Quantitative Aptitude PYQs

Concept:

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{favourable\;outcomes}}}}{{{\rm{total\;outcomes}}}}\)

Note:

Combination is the number of ways in which we can select a group of objects from a set. For example, selecting r things from set of n things is given as ⁿC_r.

 

Calculation:

Number of ways for selecting one man and one woman = ²C₁ × ²C₁

Number of ways for selecting two persons from group of four = ⁴C₂

Probability of committee having one man and one woman \(= \frac{{{{\rm{\;}}^2}{{\rm{C}}_1}{\rm{\;}} \times {{\rm{\;}}^2}{{\rm{C}}_1}}}{{{{\rm{\;}}^4}{{\rm{C}}_2}}}\)

\(= \frac{{2 \times 2}}{{\frac{{4 \times 3}}{2}}}\)

= 2/3

Properties of Probability:

• Probability of an event, 0 ≤ P(E) ≤ 1
• The Sum of the probability of all outcomes is equal to one. Let A and B be two events then:

 

P(A) + P(B) = 1

Calculation:

Sample space = {1, 2, 3, 4, 5, 6}

Let the probability of face coming up to be odd be p.

So, the probability of face coming up to be even be 2p.

So, we know sum of probability of all outcomes is equal to one.

P(odd face) + P(even face) = 1

⇒ P(1) + P(3) + P(5) + P(2) + P(4) + P(6)

⇒ x + x + x + 2x + 2x + 2x = 1

⇒ 9x = 1

⇒ x = 1/9

Now, the prime numbers in sample space are 2, 3, 5

So, probability that a prime number will show up = P(2) + P(3) + P(5)

= 2x + x + x

= 4x

= 4/9

Concept:

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{number\;of\;favourable\;outcomes}}}}{{{\rm{number\;of\;total\;outcomes}}}}\)

 

Calculation:

Let,

S: Sample space consists of non-negative number up to 50.

X: Numbers which are multiple of 3

Y: Numbers which are odd.

S = {0, 1, 2, …, 50} & n(S) = 51

X = {3, 6, 9, …, 48} & n(X) = 16

Y = {1, 3, 5, …, 49} & n(Y) = 25

Probability of an event, P(E) \(= \frac{{{\rm{number\;of\;favourable\;outcomes}}}}{{{\rm{number\;of\;total\;outcomes}}}} = \frac{{{\rm{n}}\left( {\rm{E}} \right)}}{{{\rm{n}}\left( {\rm{S}} \right)}}\)

P(X) = 16/51

P(Y) = 25/51

Hence, **both statement

(1) & (2)** are incorrect.

Concept:

Formulas used:

• P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
• P(A̅ ∩ B) = P(B) - P(A ∩ B)

 

Calculation:

Given: \({\rm{P}}\left( {\rm{A}} \right) = \frac{1}{2},{\rm{\;P}}\left( {{\rm{A}} \cup {\rm{B}}} \right) = \frac{2}{3}\) and \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{1}{6}\)

To find: P(A̅ ∩ B)

We know that, P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

So, P(B) = P(A ∪ B) + P(A ∩ B) - P(A)

\(\Rightarrow {\rm{P}}\left( {\rm{B}} \right) = \frac{2}{3} + \frac{1}{6} - \frac{1}{2}\)

⇒ P(B) = 2/6

⇒ P(B) = 1/3

 

P(A̅ ∩ B) = P(B) - P(A ∩ B)

⇒ P(A̅ ∩ B) = 1/3 - 1/6

⇒ P(A̅ ∩ B) = 1/6

Concept:

Coefficient of variation: It is defined as ratio of standard deviation to mean.

\({\rm{CV}} = \frac{{{{\rm{\sigma }}_{\rm{X}}}}}{{{\rm{\bar X}}}}\)

Range: It is difference between largest value and smallest value.

Mean deviation: Let x₁, x₂, …, x_n be n observations, then:

Mean deviation \(= \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} \left| {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right|}}{{\rm{n}}}\)

It is minimum when measure around mean.

 

Calculation:

The coefficient of variation is sometimes referred to the standard

deviation because the value of the coefficient of variation is

independent of the unit of measurement scale

So, statement (1) is incorrect.

Range is difference between largest value and smallest value. So, it defines how much the value is spread or dispread.

So, statement (2) is correct.

Mean deviation is minimum when measured around median.

So, statement (3) is correct.

Concept:

Standard deviation: Let x₁, x₂, …, x_n be n observations of X, then:

\({\rm{\sigma }} = \sqrt {\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}} \)

Calculation:

Given:

Number of observations (n) = 15, Mean (x̅) = 24 & Standard deviation (σ) = 0

So, \({\rm{\sigma }} = \sqrt {\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}} \)

\(\Rightarrow 0 = \sqrt {\frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}} \)

Squaring both sides, we get,

\(\Rightarrow 0 = \frac{{\mathop \sum \nolimits_{{\rm{i}} = 1}^{\rm{n}} {{\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)}^2}}}{{\rm{n}}}\)

\(\Rightarrow \mathop \sum \limits_{{\rm{i}} = 1}^{\rm{n}} {\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\bar x}}} \right)^2} = 0\)

⇒ (x₁ - x̅)² + (x₂ - x̅)² + … + (x₁₅ - x̅)² = 0

So, every deviation is squared and summed and summation is zero. This means every deviation is individually zero.

⇒ (x_i - x̅)² = 0 for i = 1, 2, …, 15

⇒ x_i = x̅ for i = 1, 2, …, 15

⇒ x_i = 24 for i = 1, 2, …, 15

 So, mean of smallest five observations \(= \frac{{24 + 24 + 24 + 24 + 24}}{5}\)

= 24

Concept:

• The product of regression coefficient of y on x and regression coefficient of x on y is always less than or equal to 1. So \({{\rm{b}}_{{\rm{xy}}}} \times {{\rm{b}}_{{\rm{yx}}}} \le 1\) where b_xy and b_yx are coefficient of regression.
• The signs of both coefficients should be same.

 

Calculation:

We know that the signs of both coefficients should be same. 

So option 2 and 3 can’t be answer because both coefficient are of different sign.

We will check 1 option and 4 option because both have same sign so we will check product of both coefficient which should be less than equal to 1.

Checking 4^th option

\( \Rightarrow \frac{1}{3} \times \frac{{10}}{3} \le 1\)

\( \Rightarrow \frac{{10}}{9} \le 1\)

\(\Rightarrow 1.11 \le 1\) Not true so it can’t be answer

Checking 1^st option

⇒ 1 × 1 ≤ 1

\( \Rightarrow 1 \le 1\) It is true so it is the right answer

Concept:

Conditional probability: It is the measure of probability of an event A is occurring given that another event B is already occurred, then:

\({\rm{P}}({\rm{B}}|{\rm{A}}) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{A}} \right)}}\)

Formulas used:

• P(A̅ ∩ B) = P(B) - P(A ∩ B)
• P(A̅ ) = 1 - P(A)

 

Calculation:

Given: \({\rm{P}}\left( {\rm{A}} \right) = \frac{1}{3},{\rm{\;P}}\left( {\rm{B}} \right) = \frac{1}{6}\) and \({\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) = \frac{1}{{12}}\)

To find: P(B|A̅)

We know that, conditional probability, \({\rm{P}}\left( {{\rm{A|B}}} \right) = \frac{{{\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {\rm{B}} \right)}}\)

So, \({\rm{P}}\left( {{\rm{B|\bar A}}} \right) = \frac{{{\rm{P}}\left( {{\rm{\bar A}} \cap {\rm{B}}} \right)}}{{{\rm{P}}\left( {{\rm{\bar A}}} \right)}}\)

So, P(A̅ ∩ B) = P(B) - P(A ∩ B)

⇒ P(A̅ ∩ B) = 1/6 - 1/12

⇒ P(A̅ ∩ B) = 1/12

Now, P(A̅ ) = 1 - P(A)

⇒ P(A̅) = 1 - 1/3

⇒ P(A̅) = 2/3

∴ \({\rm{P}}\left( {{\rm{B|\bar A}}} \right) = \frac{{\frac{1}{{12}}}}{{\frac{2}{3}}}{\rm{\;}}\)

⇒ P(B|A̅) = 1/8

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P( X = k) = ⁿC_k p^k (1 - p)^{(n - k)}

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

 

Properties:

• Mean of the distribution (μ_X) = n × p
• The variance (σ²_x) = n × p × (1 - p)
• Standard deviation (σ_x) = √{np(1 - p)}

 

Calculation:

Given:

Mean = n × p = 2/3    ….

(1) Variance = n × p × (1 - p) = 5/9    …. (2) Dividing equation (1) by equation (2)

\(\frac{{{\rm{np}}\left( {1 - {\rm{p}}} \right)}}{{{\rm{np}}}} = \frac{{\frac{5}{9}}}{{\frac{2}{3}}}\)

⇒ (1 - p) = 5/6

⇒ p = 1/6

Put p = 1/6 in equation (1)

n × (1/6) = 2/3

⇒ n = 4

Now, P(X = 2) = ⁴C₂ × (1/6)² × (1 - 1/6)^{(4 - 2)}

P(X = 2) = 6 × (1/6)² × (5/6)²

\(\Rightarrow {\rm{P}}\left( {{\rm{x}} = 2} \right) = \frac{{25}}{{{6^4}}}\)

\(\Rightarrow {\rm{P}}\left( {{\rm{X}} = 2} \right) = \frac{{25}}{{216}}\)

Concept:

Binomial distribution: If a random variable X has binomial distribution as B (n, p) with n and p as parameters, then the probability of random variable is given as:

P(X = k) = ⁿC_k p^k (1 - p)^{(n - k)}

Where, n is number of observations, p is the probability of success & (1 - p) is probability of failure.

 

Calculation:

Given:

Success of a ship to reach the port safely = p = 1/3

Probability that out of 5 ships, at least 4 should arrive = P(X = 4) + P(X = 5)

= ⁵C₄ (1/3)⁴ (1 - 1/3)^{(5 - 4)} + ⁵C₅ (1/3)⁵ (1 - 1/3)^{(5 - 5)}

\(= 5 \times \frac{1}{{81}} \times \frac{2}{3} + 1 \times \frac{1}{{243}} \times 1\)

\(= \frac{{10}}{{243}} + \frac{1}{{243}}\)

= 11/243

Calculation:

Let us assume that each month is equally likely as a birth month.

A: Probability that all three were born in same month.

B: Probability two of them were born in same month.

E: Probability that at least two born in same month

P(E) = P(A) + P(B)

 

So, there are 12³ choices for three birthdays all together, takin in order

For all three born in same month, there are 12 choices.

\({\rm{P}}\left( {\rm{A}} \right) = \frac{{12}}{{12 \times 12 \times 12}}\)

For 2 born in same month and the one in another; selecting any two born in any 12 months and the left one in another month.

\({\rm{P}}\left( {\rm{B}} \right) = \frac{{{{\rm{\;}}^3}{{\rm{C}}_2} \times 12 \times 11}}{{12 \times 12 \times 12}}\)

\({\rm{P}}\left( {\rm{E}} \right) = \frac{{12}}{{12 \times 12 \times 12}} + \frac{{{{\rm{\;}}^3}{{\rm{C}}_2} \times 12 \times 11}}{{12 \times 12 \times 12}}\)

\(= \frac{1}{{144}} + \frac{{33}}{{144}}\)

= 34/144

= 17/72

Concept:

Regression equation: Let x and y be two variables, then equation is given as:

\(\frac{{{\rm{x}} - {\rm{\bar x}}}}{{{{\rm{\sigma }}_{\rm{x}}}}} = {\rm{r}}\left[ {\frac{{{\rm{y}} - {\rm{\bar y}}}}{{{{\rm{\sigma }}_{\rm{y}}}}}} \right]\)

 

Calculation:

Given: x̅ = 10, y̅ = 90, σ_x = 3, σ_y = 12 and r_xy = 0.8

So, regression equation is given as,

\(\frac{{{\rm{x}} - {\rm{\bar x}}}}{{{{\rm{\sigma }}_{\rm{x}}}}} = {\rm{r}}\left[ {\frac{{{\rm{y}} - {\rm{\bar y}}}}{{{{\rm{\sigma }}_{\rm{y}}}}}} \right]\)

\(\Rightarrow \frac{{{\rm{x}} - 10}}{3} = 0.8\left[ {\frac{{{\rm{y}} - 90}}{{12}}} \right]\)

⇒ x - 10 = 0.2 × {y - 90}

⇒ x = 0.2y - 18 + 10

⇒ x = 0.2y - 8

⇒ X = 0.2Y - 8

Calculation:

Given that

\(\Rightarrow \;P\left( B \right) = \frac{3}{4},\;P\left( {A \cap B \cap {C^c}} \right) = \frac{1}{3}\) And \(P\;\left( {{A^c} \cap B \cap \;{C^c}} \right) = \frac{1}{3}\)

Form the above diagram we can write

\(\Rightarrow {\rm{P}}\left( {\rm{B}} \right) = {\rm{P}}\left( {{{\rm{A}}^{\rm{c}}}{\rm{\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right) + {\rm{P}}\left( {{\rm{A\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right) + {\rm{P}}\left( {{\rm{A\;}} \cap {\rm{B\;}} \cap {\rm{C}}} \right) + {\rm{P}}\left( {{{\rm{A}}^{\rm{c}}}{\rm{\;}} \cap {\rm{B}} \cap {\rm{C}}} \right)\)

\(\Rightarrow {\rm{P}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = {\rm{\;P}}\left( {{\rm{A\;}} \cap {\rm{B\;}} \cap {\rm{C}}} \right) + {\rm{P}}\left( {{{\rm{A}}^{\rm{c}}}{\rm{\;}} \cap {\rm{B}} \cap {\rm{C}}} \right)\)

\( \Rightarrow {\rm{P}}\left( {\rm{B}} \right) = {\rm{P}}\left( {{{\rm{A}}^{\rm{c}}}{\rm{\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right) + {\rm{P}}\left( {{\rm{A\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right) + {\rm{P}}\left( {{\rm{B\;}} \cap {\rm{C}}} \right)\)

\(\Rightarrow {\rm{P}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = {\rm{P}}\left( {\rm{B}} \right) - {\rm{P}}\left( {{{\rm{A}}^{\rm{c}}}{\rm{\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right) - {\rm{P}}\left( {{\rm{A\;}} \cap {\rm{B\;}} \cap {{\rm{C}}^{\rm{c}}}} \right)\)

Put all given values

\(\Rightarrow {\rm{P}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = \frac{3}{4} - \frac{1}{3} - \frac{1}{3}\)

\(\Rightarrow {\rm{P}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = \frac{3}{4} - \frac{2}{3}\)

\(\therefore {\rm{P}}\left( {{\rm{B}} \cap {\rm{C}}} \right) = \frac{1}{{12}}\)

Concept:

Pie Chart: A pie chart is a circular statistical graphic, which is divided into slices to illustrate numerical proportion.

Some properties:

• The arc length and central of each slide is proportional to the quantity it represents.
• The area of the pie chart denoted the total quantity.

 

Calculation:

Area of the circle = Total expenditure

Total expenditure of family A = Rs. 10000

Area of circle A = 10000

Total expenditure of family B = Rs. 8100

Area of circle B = 8100

\(\frac{{{\rm{Area\;of\;cirlce\;A}}}}{{{\rm{Area\;of\;circle\;B}}}} = \frac{{10000}}{{8100}}\)

\(\Rightarrow \frac{{{\rm{\pi r}}_{\rm{a}}^2}}{{{\rm{\pi r}}_{\rm{b}}^2}} = \frac{{100}}{{81}}\)

\(\Rightarrow \frac{{{{\rm{r}}_{\rm{a}}}}}{{{{\rm{r}}_{\rm{b}}}}} = \frac{{10}}{9}\)

⇒ r_a : r_b = 10 : 9

Concept:

Mean: Let X be a random variable with values x_i for i = 1, 2, …, n with frequencies f_i. So:

Mean \(= \frac{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{x}}_{\rm{i}}}{{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}\)

Binomial expansion formula:

• (1+ x)ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ

 

Calculation:

Given: 

x_i = i for i = 0, 1, 2, …, n

f_i = ⁿC_i for i = 0, 1, 2, …, n

So, mean \(= \frac{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{x}}_{\rm{i}}}{{\rm{f}}_{\rm{i}}}}}{{\mathop \sum \nolimits_{{\rm{i}} = 0}^{\rm{n}} {{\rm{f}}_{\rm{i}}}}}\)

\(= \frac{{0 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + 1 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + 2 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {\rm{n}} \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}{{{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}\)

Now we know that,

(1+ x)ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ ----(1)

Differentiating both sides w.r.t x

⇒ n(1 + x)^n-1 = 0 + 1 × ⁿC₁x + 2 × ⁿC₂x¹ + … + n × ⁿC_nxⁿ ----(2)

Put x = 1 in equation (2)

⇒ n × 2^n-1 = 0 + 1 × ⁿC₁ + 2 × ⁿC₂ + … + n × ⁿC_n ----(3)

Put x = 1 in equation (1)

⇒ 2ⁿ = ⁿC₀x⁰ + ⁿC₁x¹ + ⁿC₂x² + … + ⁿC_nxⁿ ----(4)

 

So, \(\frac{{0 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + 1 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + 2 \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {\rm{n}} \times {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}}{{{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_0} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_1} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_2} + \ldots + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{n}}}}} = \frac{{{\rm{n}} \times {2^{{\rm{n}} - 1}}}}{{{2^{\rm{n}}}}}\) [Using equation (3) & (4)]

\(= \frac{{\rm{n}}}{2}\)

Concept:

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{favourable\;outcomes}}}}{{{\rm{total\;outcomes}}}}\)

 

Calculation:

Probability of knowing correct answer = p

Probability of guessing the correct answer \(= \left( {1 - {\rm{p}}} \right) \times \frac{1}{{\rm{m}}}\)

So, probability of answering correctly \(= {\rm{p}} + \frac{{1 - {\rm{p}}}}{{\rm{m}}}\)

Probability of answering correctly considering that, he or she knows the correct answer \(= \frac{{\rm{p}}}{{{\rm{p}} + \frac{{1 - {\rm{p}}}}{{\rm{m}}}}}\)

\(= \frac{{{\rm{mp}}}}{{{\rm{mp}} + 1 - {\rm{p}}}}\)

\(= \frac{{{\rm{mp}}}}{{1 + \left( {{\rm{m}} - 1} \right){\rm{p}}}}\)

Concept:

Arithmetic mean: It is sum of all observations upon number of observations.

If x₁, x₂, x₃, …., x_n are the observations, then mean is given as:

\({\rm{\bar X}} = \frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + \ldots + {{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}\)

Geometric mean: It is the n^th root of product n numbers. Let x₁, x₂, …, x_n be n values of X, then:

Geometric mean \({{\rm{\bar X}}_{{\rm{geom}}}} = \sqrt[{\rm{n}}]{{\mathop \prod \limits_{{\rm{i}} = 1}^{\rm{n}} {{\rm{x}}_{\rm{i}}}}}\)

Calculation:

Given: x₁ & x₂ are positive.

A.M. - G.M. > 1

So, \(\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2} - \sqrt {{{\rm{x}}_1}{{\rm{x}}_2}} > 1\)

\(\Rightarrow {\rm{\;}}\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2} > \sqrt {{{\rm{x}}_1}{{\rm{x}}_2}} + 1\)

⇒ x₁ + x₂ > 2√(x₁x₂) + 2

⇒ x₁ + x₂ - 2√(x₁x₂) > 2

\(\Rightarrow {\left( {\sqrt {{{\rm{x}}_1}} - \sqrt {{{\rm{x}}_2}} } \right)^2} > 2\)

\(\Rightarrow \left| {\sqrt {{{\rm{x}}_1}} - \sqrt {{{\rm{x}}_2}} } \right| > \sqrt 2 \)

Concept:

• \({{\rm{\sigma }}^2} = \mathop \sum \limits_{i = 1}^N \frac{{{{\left( {{x_i} - \;\mu } \right)}^2}}}{N}\)
• \(\mu = \sum \frac{{{x_i}}}{N}\)

Where, σ² is variance, μ is Mean and N is Total numbers of observation

• The coefficient of variation expresses the standard deviation as a percentage of the sample mean. \(\Rightarrow {\bf{CV}} = \frac{{\bf{\sigma }}}{{\bf{\mu }}} \times 100\)

Where, σ is standard deviation and μ is mean of population

Calculation:

1. Variance is unaffected by change of origin and change of scale.

• Variance is independent of change of origin as the change in origin is uniformly added to all the values and hence the mean also and hence, when \({\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\mu }}} \right)^2}\) is calculated, it remains unaffected.

For example: x₁ + 5, x₂ + 5 ……… x_n+5 than mean and x_i both will change in same manner so while we calculate \({\left( {{{\rm{x}}_{\rm{i}}} - {\rm{\mu }}} \right)^2}\) it won’t change.

• But, change of scale alters all values unevenly and hence, variance changes.

For example: 2x₁, 2x₂ …………. 2x_n than mean and x_i both will change so variance also change.

So that first statement is not completely right

2. Coefficient of variance is independent of the unit of observations.

• The coefficient of variation expresses the standard deviation as a percentage of the sample mean. \(CV = \frac{\sigma }{\mu } \times 100\)

Both σ, μ have same unit so it gets cancel out.

So, CV is independent of the unit of variance 2^nd statement is right.

Concept:

Condition of triangle: Let a, b, c be the length of sides of the triangle, then:

Sum of smaller sides should be greater than third side, i.e., a + b > c

 

Probability: It is a measure of likelihood of an event to occur.

Probability of an event to occur, P(E) \(= \frac{{{\rm{favourable\;outcomes}}}}{{{\rm{total\;outcomes}}}}\)

Note:

Combination is the number of ways in which we can select a group of objects from a set. For example, selecting r things from set of n things is given as ⁿC_r

 

Calculation:

Given: five length of sticks 1, 3, 5, 7, 9 feet

Condition of a triangle to be formed, if a, b, c be length of sides of triangle then:

a + b > c

Total number of cases = ⁵C₃ = 10

Cases from which triangles can be formed:

1. 3 + 5 > 7
2. 3 + 7 > 9
3. 5 + 7 > 9

 

Number of favourable outcomes = 3

So, probability that triangle will be formed = 3/10 = 0.3

Concept:

Coefficient of Correlation: The geometric mean between two regression coefficient is equal to coefficient of correlation i.e.,

\({\rm{r}} = \sqrt {{{\rm{b}}_{{\rm{yx}}}}{{\rm{b}}_{{\rm{xy}}}}} \)

Some other properties of regression coefficient:

• The regression coefficient of y on x is represented by b_yx and x on y as b_xy.
• Both regression coefficient must have same sign.
• If one regression coefficient is greater than unity than others will be lesser than unity.
• Arithmetic mean of both regression coefficient:
• \(\frac{{{{\rm{b}}_{{\rm{yx}}}} + {{\rm{b}}_{{\rm{xy}}}}}}{2} \ge {\rm{r}}\)

 

Calculation:

Given: b_yx = 0.2 & b_xy = 1.8

To find: coefficient of correlation (r)

So, coefficient of correlation = r

\(= \sqrt {0.2 \times 1.8} \)

= √(0.36)

= 0.6