SSC CGL Tier-II (JSO) 2017 Official Paper ( Held On: 22 Feb 2018 ) (Previous Year Paper)

Given

Mid values = 30, 39, 48, 57, 66

Formula used

Mid value of class interval = (L₁ + L₂)/2

L₁ = Lower limit

L₂ = Upper limit

Calculation

For Second class mid value would be (L₁ + L₂)/2

Here L₁ and L₂ are Lower value and upper value of class interval

 Mid point of 30 and 39 = (L₁ + L₂)/2

⇒ (30 + 39)/2 = 34.5

⇒ Mid point of 39 and 48

⇒ (39 + 48)/2 = 43.5

⇒ Lower limit (L₁ = 34.5)

⇒ Upper limit (L₂ = 43.5)

∴ The second class  of this distribution is 34.5 – 43.5

Shortcut Trick

By options

1 – 34 – 44

Class size  44 – 34 = 10

39 – 30  =9

So size gap does not match so option 1 is wrong

2 – 34.5 – 43.5

Class size = 43.5 – 34.5 = 9

The mid point gap

39 – 30 = 9, 48 – 39 = 9, 57 – 48 = 9, and 66 – 57 = 9

So option (2) match with Class Size so option (2) is correct

Given

Calculation

Here, total frequency = 23

Cumulative frequency = 61

∴ Frequency distribution is less than cumulative frequency

Important Points

 Cumulative frequency = it is the sum of the class and class below it in a frequency distribution

Calculation

the number of road accident patients treated less than 4 = 11 + 16 + 21 + 23 = 71

Given

Total expenses = 504 lakhs

Raw material had a central angle 45° 

Concept used

In the pie chart overall angle is 360°

Calculation

The total expenses = 504 lakhs

⇒ 360° = 504 lakhs

⇒ 1° = 1.4 lakh

⇒ 45° = 63 lakh

∴ the company's expense on raw materials (in Rs lakhs) - is 63 lakhs

Calculation

No of employees whose salary lies from 0 - 10,000 = 235

No of employees whose salary lies from 10,001 - 20,000 = 105

No of employees whose salary lies from 20,001 - 30,000 = 50

Total employs whose salary lies from 0 - 30,000 = 235 + 105 + 50

⇒ 390

Total number of employs = 450

⇒ 390/450 = 0.867

∴ 86.7% employees get salaries less than Rs 30,000

Given

Frequency density is actually the frequency per unit width.

Frequency of class interval 20 - 40 = 14

Formula used

\(Frequency ~density = \frac{Frequency}{(upper limit – lower limit) } \)

Calculation

$⇒\frac{14}{40-20}$

$⇒\frac{14}{20}=0.7$

∴ Frequency density is 0.7

Calculation

We divide these runs into class interval and frequencies

∴ There are (2 + 3 + 2 + 4) = 11 times he scored less than 25

Given

Observations = 3, 6, 10

Formula used

Harmonic mean = n/Σ1/x_i where i = 1, 2, 3, 4 ------

n = number of observation

x = observations value

Calculation

x_i = 3, 6, 10

⇒ Σ1/xi  = 1/3 + 1/6 + 1/10 = 3/5

⇒ number of observation = 3

⇒ Harmonic mean  = 3/3/5 = (3 × 5)/3

∴ Harmonic mean = 5

Important Points

The harmonic mean it is defined as the reciprocal  of the arithmetic mean of the observations of a set

Harmonic mean = n/Σ1/xi where i = 1, 2, 3, 4 ------

n = number of observation

x = observations value

Geometric mean = it  is defined as the n-th root of the product of the n observations.

GM = n(√x₁ × x₂--------x_n)

Given

Mean = 21.5

Total frequency = 40

Formula

Mean (x̅) = Σfx/Σf

Calculation

Hear x = (upper limit + lower limit)/2

⇒ Let f₁ = x and f₂ = y to solve question easily

⇒ x + 12 + y + 8 = 40

⇒ x + y = 20    ----(I)

⇒ Σfx = 460 + 5x + 25y

⇒ Mean = Σfx/Σf

⇒ 21.5 = (460 + 5x + 25y)/40

⇒ 460 + 5x + 25y = 860

⇒ 5x + 25y = 400

⇒ x + 5y = 80    ----(ii)

subtract equation (ii) by (i)

⇒ 4y = 60

⇒ y = 15

⇒ x = 20 - 15 = 5

∴ f₁  = x = 5 and f₂ = y  = 15

Important Points

X = mid value = (lower limit + upper limit)/2

This class interval is continuous class interval

Formula used

Median = L + (N/2 - C)/F × H  for group data

L = lower class limit of the median class = 40

N = total frequency = 120

C = cumulative frequency of the pre median class = 37

F = frequency of the median class = 35

H = width of median class = 20

Calculation

Median = 40 + (120/2 - 37)/35 × 20

⇒ 40 + 60 - 37)/35 × 20

⇒ 40 + 13.14

∴ Median = 53.14

Important Points

Median =  Median is that value of the variable which divides the whole distribution into two equal parts

Median for ungrouped data = (n + 1)^th/2 when n = odd

[(n/2)^th + (n/2 + 1)^th]2 observation when n = even

Calculation

Mean = Σfx/N

⇒ Mean = 3800/400

∴ Mean is 9.5

Important Points

x =  (Upper limit + lower limit)/2 

It is continuous data

Formula used

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

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 30 ,the class corresponding to this 20-30

So modal class is 20 - 30

Now lower limit of modal class (l) = 20

Frequency (f1) of modal class = 30

frequency (f0) of preceding the modal class  = 25

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

Class size = 10

⇒ Mode = 20 + ([30 - 25)/(2 × 30 - 25 - 20)] × 10

⇒ Mode = 20 + 10/3

∴  Mode is 23.33

Important Points

Another formula to calculate mode = 3 Median - 2 Mean

Given

Number = 5, 6, 4, and 2

Formula used

σ = √[x – x̅)²/n]

x̅ = (x₁ + x₂ + x₃ + x₄)/4

σ = Standard deviation

x̅ = Mean

Calculation

Mean = x̅ = (5 + 6 + 4 + 2)/4 = 4.25

⇒ σ = √(8.75)/4

⇒ √2.1875)

∴ Standard deviation of these values is 1.479

Standard deviation = It is the square root of the arithmetic mean of the square of deviation of the items from their mean value

σ = √[x – x̅)²/n]

`

Given

Mean of certain number = 20

Mean of square of same observation = 500

Formula used

σ = √[(∑x²/n –(∑x/n)²]

(∑x/n) = mean

∑x²/n = mean of square

Calculation

Standard deviation is given by

⇒ √[(∑(x²/n –(∑x/n)²] = √[500 – (20)²]

⇒ √(500 – 400)

∴ Standard deviation is 10

Formula 

Coefficient of range = (maximum range - minimum range)/(maximum range + minimum range)

Range = Maximum value - minimum value

Calculation

Maximum range = 12 and minimum range = 3

⇒ Coefficient of range = (12 - 3)/12 + 3

∴ Coefficient of range = 0.6

Given

n₁ = 60

n₂ = 40

x̅₁ = 45

x̅₂ = 55

σ²₁ = 4

σ²₂ = 9

Formula used

Combined mean = X̅ = (n₁ × x̅₁ + n₂ × x̅ ₂)/(n₁ + n₂)

Variance of combined mean = n₁(σ²₁ + d²₁) + n₂(σ²₂ + d²₂)/(n₁ + n₂)

n₁ and n₂ are group 1 and group 2 observation

x̅₁ and x̅₂ are mean of group 1 and group 2

σ²₁ and σ²₂ are variance of group 1 and group 2

d₁ = x̅₁ – X̅

d₂ = x̅₂ – X̅

Calculation

The combined mean is X̅

⇒ (60 × 45 + 40 × 55)/(60 + 40)

⇒ (2700 + 2200)/100

∴ Combined mean is 49

⇒ d₁ = 45 – 49

⇒ d₁ = -4

⇒ d²₁ = 16

⇒ d₂ = 55 – 49 = 6

⇒ d²₂  = 36

The combined variance of both group

⇒ 60(4 + 16) + 40(9 + 36)/(60 + 40)

⇒ (60 × 20 + 40 × 45)/(100)

⇒ (1200 + 1800)/100

⇒ 3000/100

∴ The combined variance is 30

Given

Arithmetic mean =  (μ) = 16

variance = σ² = 4

Formula

Standard deviation = σ = √variance

CV = Coefficient of variation = σ/μ

σ = Standard deviation

μ = Mean

Calculation

Standard deviation =  σ = √4 = 2

⇒ CV = (σ/μ) × 100

⇒ CV = (2/16) × 100

∴ Coefficient of variation is 12.5

Explaination

We all know that a measure of skewness is obtained by the use of the third moment about the mean (Third central mean)

Skewness is measured by the term β₁

⇒ μ²₃/μ²₂

The first moment about the origin is the measure of the mean

The second moment about the mean measures the variance 

The third moment about the mean measures the skewness

Calculaton

Moment about mean = The rth moments of a variable about the mean μ_r  denoted by μ =(1/N)[∑fi(xi - μ_x )r 

⇒ Put r = 2

∴ ∑fi(xi - μx )² represents the 2^nd moment around the mean

Given

Q₁ = 16

Q₃ = 40

M_d = 24

Formula 

absolute measure of skewness = = Bowley's measure of skewness S_q = (Q₃ - 2Md + Q₁)(/Q₃ - Q₁)

Q₁ = Upper quartile

Q₃ = Upper quartile

M_d = Median

Calculation

Bowley's measure of skewness = (40 + 16 - 2 × 24)/(40 - 16)

⇒ (56 - 48)/24

∴ Bowley's measure of skewness is 8/24 or 0.33

Bowley's measure of skewness = This measure is based on quartiles. For a symmetrical distribution, Q1 and Q3 are equidistance from median

⇒ absolute measure of skewness =  (Q3 - Md) - (Md - Q1)

⇒ A relative measure of skewness is called Bowley's coefficient and it is denotted by SQ

⇒ SQ = [(Q3 - Md) - (Md - Q1)]/[(Q3 - Md) + (Md - Q1)]

⇒ SQ = (Q3 - 2Md + Q1)(/Q3 - Q1)

1 - Given

Pearson’s coefficient of skewness = 0.25

Standard deviation = 6

Mode = 18

Formula

S_K = (Mean – Mode)/Standard deviation

S_k = Karl pearson’s coefficient

Calculation

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

Let mean be x

⇒ S_k = (Mean – Mode)/Standard deviation

⇒ 0.25 = (x – 18)/6

⇒ x = 0.25 × 6 + 1

⇒ x = 19.5

∴ Mean is 19.5

Calculaton

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

⇒ S_k = (Mean – Mode)/Standard deviation

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

If S_k > 0 the distribution is positive skewed

If S_k < 0 the distribution is negative skewed

⇒ To obtained the  S_k > 0 we have one condition

∴  Mean > Median > Mode

Explaination

Kurtosis measure the shape of a distribution or in other word kurtosis measures the peakness of its frequency distribution

The measures of kurtosis is given by β₂ = μ₄/μ²₂

⇒ β₂ =3 for mesokurtic

⇒ β₂ > 3, the curve is more peaked than mesokurtic and called leptokurtic

⇒ β₂ < 3, the curve is less peaked than mesokurtic and called platykurtic.

∴ For normal distribution the kurtosis β₂ =3

Explaintaion

Kurtosis measure the shape of a distribution or in other word kurtosis measures the peakness of its frequency distribution

The measures of kurtosis is given by β₂ = \Meu₄/meu²₂

⇒ β₂ =3 for mesokurtic

⇒ β₂ > 3, the curve is more peaked than mesokurtic and called leptokurtic

⇒ β₂ < 3, the curve is less peaked than mesokurtic and called platykurtic.

From this diagram it is shown that character of a leptokurtic distribution are

⇒ Positive excess kurtosis

⇒ Slender in shape

⇒ Fatter tails

∴ Option 1 is correct

Explanation 

Kurtosis measure the shape of distribution or in other word kurtosis measures the peakness of its frequency distribution

∴The measures of kurtosis is given by:

$\beta_2=\frac{\mu_4}{\mu_2^2}$

If the curve has much flatness then the normal curve is called a Leptokurtic curve

If the curve has lesser flatness than the normal curve then it is called a platykurtic curve

If the curve is equally flat as a normal curve then it is called a Mesokurtic curve

Given

b_xy = 0.5

b_yx = 405

Formula

The correlation coefficient is the geometric mean of two regression coefficient

r = √(b_xy × byx)

r = correlation coefficient

Calculation

r = √(0.5 × 4.5)

r = 1.5

∴ Correlation coefficient is 1.5

Calculation

∴ The Karl Pearson's coefficient of correlation is denotted by r(x, y) is defined as

r(x, y) = cov(x,y)/√σ_x² × √σ_y²

⇒ cov(x,y)/σ_x × σ_y

⇒ cov(x,, y)/S_xS_y

Important Points

Cov(x, y) = 1/n[∑(X_i – X̅)(Y_i – Y̅)

Where

We know that covariance = σ_xy = 1/n[∑(X_i – X̅)(Y_i – Y̅)

σx, σy or Sx and Sy denoted Standard deviation

Explaination

Remember the result that the range of values that coefficient of correlation can take are -1 ≤ r ≤ +1

Important Points

Basically, Correlation refers to the relationship of variables . There are 3 types of correlation

1 - Positive or negative correlation

2 - Simple, partial or multiple

3 - Linear or non linear

Calculation

The spearman's rank correlation is denotted by ρ

⇒ ρ = 1 - 6∑d²/n(n² - 1)

where as,

d = difference between the ranks of paired items of X and Y variables

∑d = total of squares of rank differences

n = number of pairs if items

Important Points

The method of ascertaining the coefficient of correlation by ranks was discovered by spearman. This method is based on the ranks of the variables

This method is applicable only to individual observations

In it biggest item gets first rank and so on.

Given

Coefficient of correlation r = 0.5

Formula

Coefficient of determination = r²

Coefficient of non determination = 1 - coefficient of determination

Calculation

r = 0.5

⇒ r² = 0.25

∴ The coefficient of determination = 0.25

∴ The coefficient of non determination is s1 - 0.25 = 0.75

1 - Given

X = 3u + 1

Y = 4v + 2

Regression coefficient between x and y = (b_uv) = 1.2

Formula

Regression coefficient are independent of origin but depend on scale

If u = ax + b and v = cy + d

Then b_uv = (a/c) × b_xy

Calculation

According to the question

x = 3u + 1

⇒ u = x/3 – 1/3

Similarly

v = y/4 – 1/2

By using b_uv = (a/c) × b_xy

⇒ 1.2 = (1/3)/(1/4)b_xy

⇒ 1.2 = (4/3)b_xy

⇒ b_xy = 3 × 1.2/4

∴ Regression coefficient between x and y is 0.9

Given              

Formula

Correlation coefficient = n∑xy – (∑x × ∑ y)/√{[n∑x² – (∑x)²] × [n∑y² – (∑y)²}

n∑xy = arithmetic mean of the summition of product of x and y

x and y = arithmetic average of x and y series

n∑x² and n∑y² = arithmetic mean of sum of sauare of item of x and y

Calculation

Putting these value in formula

⇒ r = 5 × 93 – 21 × 17/√{[5 × 109 – (21)²] × [5 × 93 - (17)²}

⇒ r = 108/√[104 ×  176]

⇒ r = 108/135

⇒ r = 0.8

∴ The correlation coefficient between x and y are 0.8

Formula

Spearman’s rank correlation coefficient is denoted by ρ

ρ = 1 – [6∑d²/n(n² – 1)]

d = difference between rank of paied items  of x and y variables

∑d² = total of squares of rank differences

n = number of pairs of items

Given

∑d² = ∑(R₁ – R₂)² = 21

n = 6

Calculation

ρ = 1 – [6 × 21/6(6² – 1)

ρ = 1 – 21/35

ρ = 14/35

∴ The rank correlation coefficient is 0.4

Given

Regression line of y on x

4y = 15 + 6x

y = 15/4 + (6/4)x

Coefficient of correlation between x and y = ρ_xy = 0.9

Formula

If the regression coefficient of x on y be b_xy, then

ρ_xy = √(b_yx × b_xy)

Calculation

b_xy = 6/4 = 1.5

⇒ 0.9 =√(1.5 × b_xy)

⇒ b_xy = (0.9 × 0.9)/1.5

⇒ b_xy = 0.54

∴ The regression coefficient of x on y is 0.54

Given 

Correlation coefficient between x on y = 0.75

Standard deviation of x and y = 5 and 6.4

Formula

Correlation coefficient ‘r’ = Cov(x,y)/S_xS_y

Cov(x,y) = Coveriance of x and y

S_x = Standard deviation of x

S_y = Standard deviation of y

Correlation coefficient of y on x = b_yx = Cov(x,y)/S_x²

Calculation

0.75 = Cov(x,y)/5 × 6.4

⇒ Cov(x,y) = 0.75 × 5 × 6.4

⇒ Cov(x,y) = 24

⇒ b_yx = 24/5²

∴ b_yx = 24/25

∴ correlation coefficient of y on x is 0.96

Given

Red ball = 4 and blue balls = 6

Formula

P(E) = P(E₁) + P(E₂)

ⁿC_r = n!/r!(n – r)!

Calculation

Probability when 2 red balls are drawn then the probability

⇒ P(E₁) = ⁴C₂/¹⁰C₂

Probability when 2 blue balls are drawn

P(E₂) = ⁶C₂/¹⁰C₂

⇒ P(E) = = ⁴C₂/¹⁰C₂ + ⁶C₂/¹⁰C₂

⇒ (⁴C₂ + ⁶C₂)/¹⁰C₂

⇒ [4!/2!(4 – 2)! + 6!/2!(6 – 2)!]/10!/2!(10 – 2)!

⇒ (6 + 15)/10!/2! × 8!

⇒ (21 × 2)/(10 × 9)

⇒ 42/90

⇒ P(E) = 7/15

∴ The probability of picking 2 balls of the same order is 7/15

Given

Probability that patient is alcoholic P(A) = 0.06

Probability that a patient has lever disease P(L) = 0.12

Probability that the patient is alcoholic given that the patient has lever disease = P(A/B) = 0.08

Formula

Probability that the patient has lever disease given that he or she is alcoholic

P(L/A) = P(A ∩ L)/P(L)

P(A ∩ L) = P(A/L) ×  P(L)

P(L/A)_ = P(A ∩ L)/P(A)

Calculation

P(A ∩ L) = 0.08 × 0.12

P(L/A)_= 0.08 × 0.12/0.06

⇒ 0.16

∴ The probability that the patient has liver disease given that he or she is alcoholic is 0.16

Calculation

The numbers which are multiples of 3 from 1 to 500 are

⇒ 3, 6, ------498

⇒ 498 = 3 + (n – 1) × 3

⇒ n – 1 = 495/3

⇒ n = 166

The numbers which are divisible by 7

7, 14, ---------497

⇒ 497 = 7 + (n – 1) × 7

⇒ n – 1 = 490/7

⇒ n = 71

⇒  Multiple of both (3 and 7)  = 23

The probability that the selected randomly from 1 to 500 is a multiple of 3 or 7 will be

⇒ (166 + 71)/500 – 23/500

⇒ 214/500

∴ The probability that a number selected randomly from 1 to 500 is multiple of 3 or 7 is 0.428

Given

P(A) = 2/3

P(B) = 5/8

P(Only A occurring) = 1/6

Calculation

P(A ∩ B) = P(A) – P(Only A occurring)

⇒ 2/3 – 1/6

P(A ∩ B) = ½

Probability of only B occurring = P(B) - P(A ∩ B)

⇒ 5/8 – 1/2

∴ Probability of only B occurring is 1/8

Given:

P(A) = 3/4

P(B) = 9/20

P(A ∩ B) = 1/4

Calculation:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 3/4 + 9/20 – 1/4

= 19/20

P(A'/B') = P(A' ∩ B')/P(B')= 1 – P(A ∪ B)'/[1 – P(B)]

= [1 – 19/20]/[1 – 9/20]

=1/11

∴  Probability P(A'/B') is 1/11.

Given

Odds in favour of A and B are 1 : 1 and 3 : 2

P(A) = 1/2

P(B) = 3/5

Calculation

The event are independent

⇒ P(A ∩ B) = P(A) × P(B)

⇒ 1/2 × 3/5 = 3/10

The probability that only one of the two events occurs

⇒ P(A) – P(A ∩ B) + [P(B) - P(A ∩ B)]

⇒ (1/2 – 3/10) + [3/5 – 3/10]

⇒ 1/2 + 3/5 – 6/10

∴ The probability that only one of the two events occurs is 1/2

Explanation:

The total number of 5 digit numbers that can be formed by using the five cards

⇒ 5! = 120

Divisibility of 4:

A number is divisible by 4 if it's last two digit is divisible by 4

Now, The numbers of number formed by these 5 digits which is divisible by 4 

Case 1. When the last two digit is 16 

The numbers of number formed = 3! = 6

Case 2. When the last two digit is 36 

The numbers of number formed = 3! = 6

Case 3. When the last two digit is 56

The numbers of number formed = 3! = 6

Case 4. When the last two digit is 96

The numbers of number formed = 3! = 6

So, The total numbers of number formed by these 5 digits which is divisible by 4 is 24

∴ The probability P(E) IS 24/120 = 0.2

Explanation

When 3 dice dice thrown the total outcomes n(s) = 216

Total outcomes having all the same digits are

{(1, 1, 1), (2, 2, 2), .....(6, 6, 6)} = 6 outcomes

Total outcomes having two same digits are

(1, 1, 2), (1, 1, 3), (1, 1, 4), (1, 1, 5), (1, 1, 6) = 5 outcomes

(2, 2, 1), (2, 2, 3) _-------(2, 2, 6)   = 5 outcomes

-------------------------------------------  = 5 outcomes

-------------------------------------------  = 5 outcomes

-------------------------------------------  = 5 outcomes

(6, 6, 1), (6, 6, 2) _-------(6, 6, 5)  = 5 outcomes

Total outcomes = 6 + 6 × 5 = 36 

∴ Probability of getting same digit of at least two dice = 36/216 = 1/6

Formula

Probability = Favourable outcome/total outcome

Calculation

Family of 3 children = G, G, G whereas G = Girls

We can select girl as ³C₁ = $\frac{3 !}{1! (3-1)!}$ = 3

No of possible ways to select the girls be $\frac{^3 C _1}{2!}$ 

⇒ Possible outcome = $\frac 32$

Total way to select girls =3! = 3 × 2

⇒ Total outcome 6

Probability = Favourable outcome/total outcome

⇒ Probability = $\frac{3/2}{6}=\frac{3}{ 2 \times 6} = \frac 14$

∴ The probability that the third child is a girl given that the other two are also girls is 1/4

Given

P(A) = 2P(B) = 6P(C)

Concept

When event are mutually exclusive and exhaustive

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

Calculation

Let P(A) be k

⇒ P(B) = k/2

⇒ P(C) = k/6

So according to the concept

⇒ k + k/2 + k/6 = 1

⇒ 10k/6 = 1

⇒ k = 3/5

∴ P(B) is k/2 = 3/(5 × 2) = 0.3

Calculation

We know expected value of random variable

f(x) = ∑x_ip_i

where I = 1, 2, 3,-----n

f(x) = P₁x₁ + P₂x₂ + P₃x₃ + P₄x₄ + P₅x₅

⇒ f(x) = 0.1 × 2 + 0.3 × 4 + 0.4 × 6 + 0.1 × 8 + 0.1 × 10

⇒ 0.2 + 1.2 + 2.4 + 0.8 + 1

∴ f(x) = 5.6

Calculation

Total number of bulbs = 100

Defective bulb = 30

Probability of defective bulb = p = 30/100

⇒ p = 3/10

⇒ q = 1 – p

⇒ q = 1 – 3/10 = 7/10

When 5 bulb are chosen random, n = 5

P(X = x) = ⁿc_xp^xq^{n – x}

⇒ ⁵c_xp^xq^{5 – x}

Getting more than 2 defective bulbs, x = 3, 4, 5

⇒ ⁵c₃(3/10)³(7/10)² + ⁵c₄(3/10)⁴(7/10) + ⁵c₅(3/10)⁵(7/10)^{5 – 5}

⇒ 0.1323 + (5 × 81 × 7)/100000 + (243/100000)

⇒ 0.1323 + 0.02835 + 0.00243

⇒ 0.163

∴ Probability of getting more than 2 defective bulbs if 5 are chosen randomly from the box is 0.163

Given

Mean of binomial distribution = np = 5

Variance of binomial distribution = npq = 4

Calculation

npq = 4 and np = 5

⇒ 5q = 4

⇒ q = 4/5

⇒ p = 1 – q

⇒ p = 1 – 4/5

∴ Probability of success = 1/5 = 0.2

Given

Mean of poisson distribution = 9

Calculation

In poisson distribution mean is equal to variance

∴ Variance is 9

Given

P(3) = P(4)

Concept

In poisson distribution = it is also a discrete distribution. Poisson distribution fulfil the limitation of nomal distribution

Poisson distribution is given by f(x) = (e^-λ × λ^x)/x!

Where x = 0, 1, 2, 3---- and λ = mean

Calculation

Here x = 3 and 4

⇒ (e^-λ × λ³)/3! = (e^-λ × λ⁴)/4!

⇒ λ  = 4!/3!

⇒ λ = 4

∴ mean of x in poisson distribution is 4

Calculation

P(- 1 < z < 1) = F(1) – F( - 1)

⇒ F(1) – [1 – F(1)]

⇒ 0.84 – [1 – 0.84]   (F(1) = 0.84

⇒ 0.84 – 0.16

∴ For standard normal distribution P(-1 < z < 1) is 0.68

Alternate Method

• Normal (Gaussian) Distribution:
• The normal distribution is another limiting form of the binomial distribution under the following conditions:
• The number of trials 'n' is definitely large.
• Neither p nor q is very small.(p = event will happen, q = event will not happen)
• ​The normal probability density function with mean μ and standard deviation σ is given by 
• $\frac{1}{{\sigma \sqrt {2\pi } }}{e^{\left( {\frac{{ - {{\left( {X - μ } \right)}^2}}}{{2{\sigma ^2}}}} \right)}}$​--------1
• The maximum probability occurs at the point x=μ and is given by :
• $\frac{1}{{\sigma \sqrt {2\pi } }}$--------2
• From equation 2 it signifies that smaller the values of standard deviation produce larger values of maximum probability this is called a measure of the precision of data or precision index.

​

Given

μ = mean = 20

σ2 = variance = 25

Formula

P(z ≤ 1) = 0.841

P(z ≤ 2) = 0.977

In normal distribution

z = (x - μ)/σ

Calculation

According to question, here x = 25

Standard deviation = √σ = 5

⇒ z = (25 - 20)/5

⇒ z = 1

⇒ P(x ≥ 25) = P(z ≥ 1)

⇒ 1 - P(z ≤ 1)

⇒ 1 - 0.841

⇒ 0.16

∴ The value of P(x ≥ 25) is 0.16

Given

10% people of the village are affected with viral disease

error margin = 5% or 5/100

Formula

Error margin  ≤ z × √(pq/n)

p denotes the probability of a success on one trial and q denotes the probability of a failure on one trial

n = size of the sample

Calculation

p = 10% = 10/100

⇒ p = 0.1

We know that q = 1 - p

⇒ q = 1 - 0.1

⇒ q = 0.9

According to formula

⇒ 5/100 ≤ 1.96 × √(0.1 × 0.9/n)

⇒ 5/100 = 1.96 × 0.3/√n

⇒ 5√n = 196 × 0.3

⇒ √n = 58.8/5

⇒ n = 138

∴ Size of the sample should taken as 138

Additional Informationz - Values for Selected (Percentage) Confidence Levels:

Given

n = 100

p = 10/100 = 0.1

q = 1- p = 1 - 0.1 = 0.9

Formula

Standard error of estimation = SE_p = √[(pq)/n]

Calculation

SE_P = √(0.1 × 0.9)/100]

⇒ 0.03

∴ Standard error of estimation is 0.03

Given

Standard deviation σ = 2 cm

z = 2.58 for 99% confident level

Formula

Standard error = σ/√n

Calculation

SE= σ/√n

⇒ 2.58 × σ/√n  ≤ 0.43 here z = 2.58

⇒ √n = (2.58 × 2)/0.43

⇒ √n = 12

⇒ n = 144

∴ The size of the sample in order to be 99% confident that the error of her estimates of mean does not exceed 0.43 cm is 144

Explanation

When the standard deviation of the population is known( then for the large samples

Standard error = s/√n

s = standard deviation for large sample

n = number of obserbvations

Given

Total items = 15

Select = 2

Formula

ⁿC_r = n!/r!(n - r)!

Calculation

The number of such samples = ¹⁵C₂ = 15!/2!(15 - 2)!

⇒ 15 × 14 × 13!/2 × 13!

⇒ 15 × 7 = 105

∴ The number of all such samples is 105.

Given

Sample size = 50 mm

Mean = x̅ = 40 mm

Standard deviation = σ = 21 mm

Concept

For the population means μ (or expected length; the standard is normal and it is given by

x̅ ± 1.96 σ/√n

Calculation

40 ± (1.96 × 21)/√50

⇒ 40 ± 5.88

⇒ 34.12 or 45.88

∴ The limits which have a 95% chance of including the expected length of the leaf was  34.12 or 45.88

Given

Confidence level limits = 94.12 - 105.88

Sample size = 100

Concept 

p ± 1.96 SE

SE = 1.96 σ/√n

Calculation

SE = 1.96 σ/√100

⇒ SE = (1.96 × σ)/10

⇒ SE = 0.196 σ 

⇒ p + 1.96 σ = 105.88

⇒ 100 + 0.196σ = 105.88

⇒ 0.196 = 5.88

⇒ σ = 5.88/0.196

∴ The standard deviation is 30

Given

n = 50

S = 8

n = sample size

S = Standard deviation

Formula

S = ρ/√n

Calculation

We know that from the upper formula, ρ = S√n

⇒ 8 × √50

According to the question, if the sample size increased by 200

⇒ new n = 200

⇒ Standard error = ρ/√n

⇒ (8 × √50)/√200

⇒ 4

∴ The standard error is 4 if sample size was increased to 200.

Additional Information 

• Standard Deviation (S. D): 
• A standard deviation denoted by σ is a measurement of the spread of a dataset around its mean.
• Low standard deviation means data are clustered around the mean, and high standard deviation indicates data are more spread out.
• ​Standard Error (S. E): 
• The standard error of the mean, or simply standard error, indicates how different the population mean is likely to be from a sample mean.
• It tells you how much the sample mean would vary if you were to repeat a study using new samples from within a single population.

Given

Number of defective bulbs = 15

Total sample = 200

Calculation

The estimation of sample proportion = Number of defective bulbs/(Total sample - Number of defective bulbs)

⇒ 15/(200 - 15)

⇒ 15/185

⇒ 0.0816

∴ The estimate of the sample proportion of defective articles is 0.0816

Given

Total manufactures bulbs = N = 10,000

Sample = n = 500

Defective pairs = 4%

Calculation

Here

⇒ p = 4% = .04

⇒ q = 1 - p

⇒ q = 1 - 0.04

⇒ q = .96

⇒ Estimation of error = SE = √pq/sample

⇒ √(.04 × .96)/500

⇒ SE = .0087

⇒ Z_α = 1.96 (Given value)

⇒ SE × Zα = .0087 × 1.96

⇒ 0.0167

⇒ p ± SE × Zα 

⇒ 0.4 ± 0.167

⇒ 0.4 + 0.167 = 0.0567

⇒ 0.4 - 0.167 = .0233

⇒ Number of bolts can be expected to be spoiled = 10000 × 0.0567

⇒ 567

⇒ Number of bolts can be expected to be spoiled = 10000 × 0.0233

⇒ 233

⇒ the interval of a number of bolts that would be of unacceptable quality produced per day at 95% confidence is 233, 567

Calculation

n = sample size = 15

Mean = x̅ = 65 lakhs

Standard deviation = S = 6 lakh

Concept

Confidence interval = x̅ - t_α/2)S/√n < μ < x̅ + tα/2)S/√n

t(_α/2) = t - value with (n - 1, degrees of freedom leavin an area of α/2 to the right

t_(α/2) = 2.98 ( Given in question)

Calculation

According to question, putting the values

⇒ 65 × 2.98 × 6/√15 < μ < 65 + 2.98 × 6/√5

⇒ 65 - 17.88/3.873 < μ < 65 + 17.88/3.873

⇒ 65 - 4.62 < μ < 65 + 4.62

⇒ 60.38 < μ <  69.62

∴ μ = { 60.22, 69.77}

Explaination

Standard error of the mean = SE_x- = σ_ρ/√n

Hence SE_x- decreases when population standard deviation(σ_ρ) decreases or if sample size (n) increase

Explaination

According to the definition and the adjacent figure a confidence interval increases on width as the level of confidence increases

Formula

Z- score = ( x - μ)/σ

μ = mean

σ = Standard deviation

Standard deviation = √(variance)

x = 125

μ = 121

Variance = 4

Calculation

Standard deviation = √4

⇒ 2

Z- score = ( 125 - 121)/2

⇒ 4/2

∴ Z - score is 2

Explaination

In statistical in inferences, according to testing of hypothesis, rejecting a null hypothesis whenit is true while type 11 error is incorrectly retaining of a false hypothersis. So, second option is the correct answer.

Given

Grand mean = (16/3 + 14/3 + 16/3)/3 

⇒ 46/9

Variance between the sample

Total sum of squares

⇒ (3 × 4/81 + 3 × 16/81 + 3× 4/81)

⇒ 72/81 = 8/9

⇒ For sample 2 = (3 × 4/82 + 3 × 16/81 + 3 × 4/81)

⇒ 72/81 = 8/9

⇒  For sample 3 = (3 × 4/82 + 3 × 16/81 + 3 × 4/81)

⇒ 72/81 = 8/9

⇒ The degree of freedom of treatment is 8/9/8/9 = 1

⇒ The degree of freedom of treatment is 8/9/8/9 = 1

∴ the degree of freedom is 1 + 1 = 2

Explaination

When it is believed that two independent factors might have as effect on the response variables of interest. It is possible to design the test so that an analysis if variance can be used to test for the effect of the factors simultaneously with the two factors analysis of variance, we can test two sets hypothesis with the same data at the same times.

∴ Option 3 is correct

Given

Confidence level = 90%

Calculation

⇒ One level alpha = 100 - 90 

⇒ 10%

∴ Two level alpha 10%/2 is  5%

Explanation

Smaller the value of α level, the confidence level will be high and hence smaller will be the level of null hypothesis

∴ Smaller will be the area of rejection

Formula

MSE = SSE/Degree of freedom

MSE = Variance within the samples

SSE = Sum of the squares of the error

Calculation 

For one - way classification model degree of freedom = n - m

n = Number of data

m = number of sample

∴ MSE = SSE/(n - m)

Given

S₁² = 9

S₂² = 4

S₁² = mean square for treatment

S₂² = Mean square for error

Formula

F - statistic = S₁²/S₂²

Calculation

F - statistic = 9/4

∴ F - statistic is 2.25

Calculation

Degree of freedom df₁ = (c - 1)

⇒ (4 - 1)

∴ Degree of freedom df1 = 3

⇒ Degree of freedom df₂ = c(r - 1)

⇒ 4(5 - 1)

∴ Degree of freedom df₂ = 16

∴ The degrees of freedom dF1 and dF2 for the critical value of F are  3 and 16

Concept

In one way classification model of ANOVA MSC = SSC/C - 1

MSC = mean sum of square between samples

SSC = sum of squares between samples

c = number of populations

Using the formula MSC = SSC/(C - 1) we can write

∴ MSTR = SSTR/(K - 1)

Concept

From annual trend equations, we can obtain monthly trend equations without any loss in accuracy. When Y units are annual totals then an

annual trend equation can be converted into an equations of monthly totals by dividing the computed constant a by 12 and value of b by 144

because the data are sums of 12 months hence, 'a' and 'b' must be divided by 12 and 'b; is again divided by 12 so that the time units (X will

be months as well as, 'b' would give monthly increments)

∴ The monthly trend equation becomes y = (a/12) + (b/144)X

Explanation

The second degree trend equation is given by Y = a + bx + Cx²

∴ The shape of the second degree trend equation is barabolic

Explanation

With 2010 as origin the trendline equation is given Y = 16.25 + 1.25x

Now if we take 2015 as the origin then the difference is 2015 - 2010 = 5

∴ The equation of the trendline will be Y = 16.25 + 1.25(x + 5)

∴  The equation of the trendline will be Y = 22.50 + 1.25x

Explanation

For the sales Y, the trend equation is given by

⇒ y = 500 + 20t - t²

⇒ dy/dt = 20 - 2t

For maximum sales dy/dt = 0

⇒ 20 - 2t = 0

⇒ t = 10

∴ The maximum sales will be achieved when time is 10 years

Explanation

The trend equation for output of a copper mine with 2010 as origin is given as

⇒ Y = 500 + 0.75t - 2.5t²

Now, new origin 2020, then trendline

⇒ replace t = t + 10

⇒ Y = 500 + 0.75(t + 10) - 2.5(t + 10)²

⇒ Y = 500 + 0.75t + 7.5 -2.5(t² + 100 +  20t)

⇒ Y = 500 + 0.75t + 7.5 - 2.5t² - 250 - 50t

⇒ Y = 257.5 - 49.2t - 2.5t²

∴ This is origin 2.20 trendline. So to calculate production at this origin take t = 0

⇒ Y = 257.5 tonnes

Explanation

The fitted annual sales trend is Y_t = 180e^0.2t

The t is always positive

⇒ e^x = 1 + x + x²/2! + x³/3! + -------

∴ Y_t is rising by a increasing absolute amount

Explanation

Seasonal variations are those periodic movements in business activity which occur regularly every year and have their origin in the nature of the year itself

∴ Those variations appear during a period of 12 months they can be predicted fairly accurately

∴ Option (4)  is correct option

Shortcut Trick

Identify the period of moving average by calculating the sum for 2010.

Sum (2010) = 23.6 × n. Checking n = 5: (22+23+24+21+24) = 114. Average = 114 ÷ 5 = 22.8 (No).

Checking 5-year moving average centered at 2010: (24+21+24+27+22) = 118. Average = 118 ÷ 5 = 23.6 (Matches).

For 2011 average, slide the window by one year: 2009 to 2013.

New Sum = 118 − 2008 value (24) + 2013 value (30) = 118 + 6 = 124.

New Average = 124 ÷ 5 = 24.8.

∴ The correct answer is 24.8.

Alternate Method

Given: Moving average for 2010 is 23.6. We need to find the moving average for 2011.

Formula Used: Moving Average = Sum of values in period ÷ Number of years (n)

⇒ Moving average for 2010 (5-year centered) = (Sales₂₀₀₈ + Sales₂₀₀₉ + Sales₂₀₁₀ + Sales₂₀₁₁ + Sales₂₀₁₂) ÷ 5

⇒ Sum for 2010 = 24 + 21 + 24 + 27 + 22 = 118

⇒ Average for 2010 = 118 ÷ 5 = 23.6 (Given condition satisfied)

⇒ For the year 2011, the 5-year period is 2009 to 2013.

⇒ Sum for 2011 = Sales₂₀₀₉ + Sales₂₀₁₀ + Sales₂₀₁₁ + Sales₂₀₁₂ + Sales₂₀₁₃

⇒ Sum for 2011 = 21 + 24 + 27 + 22 + 30 = 124

⇒ Moving Average for 2011 = 124 ÷ 5

⇒ Moving Average for 2011 = 24.8

∴ The correct answer is 24.8.

Additional Information

Simple Moving Average (SMA)

A calculation used to analyze data points by creating a series of averages of different subsets of the full data set to smooth out short-term fluctuations.

Odd vs Even Period

For odd periods (3, 5, 7), the average is placed against the middle year. For even periods (4, 6), a centered moving average is often required to align with a specific year.

Trend Analysis

Moving averages are primarily used in time series analysis to identify long-term trends by eliminating seasonal or irregular variations.

Explanation

The trend equations for the annual sales is given as Y = 500 + 6t

Compare this equation with linea trend equation y = a + bt

b = The average change in the dependent variable for each unit change in time

∴ Increase in sales per year = 6 lakhs

Given

The linear trend equation Y = a + bt

Calculation

After differentiate

dy/dt = b