# Ex 13.4, 15

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 13.4, 15 In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0 if he opposed, and X = 1 if he is in favour. Find E(X) and Var (X). Given that X = 0 is Members oppose X = 1 is members favour proposal Given, 70% of members favour proposal So, P(X = 1) = 70% = 0.7 and 30% of members oppose proposal So, P(X = 0) = 30% = 0.3 ∴ Probability distribution is The expectation value E(x) is given by : E 𝑿 = 𝑖 = 1𝑛𝑥𝑖𝑝𝑖 = 0 × 0.3 + 1 × 0.7 = 0.7 The variance of x is given by : Var 𝑋=𝐸 𝑋2− 𝐸 𝑋2 So, finding 𝐸 𝑋2 E 𝑿𝟐= 𝑖 = 1𝑛 𝑥𝑖2𝑝𝑖 = 02 × 0.3 + 12 × 0.7 = 0 + 0.7 = 0.7 Now, Var 𝒙 = 𝐸 𝑥2− 𝐸 𝑥2 = 0.7 – 0.72 = 0.7 1−0.7 = 0.7 0.3 = 0.21 Hence the expectation E(x) = 0.7 & variance var(x) = 0.21

Example 7
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Ex 13.4, 15 Important You are here

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Ex 13.5, 4 Important

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Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.