

Formula

β₂ = (μ₄)/(μ₂)²

μ₂ and μ₄ are moments about the mean

β₂ = Coefficient of kurtosis

Calculation

According to formula

⇒ 3 = 18.75/(μ₂)²

⇒ μ₂² = 18.75/3

⇒ 6.25

∴ The value of μ₂ = 2.5

Explanation

In an analysis of variance, all the sum of squares are independent of the change of origin

One way analysis of variance or one way classification table is as follows

SST = Total sum of the square

SST = Sum of the square of the column

SSE = Sum of the square of the Error

MSC = Variance between samples

MSE = Variance within the samples

Explanation

The mean deviation from the median calculates the arithmetic average of the deviations of the observed values from median of the observed values.

The mean deviation is minimum when the deviation is taken from median as the median is the middlemost value of the series and disperse the whole series into minimal value

∴ The mean deviation from the average A is minimum if A represents Median

 

Mean deviation = 1/n(∑x - M)

M = Median or mean 

x = values of the observations

Coefficient of mean deviation = Mean deviation/(Mean or median or mode)

The mean deviation about median is minimum

The absolute mean deviation about the mean is independent of change of origin.

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.In Histogram we can represent time series.

Straight-line = If an equation can be rearranged into the form y = m x + c , then its graph will be a straight line

Historigram = A historigram is a graphical representation of a time series that reveals the changes that occurred at different time periods

∴ Historigram is the correct option

Explanation

The approx. percentage is 95% of the data fall within two standard deviation of the mean

Now the approx. 99,7% of the data  fall within three standard deviation of the mean

Given

In Poisson distribution

P(X = 0) = 0.6

Formula

Poisson distribution is given by

f(x) = e^-λλ^x/x!

Calculation

P(X = 0) = e^-λλ⁰/0!

⇒ 0.6 = e^-λ

⇒ 1/e^λ = 6/10 = 3/5

⇒ e^λ = 5/3

Taking log on both side

⇒ log_ee^λ = log_e(5/3)

∴ λ = Log_e(5/3)

Given

∑x = 0

∑x² = 28

∑xy = 56

Formula

a = ∑y/n

b = ∑xy/∑x²

Explanation

Y = a + bx

⇒ a = 616/7

⇒ a = 88

⇒ b = 56/28

⇒ b = 2

Y = 88 + 2x

⇒ X = 2008 – 2001 = 7

Y = 88 + 2 × 7

∴ estimate of the production for 2008 is 102

Given

F(x) = 4x³

Formula

For interquartile range = Q₃ – Q₁

Q₃ = 75/100

Q₁ = 25/100

Calculation

**\mathop \smallint \nolimits_o^{Q3} **4x³dx = 75/100

⇒ ¾ = 4 (X⁴/4)^Q3₀

⇒ ¾ = (Q⁴₃ – 0))

⇒ ¾ = Q⁴₃

∴ Q₃ = (3/4)^1/4

For Quartile (Q₁),

⇒ 25/100 = \mathop \smallint \nolimits_o^{Q1} 4x³dx

⇒ ¼ = 4(x⁴/4)^Q1₀

⇒ ¼ = (Q₁⁴ – 0)

⇒ ¼ = Q₁⁴

∴ Q₁ = (1/4)^1/4

∴ Interquartile range Q₃ – Q₁ IS (3/4)^1/4  - (1/4)^1/4

Given

Variance = 2

Mode = 24

Arithmetic mean = 25

Formula

Var(X) = (1/n)∑(x_i - x̅)²

Calculation

Var(X) =  (1/n)∑(xi - x̅)²

⇒ 2 = (1/n)∑(x_i - 25)²

⇒ 2n = ∑x_i² + 625n - 50∑x_i

⇒ 2n = ∑x_i² + 625n - 50(∑x_i/n) × n

⇒ 2n = ∑x_i² + 625n - 50n × 25       {∑x_i/n = mean = 25}

⇒ 2n = ∑x_i² - 625n

⇒ ∑x_i² = 627n

Mean square deviation = MSD = (1/n)∑(x_i - Z)²  here Z = Mode

⇒ MSD = (1/n)∑(x_i - 24)²

⇒ (1/n)∑(x_i² + 576n - 48x_i)

⇒ (1/n)∑x_i + 576n - 48 (∑x_i/n) × n)

⇒ (1/n)(627n + 576 - 48 × 25)

⇒ (1/n)( 627n - 624n)

⇒ (1/n)(3n)

∴ The value of mean square deviation from mode is 3

Calculaton

For the positive skewed or right skkewed distribution the frequency curve has longer tail towards the right

Karl pearson’s measure of skewness is based upon the divergence of mean from mode in a skewed distribution

⇒ Sk = (Mean – Mode)/Standard deviation

The sign Sk gives the direction and its magnitude gives the extent of skewness

If Sk > 0 the distribution is positive skewed or right skewed

If Sk < 0 the distribution is negative skewed

⇒ To obtained the  Sk > 0 we have one condition

∴  Mean > Median > Mode

Given

Two series of data are given

N₁  = 5 and N₂ = 54

x̅₁ = 63 and x̅₂ = 4

σ₁ = 9 and σ₂ = 6

X̅ = (N₁x̅₁ + N₂x̅ ₂)/(N₁ + N₂)

X̅ = Combined mean of both group

x̅₁ and x̅₂ are mean of two series

N₁ and N₂ are group of male and female

σ₁ and σ₂ are the standard deviation

Formula

Variance of (X₁ + X₂) = [N₁(σ²₁ + D²₁) + N₂(σ²₁ + D²₂)]/(N₁ + N₂)

D₁ = x̅₁ – X̅

D₂ = x̅₂ – X̅

Calculation

Combined mean = (N₁x̅₁ + N₂x̅ ₂)/(N₁ + N₂)

⇒ X̅ = (5 × 63 + 4 × 54)/(5 + 4)

⇒ X̅ = 531/9

∴ Combined mean = 59

⇒ D₁ = 63 – 59 = 4

⇒ D₂ = 54 – 59 = -5

⇒ Var(male and female combined) = [5(9² + 4²) + 4(6² + (-5)²]/(5 + 4)

⇒ 729/9

∴ The combined variance is 81

 

We define variance as the square of the standard deviation  and it is denoted by σ²

Basically the mathematical formula of Combined variance is given by

Combined Variance = [N₁(σ²₁ + D²₁) + N₂(σ²₁ + D²₂)]/(N₁ + N₂)

N₁ and N₂ are No of observations in series 1 and 2

σ₁² and σ₂² are variance of  first series and second series

d₁ and d₂ are combined mean  - mean of first and second series