SSC CGL Tier-II (JSO ) 2016 Official Paper ( Held On : 2 Dec 2016 ) (Previous Year Paper)

Explanation

In the determination of median graphically we adopt any of the two methods. In both the methods we draw ogives as per requirement.

To calculate mode graphically a histogram of the given data is drawn at first.

∴ For determination of mode and median graphically, one considers Histogram and Ogive

Bar diagram = A bar chart or bar graph is a chart or graph that presents categorical data with rectangular bars with heights or lengths proportional to the values that they represent.

We can represent Mean on Bar graph

Ogive = An ogive (oh-jive), sometimes called a cumulative frequency polygon, is a type of frequency polygon that shows cumulative frequencies. In other words, the cumulative percents are added on the graph from left to right.

We represent partition value on ogive curve like median

Line graph = line graph is a type of chart used to show information that changes over time. We plot line graphs using several points connected by straight lines. We also call it a line chart

Given

f(x) = { ke^-3x, x > o

          { 0, elsewhere

Concept used

$\smallint \limits_{ - \infty }^\infty f\left( x \right)dx$ = $\smallint \limits_{ - \infty }^0 f\left( x \right)dx$ + $\smallint \limits_0^\infty f\left( x \right)dx$ = 1

Calculation

According to given part

⇒$\smallint \limits_{ - \infty }^0 f\left( x \right)dx$ = 0

⇒ 0 + $\smallint \limits_0^\infty f\left( x \right)dx$ = 0

⇒ ∫ke^-3xdx = 1

⇒ k[-e^-3x/3]

⇒ -k/3[e^-∞- e⁰] = 1

⇒ -k/3(0 – 1) = 1

⇒ k./3 = 1

∴ The value of k = 3 for PDF

Explanation

In the constructing of cost of living index, the commodities are selected by judgement/purposive sampling. In this method of sampling the choice of the sample items depends exclusively on the judgement of the investigator

Simple random sampling = It is a sampling technique where every item in the population has an even chance and likelihood of being selected in the samplr

In it selection of items completely depends on chance or by probability  and each item has equal chance to be selected

It is also known as the method of chances

Stratified random sampling = With this sampling, the researcher divides the population into separate groups called strata. Then, a probability sample (often a simple random sample) is drawn from each group.

For example, Dividing the entire state into 3 - 4 geographical regions and then selecting a required sample of districts from each of these regions**.**

Systematic sampling = In this sample members from a large population are selected according to a random starting point but with a fixed, periodic interval and this interval is called sampling interval(k)

The sampling interval = Population size/sample size = N/n

Explanation

If x and y are uncorrelated variables then there can not be any line of regression between them drawn and therefore there is no line of regression relationship between them

∴ Option 1 is correct

Important Points

The line of regression = The line  y = a + bx  which is fitted  to a set of n points (x_i, y_i) by the method of least square called line of regression of y on x. similarly if regression x = a + by is fitted to (x_i, y_i) is called line of regression of x on y

The coefficient of correlation is the GM between the regression coefficient

Explanation

I understand that when sampling from a finite population and our sample size is more than 5% of the population, we need to make a correction on the sample's mean and standard error using this formula:

⇒ FPC = N - n/(N - 1)

Where N is the population size and n is the sample size.

∴ In simple random sampling of n units from a population of N units the quantity [1 - (n / N)] is called the finite population correction factor

Formula

Coefficient of range = (L - S)/(L + S)

L = Largest number

S = Smallest number

Calculation

Coefficient of range is denoted by R

⇒ Largest number of series = 84

⇒ Smallest number of series = 15

⇒ R = (84 – 15)/(84 + 15)

⇒ R = 69/99

∴ Coefficient of range is 0.696

Important Points

The rangeis the simplest measure of dispersion. We can define the range as the difference between the highest and the lowest value in a set of observation

For example the range of the series 1, 2, 4, 6, 8, 9, 10, 12

Largest number = 12

Lowest number = 1

Range = L - S = 12 - 1 = 11

Coefficient of range = (L - S)/(L + S)

Coefficient of this range = 11/13

Explanation

When n units are selected with SRSWR, the total number of possible samples are Nⁿ

Important Points

Simple random sampling = It is a sampling technique where every item in the population has an even chance and likelihood of being selected in the samplr

In it selection of items completely depends on chance or by probability  and each item has an equal chance to be selected

It is also known as method of chances

The simple random sampling with replacement = It is called SRSWR, in it a unit is selected from the sampling frame. The unit is replaced back and the next unit is selected. This process is repeated till a sample of the desired size is selected.

When n units are selected with SRSWR, the total number of possible samples are Nⁿ

The probability of drawing a sample is 1/Nⁿ.

Given

∑ p₁q₀ = 231,

∑p_oq_o = 228,

∑p₁q₁ = 295,

∑p_oq₁ = 283,

Formula

P**₀₁ =** [∑p₁q₁/∑p_oq₁]

P₁ = Price in current year

q₁ = quantity in current year

p_o = price in base year

q₀ = quantity in base year

Formula

According to the formula

P₀₁ = (295/283) × 100

⇒ 104.24

∴ The value of paasche’s index is 104.24

The general formula for Paasche's index number is

P = [∑(p_nq_n/∑p_oq_n] × 100

Where

q_n= are quantities bought or sold in the current year

Given

Population N = 10

Sample n = 2

Concept

The probability of selecting sample in SRSWOR is 1/^Nc_n

^NC_n = N!/n!(N - n)!

Calculation

The required probability is 1/¹⁰c₂

⇒ 1/(10!/(2!(10 -2)!

⇒ 10!/2!/8!

⇒ (8! × 2!)/10!

⇒ 1/45

∴ the probability of selecting a specified sample of 2 units is 1/45

 When we select as SRSWR the probability of selecting sample is 1/Nⁿ

Explanation

The probability of selecting the specified member from a population size of N will be 1/N

∴ The probability of not selecting the specified member = 1- 1/N

⇒ (N – 1)N

∴ The probability of not including the specified member in the choosen sample = (1 – 1/N)ⁿ

∴ The probability of always including the specified member in the specified sample is 1 – (1 – 1/N)ⁿ

The simple random sampling with replacement = It is called SRSWR, in it a unit is selected from the sampling framE. The unit is replaced back and the next unit is selected. This process is repeated till a sample of the desired size is selected.

When n units are selected with SRSWR, the total number of possible samples are Nⁿ

Given

 person would earn Rs. 10,000 in base period.

the cost of living index increases by 20% over base period.

Calculation

The cost of living index increases by 20%

For maintaining the same living standard the employee should get a monthly income of

⇒ 10000 × 20/100

∴ Rs.2000

Then the employers of that person increase his salary by Rs. 1500

∴ 2000 – 1500 = Rs.500

∴ He should claim Rs. 500/- as Dearness Allowance

The consumer price index = (cost of goods and service in given year)/(cost of goods and services in base year) × 100

Given

The correlation coefficient between two variables X and Y = 0.4

Concept used

The correlation coefficient (r) is independent of origin and scale and depend on the sign of variables

Calculation

The correlation coefficient between the two variables is the measure of the slope between the variables in the regression graph. It is given that the correlation coefficient between X and Y is 0.4 and the correlation coefficient is independent of change of origin and scale but it depends on variables 

∴ The correlation coefficient between 2X and (-Y) is - 0.4

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

The regression coefficient is independent of the change of origin. But, they are not independent of the change of the scale. It means there will be no effect on the regression coefficient if any constant is subtracted from the values of x and y

Explanation

The cost of living also known as consumer price index is the country’s principal measure of price change.

⇒ Cost of living index is denoted by CLI

⇒ CLI = ∑I_w/∑w

Where as, I = Relative price

W = Weight

∴ Each I in the group increases K times. If group indices increase k times  so also CLI does

CLI in terms of purchasing power

Purchasing power of money = (1/CLI) × 100

Given

Total number of students to be selected = 60

Size of the population = 180

Calculation

n₁ = (60/480) × 120

⇒ 15

n₂ = (60/480) × 200

⇒ 25

n₃ = (60/480) × 160

⇒ 20

∴ The value of n₁, n₂, and n₃ are 15, 25, 20

Proportional stratified sampling = Proportional stratified sampling also called proportionate stratification and stratified sampling with proportional allocation) is one of the most commonly applied forms of stratified sampling. In proportional stratified sampling, the researcher creates a stratified sampling frame, determine the sample sizes for the strata such that the sample size are proportional to the population strata sizes and then selects  a random sample of the appropriate size from each stratum. As a result, the sample strata will be proportional to their  sizes in the population and random in every other respect

Explanation

The consumer index number (CPI) or cost living index number are helpful in wage negotiation, formulation of income policy, price policy, rent control, taxation and general economic policies formulation**.**

⇒ The CPI is used in calculating the purchasing power of money and real wages

⇒ The purchasing power of money = (1/cost of living index)

⇒ Real wages wage = ( Money wage/cost of living index) × 100

∴  The relationship between the real wage and money wage is given by $Real wage = \dfrac{Money:wage}{Cost:of:living:index}\times 100$

Explanation

Qualitative classification is defined as the data that approximates and characterizes. Qualitative data can be observed and recorded. This data type is non-numerical in nature. This type of data is collected through methods of observations, one-to-one interviews, conducting focus groups, and similar methods.

Chronological classification = In such a series, time is the main factor according to which data is arranged. For example the production of milk in the country over a period of time

Geographical classification = Place, location or geographical division is the main factor in such a series. For example, a number of listed consumer's goods companies in different countries

Quantitative classification = When data is classified according to measurable characteristics such as height and weight it is quantitative classification of data. Quantitative data are expressed in terms of number. For example, classification of data based on the height of students in a class

∴ When the collected data is grouped with reference to time, we have Chronological classification

Explanation

Suppose that a simple random sample of size n is selected and n₁ unit in this sample belong to the class C.

Lert p = n₁/n be the sample proportion of units in the class C.

⇒ y̅  = n₁/n = p

⇒ S²_y = npq/(n – 1)

We have two cases

(i)- SRSWOR

When p is an unbiased estimator of P

⇒ Var(p) = [(N – n)/(N – 1)] × PQ/n

⇒ Est. Var(p) = [(N – n)/N] × pq/(n – 1)

⇒ rse(p) = √Var(p)/P

⇒ √[(N – n)/(N – 1)] × √)Q/Pn

∴ Var(p) = [(N – n)/(N – 1)] × PQ/n

Important Points

For SRSWR

p is an unbiased estimator of P

(ii) - Var (p) = PQ/n

Est Var(p) = PQ/(n - 1)

rse(p) = √Var(p)/P = √Q/Pn

Given

X̅ = mean of sales = 40

Y̅ = mean of advertisement cost = 6

Formula

Regression equation of X and Y = X - X̅ = r(x/y)(Y - Y̅)

Calculation

⇒ X - 40 = 0.09(10/1.5)(Y - 6)

⇒ X - 40 = (9/1.5)(Y - 6)

⇒ X - 40 = 6(Y - 6)

⇒ X - 40 = 6Y - 36

⇒ X = 6Y + 4

⇒ X = 6(10) + 4

∴ The likely sales for a proposed advertisement expenditure of Rs. 10 crore is 64 crore

Explanation

Systematic sampling = In this sample members from a large population are selected according to a random starting point but with a fixed, periodic interval and this interval is called sampling interval(k)

The sampling interval = Population size/sample size = N/n.

Quota sampling = It is non prbbaility version of stratified sampling. In this assembled sample has the same proportions of individuals as the entire population with respect to known characteristics, traits.

Multi stage sampling = It is the taking of samples in stages using smaller and smaller sampling units at each stage. It can be a complex form of cluster sampling.

Area sampling = It is sometimes referred to as block sampling, especially in the United States. It is also a form of cluster sampling. Evaluation. Area sampling is used a great deal in countries where there are no adequate population lists, which are replaced instead by maps.

∴ Option 3 is correct

Explanation

Mean of x₁, x₂, ---------,x_n = x̅

⇒ Mean of -x₁, -x₂ -------- -x_n

⇒ Mean of -(x₁ + x₂ --------x_n)/n

⇒ Mean of new observation = - x̅

Standard deviation = √[(∑(x_i - x̅)²/N]

(x_i - x̅)² is always positive 

∴ Mean and standard deviation is - x̅  and σ

Explanation

The error arising out due to drawing inference about the population on the basis of few observations(sampling) is termed sampling error. Such errors are non existent in complete enumeration survey.

∴ The standard error of the sample mean will be zero if n is N

Simple random sampling = It is a sampling technique where every item in the population has an even chance and likelihood of being selected in the sample

In it selection of items completely depends on chance or by probability  and each item has equal chance to be selected

It is also known as method of chances

The simple random sampling with replacement = It is called SRSWR, in it a unit is selected from the sampling fram E. The unit is replaced back and the next unit is selected. This process is repeated till a sample of the desired size is selected.

When n units are selected with SRSWR, the total number of possible samples are Nⁿ

The probability of drawing a sample is 1/Nⁿ.

Given

σ_x = √16 = 4

r = 0.48

Covariance = ∑xy/N = 36

Fomula

Covariance = ∑xy/N

r = ∑xy/N.σ_x × σ_y

Calculation

According to question

⇒ 0.48 = 36/4 × σ_y

⇒ σ_y = 9/0.48

∴ The standard deviation of y(σ_y) is 18.75

Formula

λ  = N/n

Calculation

N = (4 + 5 + 6 + 6 + 6 + 6 + 6 + 6 + 6 + 7) = 58

N = 10

⇒ λ = 58/10

∴ Unbiased estimate of λ is 5.8

Formula

Average = Sum of observation/number of observations

Calculation

Let the average expenditure be Rs.x

Total expenditure be 35x

When the number of students increased by 7, the average expenditure decreases by 1 and the total per day expenses increases by 42

According to question

⇒ 42(x - 1) - 35x = 42

⇒ 7x - 42 = 42

⇒ 7x = 84

⇒ x = 12

∴ The total expenditure of the hostel is 12 × 35 is Rs. 420

Given

X̅ = Sample mean = 20

N = sample size = 81

T0.025 = 1.96 = Z = Upper critical valuye

Formula

X̅ ∓ (z × σ/√N)

Calculation

According to the formula

⇒ 20 ∓ [1.96 × 0.90/√81]

⇒[ 20 - 1.96 × 0.90/√81] [20 + 1.96 × 0.90/√81]

⇒ [20 - 1.96(0.1)][20 + 1.96(0.1)]

⇒ [20 - 0.196][20 + 0.196]

⇒ [19.80][20.20]

∴ 95% confidence interval for μ is [19.80][20.20]

Formula

The regression line of y on x is given as

y - y̅ = r × σ_y(x - x̅)/σ_x

Calculation

According to the question

y - 2 = 0.8 × 9(x - 1)/3

⇒ y - 2 = 2.4(x - 1)

∴ y = 2 + 2.4(x - 1)

Formula

H_O : μ₁ = μ₂ is given by

(x̅₁ - x̅₂)/√(σ₁²/n₁ + σ₂²/n₂)

Calculation

According to the formula

⇒ (x̅₁ - x̅₂)/√(4/10 + 6/20)

⇒ (x̅1 - x̅2)/√(0.4 + 0.3)

∴ The test statistic for testing H0 : μ1 = μ2 is (x̅1 - x̅2)/√(0.7)

Explanation

We can collect the primary data from direct personal interview, indirect personal interview and schedules through enumerators

∴ All options are correct

Given

Observations = -2, 2, 5, 9, -9, 7, -7

Pdf = f\left( x \right) = \left{ {\begin{array}{*{20}{c}} {{e^{ - \left( {X - θ } \right)}},}&{x > θ }\ 0&{elsewhere} \end{array}} \right.

Formula

Likelihood function is given by

L\left( {x;θ } \right) = ;\mathop ∑ \limits_{i = 1}^n f(x_i; θ)

Calculation

L(x; θ) = \mathop ∑ \limits_{i = 1}^ne^{-(x - θ)}

Taking log on both side 

⇒ Log L(x; θ) = \mathop ∑ \limits_{i = 1}^nlog(\mathop ∑ \limits_{i = 1}^ne^{-(x - θ)}

⇒ \mathop ∑ \limits_{i = 1}^n(x_i - θ ])loge)

Log L is not differentiable w.r.t θ

According to maximum likelyhood principle

Log l has to be maximised

Log L is maximum if ∑(x_i - θ) is minimum

The mean deviation about a constant θ  = \mathop \sum \limits_{i = 1}^n(x_i - θ)

When x is odd than θ  = Median(x₁, x₂, ----x_n)

∴  Median of (-2, 2, 5, 9, -9, 7, -7) is 9

∴ Then the MLE of θ is 9

Calculation

The regression line of Y on X is Y = 30 - 0.9x

⇒ Y - 30 = - 0.9x

The regression equation line of Y on X is = y - y₁ = r(s_y/s_x)(x - x₁)

Comparing both equations, we get

⇒ r(s_y/s_x) = -0.9

⇒ r(9/2) = -0.9

⇒  r = (-0/9 × 2)/9 = - 0.2

∴ The value of the correlation coefficient rxy is -0.2

Explanation

A coin is tossed repeatedly until a head appears. If 'p' be the probability of getting a head in any toss and x be the number of tosses required to get the

first head then x is an unbiased estimator of 1 / p

Calculation

According to the question

A < x̅ 

If $P=\dfrac{\sum^n_{i=1}(x_i-A)^2}{\sum^n_{i=1}(x_i-\bar x)^2}$ > 1

Explanation

⇒ The H_o is rejected at level α if

⇒ T > F_α ;, n - 1

∴ H0 is rejected at level α if observed T > χ2α,n-1

Given

r12 = r13 = r23 = r

Formula

r_12.3 = (r₁₂ - r₁₃.r₂₃)/√[(1 - r₁₂²)(1 - r₂₃²)

Calculation

According to question

⇒ (r - r × r)/√(1 - r²)(1 - r²)

⇒ (r - r²)/(1 - r²)

⇒ r(1 - r)/(1 - r)(1 + r)

⇒ The value of r_12.3 is r/(1 + r)

Explanation

An observation is drawn from an experimental distribution with mean θ, if the observation is found to be 2, and we test H0: θ >1 against H1: θ >1 with

level e-3 then H_o will be accepted

Given

The average of 25 number = 36

The average of first 13 number = 32

Sum of last 13 number = 39

Calculation

Sum of 25 numbers = 25 × 36 = 900

⇒ Sum of first 13 number = 13 × 32 = 416

⇒ Sum of last 13 number = 13 × 39 = 507

⇒ The value of 13^th number = (416 + 507) - 900

∴ The value of 13^th number is 23

Formula

MSE = SSE/r(c - 1)

MSE = Variance within the sample

SSE = Sum of the squares of the error

Calculation

According to the question

⇒ MSE = 399.6/3(10 - 1)

⇒ MSE = 399.6/(3 × 9)

⇒ MSE = 399.6/27

∴ The MSE for the analysis would be 14.8

Formula

The spearman rank correlation coefficient = r = 1 -(6∑d²)/n(n² - 1)

Calculation

According to the formula

⇒ r = 1 - [(6 × 70)/6(6² - 1)]

⇒ r = 1 - [(6 × 70/210]

⇒ r = 1 - 6/3

∴ Spearman rank correlation coefficient r is - 1

Given

F = 1.5

SST = 18

N = 3

Formula

Total mean square = SST/(N - 1)

SST = Total sum of squares

N = number of observations

Calculation

Total mean square = 18/(3 - 1)

⇒ Total mean square = 18/2

⇒ Total mean square = 9

∴ Mean sdquare between classes is 9/3 = 3

Explanation

The standard deviation of whose batsman is less will be more more consistent

The standard deviation of A is least

∴ The A is the most consistent

Explanation

The residual or error is given by the ratio of sum of square due to error to the number of degree of freedom of freedom for residual

∴ MSE = SSE/(p - 1)(q - 1)

Important Points

MSE = Variance within the sample

SSE = Sum of squares of the error

Explanation

The simplest device for ascertaining whether two variables are related is to prepare a dot chart called scatter diagram. When this method is used the given data are plotted on a graph paper in the form of dots.

∴ If a data set contains n paired values on two variables x(independent) and y(dependent), then their plot is called Scatter diagram

Formula

Degree of freedom = (k - 1), k(c - 1)

Calculation

Here, k = 5, c = 20

⇒ The degree of freedom associated within F = (5 - 1), 5(20 - 1)

⇒ 4, 95

∴ The degree of freedom associated with F is 4, 95

Explanation

Secondary data = Secondary data is data collected by any party other than the researcher, including administrative data from programs, geodata from specialized sources, and census or other population data from governments.

∴ The data taken from the publication "sankhya" will be considered as Secondary data

Formula

If θ be the angle between the regression lines then,

Tan θ = [(1 - r²)/r] × (σ_x ×  σ_y)/(σ_x² + σ_y²)

Explanation

According to question

⇒ σ_x = σ_y

⇒ Tanθ = [(1 - r²/r] × (σ_x²/(2σ_x)

⇒ Tanθ = [(1 - r²/r] × 1/2

⇒ Tan θ = (1 - r²/2r

⇒ Tanθ = Sinθ/cosθ

⇒ Tanθ = Sinθ × cosθ 

⇒ Sinθ = Tanθ /secθ 

⇒ Sinθ = Tanθ/√(1 + tan²θ)

⇒ Sinθ = [(1 - r²/r]/√(1 + (1 - r²/2r)²)

⇒ Sinθ = (1 - r²)/√[(1 - r²)² + 4r²]

∴ $sin \theta=\sqrt{\dfrac{1-r^2_{xy}}{1+r^2_{xy}}}$

Explanation

The statement which is wrong about the variation is Chance causes variations are preventable

Explanation

The factor means for factor (F) = MST/MSE

Important Points

Where as, MTS = Mean total sum of square

MSE =  Mean sum of squares

Explanation

Kurtosis means ''bulginess''. Kurtosis refers to the degree of flatness or peakedness in the region about the mode of frequency curve. The degree of kurtosis of a distribution is measured relative to the peakedness of normal curve. In other words measures of kurtosis tell us the extent to which a distribution is more peaked or flat topped than the normal curve

∴ This feature of the frequency distributions in which they are different are called kurtosis

Explanation

The object of the F- test is to find out whether the two independent estimates of population variance differ significantly or whether the two samples may be regarded is drawn from the normal populations having the same variance

∴ In ANOVA for testing the equality of group means, one conducts F- test

Given

P(A ∩ B) = 1/2,

P(Ac ∩ Bc) = 1/2 

2P(A) = P(B) = p

Formula

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Calculation

According to the question

⇒ P(A ∪ B) = p/2 + p - 1/2

⇒ P(A ∪ B) = 3p/2 - 1/2

⇒ P(A ∪ B) = 1 - P(Ac ∩ Bc) 

⇒ P(Ac ∩ Bc)  = 1 - 1/2

⇒ P(Ac ∩ Bc)  = 1/2

⇒ 3p/2 - 1/2 = 1/2

⇒ 3p/2 = 1

∴ The value of p is 2/3

Explanation

Seasonal indices = Seasonal variation is measured in terms of an index, called a seasonal index. It is an average that can be used to compare an actual observation relative to what it would be if there were no seasonal variation. An index value is attached to each period of the time series within a year.

∴ Ratio to trend method for seasonal indices provide good results if:the periods are of short duration

Formula

Average = sum of observation/number of observations

Calculation

Let the total age of 25  students be y months and average be x months

⇒ x = y/25     ------(i)

Now 1 student aged 22years or 22 × 12 months is replaced by student of λ years

The average of the 25 students decreases by 3 months then,

⇒ x - 3 = (y - 22 × 12 + λ )/25

⇒ x = 3 + (y - 22 × 12 + λ)/25      ------(ii)

From eq (i) and (ii) we get,

⇒ y/25 = 3 +(y - 22 × 12 + λ)/25

⇒ y = 75 + y - 264 + λ 

⇒ λ = 189 months

∴ The age of new student is 15 years 9 months

Explanation

When we ignore residual errors in performing ANOVA then SSE will definitely increases

Formula

The yearly trend equation is in form of

y = a + bx

Then monthly trend line is obtained by dividing a by 12 and b by 144

Calculation

The yearly trend equation = y = 148.8 + 7.2x

⇒ The monthly line equation = y = 148.8/12 + 7.2y/144

∴ The monthly equation line is Y = 12.4 + 0.05X

Given

P(A1) = 1 / 3,

P(A2) = 1 / 4

P(A3) = 2 / 5

Calculation

If P(A₁) = 1/3

⇒ P(A̅₁) = 1 -1/3 = 2/3

⇒ P(A̅₂) = 1 - 1/4 = 3/4

⇒ P(A̅₃) = 1 - 2/5 = 3/5

Now the probability that exactly one of the events occur  = P(A1) × P(A̅2) × P(A̅3) + P(A̅1) × P(A2) × P(A̅3) + P(A̅1) × P(A̅2) × P(A3)

⇒ 1/3 × 3/4 × 3/5 + 2/3 × 1/4 × 3/5 + 2/3 × 3/4 × 2/5

⇒ 3/20 + 2/20 + 4/20

∴The probability that exactly one of the events occur is 9/20

Given

the trend line with 2008 as origin is Y = 20.6 + 1.68X

Calculation

When we take 2004 as the origin then we go down by (2008 - 2004) = 4 years

⇒ y = 20.6 + 1.68(x - 4)

⇒ y = 20.6 + 1.68x - 6.72

⇒ y = 13.88 + 1.68

∴ the trend line with 2004 as origin is y = 13.88 + 1.68

Explanation

The first central moment about mean is Zero

Second Central moment is the variance.

Second and Central Moments are needed to find skewness.

Second and fourth central moments are needed to find the kurtosis.

∴ Option 1 is wrong statement so option 1 is answer

Formula

The equation of exponential trend line is given by

y = ab^x

Calculation

The exponential trend is given by

⇒ y = 20 × (1.5)^x

When shift the origin from 2004 to 2006 then

⇒ y = 20 × (1.5)^{(X + 2)}

⇒ y = 20 × (1.5)^x × (1.5)²

⇒ y = 20 × (1.5)^x × 2.25

∴ The equation of the exponential trend with 2006 as origin is: y = 45(1.5)^x

Calculation

Total number of shoes = 16

8 are from the left leg and 8 from the right leg

When we select 4 shoes at random

Total number of ways of choosing shoes = ¹⁶c₄

Probability of choosing shoes are all of right leg or left leg = [2 × ⁸c₄]/[¹⁶c₄]

⇒ [2 × 8!/4! × 4!]/16!/4! × 12!]

⇒ [2 × 8!/4! × 4!]/12! × 4!/16!]

⇒ 2 × 8! × 12!/4! × 16!]

⇒ (2 × 8 × 7 × 6 × 5 × 4! × 12!)/(4! × 16 × 15 × 14 × 13 × 12!)

∴ The required probability is 1/13.

Explanation

Seasonal variations are those periodic movements in business activities which occur regularly every year and have their origin their origin in the nature of  the year itself. The variations repeat during a period of 12 months.

The term cyclic refers to the recurrent variations in time series that usually last longer than a year and are regular neither in amplitude nor in length

∴ First option is correct one

Explanation

Secular trends refers to the general tendency of the data to grow or decline over a long period of time 

∴ First option is correct answer

Formula

Mean = ∑fx/∑f

Calculation

⇒ Mean = 3180/84

∴ The Mean is 37.86 (approx)

Explanation

The ratio to trend method = [Original data/Trend data] × 100

The seasonal indices are calculated using the following formula

⇒ [Quarterly Average/General Averages] × 100

∴ We multiply 1 so that we computed seasonal indices from ratio to trend method

Key Points

1. Equally likely events are those events which have an equal probability of occurring.

• For example: When we toss an unbiased coin, the probability of getting a heads is 1/2 and the probability of getting a tails is 1/2. So, it is an equally likely event.
• If A and B are equally likely events, then P(A) = P(B)

2. Exhaustive events are a set of events in a sample space such that one of them compulsorily occurs while performing the experiment.

• ​For example, while tossing an unbiased coin, there are two possible outcomes - heads or tails. So, these two outcomes are exhaustive events as one of them will definitely occur while flipping the coin.
• The probability of exhaustive events is equal to 1.

​3. Independent events are those events if the outcome of one event does not affect the outcome of another.

• If A and B are independent events, then P(A│B) = P(A)
• Using Multiplication rule of probability, P(A ∩ B) = P(B).P(A│B)
• P(A ∩ B) = P(B).P(A)

Important Points

For any two equally likely, exhaustive and independent events A and B.

• P(A ∩ B) = P(B).P(A)
• Since, P(A)= P(B) and P (A) + P(B) =1
• Therefore, P(A ∩ B)​ = 0.5 x 0.5
• P(A ∩ B) = 0.25

Thus, For any two equally likely, exhaustive and independent events A and B, P(A B) will be equal to 0.25.

Explanation

In second degree trend equation when the differences are taken of the first differences, the result obtained is a constant or nearly constant

∴ Differences should be used twice

Formula

Kurtosis = μ₄/μ₂²

μ₄ = ∑(x_i - x̅)⁴/n

μ₂ = ∑(x_i - x̅)²/n

Calculation

Mean = (-1 + 1 + 4)/4

⇒ Mean = 1

⇒ μ₄ = [(-1 -1)⁴ + ( - 1 - 0)⁴ + (1 - 1)⁴ + (4 - 1)⁴]/4

⇒ [16 + 1 + 81]/4

⇒ μ₄ = 98/4

⇒ μ₂ = [(1 - 1)² +(-1 - 0)² + (1 - 1)² + (4 - 1)²]/4

⇒ μ₂ = [4 + 1 + 9]/4

⇒ μ₂ = 14/4

⇒  μ2 = (14/4)²

⇒ Kurtosis = 98/4/(14/4)²

⇒ 98/4 × 16/196

∴ The value of kurtosis is 2

Explanation

The multiplicative model of time series: Xt = Tt x St x Ct x I_t

⇒  Tt, St, Ct, It are a trend, seasonal, cyclical, and irregular components

∴ I_t is an irregular component chosen randomly positive quantity

Calculation

According to the question A and B are independent event

⇒ P(A ∩ B) = P(A) × P(B)

⇒ P(A U B) = P(A) + P(B) - P(A ∩ B)

⇒ P(A U B) = P(A) + P(B) - P(A) × P(B)

⇒ 0.7 = 0.4 + p - 0.4p

⇒ p - 0.4p = 0.7 - 0.4

⇒ 0.6p = 0.3

∴ p is 0.5

Calculation

⇒ ∑ = ∑Y/N

⇒ 30/5 = 6

∴ Fitting a straight-line, is the trend value for the year 1992 is 6

Formula

Mean - Mode = 3(Mean - Median)

Calculation

26.8 - Mode = 3(26.8 - 27.9)

⇒ Mode = 26.8 + 3.3

∴ The value of Mode is 30.1

Explanation

From the ratio - to - trend method we can estimated seasonal index

∴ Option 3 is correct answer

Given

The probability of getting plumbing contract = P(A) = 2/3

Probability of getting not electric contract P(B̅) = 5/9

Probability of getting atleast one contract P(A U B) = 4/5

Calculation

The probability of not getting electric contract = 1 - P(B̅)

⇒ = 1- 5/9 = 4/9

Probability of getting both the contracts P(A ∩ B) = P(A) + P(B) - P(A U B)

⇒ P(A ∩ B) = 2/3 + 4/9 - 4/5

⇒ (90 + 60 - 108)/135

∴ The probability that he will get both the contracts P(A ∩ B) is 14/45

Explanation

Using moving average method the three means will be

⇒ (a + b + c)/3, (b + c + d)/3, (c + d + e)/3

⇒ (30 + 35 + 25)/3, (35 + 25 + 42)/3, (25 + 42 + 62)/3

⇒ 90/3, 102/3, 129/3

∴ The three year moving average is 30, 34, 43

Calculation

∴ The trend eliminated values are 0.93, 1.04, 1.08, 0.96, 1.0, 1.05, 0.95

Explanation

A and B are independent events 

⇒ P(A ∩ B) = P(A) × P(B)

According to question

⇒ P(A) × P(B) = 0.16   -----(A)

⇒ P(A') × P(B') = 0.36      ----(I)

⇒ [(1 - P(A)} × {1 - P(B)} = 0.36

⇒ [1 - P(B) - P(A) + P(A)P(B)] = 0.36

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

⇒ [1 - 0.36 + 0.16] = P(B) + P(A)

P(A) + P(B) = 0.8

By option elimination, we can conclude

P(A) = 0.4 and P(B) = 0.4

Explanation

Only trend can be checked by the link relatives in the time series

∴ Option 1 is correct

Calculation

Annual production of motor cyle = 1.80 lakh

⇒ 90° = 1.80

⇒ 1° = 0.02

⇒ 120° = 2.4 lacs

∴ The annual production of Bicycle is 2.4 lacs

Given

P(E ) > 0,

P(F|E) = 0.3

P(E ∩ Fc ) = 0.2

Calculation

P(F/E) = P(E ∩ F) /P(E)

⇒ P(F/E) = {P(E) -  P(E ∩ F')}/P(E)

⇒ 0.3 = P(E) - 0.2}/P(E)

⇒ 0.3 = 1 - 0.2/P(E)

⇒ 0.2/P(E) = 1 - 0.3

⇒ 0.2/P(E) = 0.7

∴ The value of P(E) is 2/7

Given

Trend equation line = 26y = 5335 + 624t

Calculation

Trend equation line = 26y = 5335 + 624t

⇒ y = 5335/26 + 624t/26

This equation is in the form of y = a + bt

The monthly rise in the production = b/12

∴ The monthly increase in production of fertilisers (in kg) is 624/26 × 12 = 2

Concept:

Let E₁, E₂,…, E_n be a set of events associated with a sample space S, where all the events E₁, E₂,…, E_n have a nonzero probability of occurrence and they form a partition of S. Let A be any event associated with S, then according to Bayes theorem,

$P(E_i/A)=\frac{P(E_i)P(A/E_i)}{\sum_{k=1}^n P(E_k)P(A/E_k)}$

for any k = 1, 2, 3, …., n

Calculation:

Let A_i ( i = 2, 3, 4, 5) be the event that urn contains 2, 3, 4, 5 white balls and let B be the event that two white balls have been drawn then we have to find P (A₅/B).

Since the four events A₂, A₃, A₄ and A₅ are equally likely we have

P (A₂) = P (A₃) = P (A₄) = P(A₅) = $\frac{1}{4}$

P(B/A₂) is probability of event that the urn contains 2 white balls and both have been drawn.

$P(B/A_2) = \frac{^2C_2}{^5C_2}= \frac{1}{10}$

Similarly,$P(B/A_3) = \frac{^3C_2}{^5C_2}= \frac{3}{10}$,

$P(B/A_4) = \frac{^4C_2}{^5C_2}= \frac{3}{5}$,

$P(B/A_5) = \frac{^5C_2}{^5C_2}= 1$