web analytics

Ex 13.4, 14 - A class has 15 students whose ages are 14, 17, 15 - Ex 13.4

Slide67.JPG
Slide68.JPG Slide69.JPG Slide70.JPG

  1. Chapter 13 Class 12 Probability
  2. Serial order wise
Ask Download

Transcript

Ex 13.4, 14 A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded. What is the probability distribution of the random variable X? Find mean, variance and standard deviation of X. Let X be the age of student selected The age of student selected can be 14, 17, 15, 21, 19, 16, 18, 20 So, X can have the value 14, 15, 16, 17, 18, 19, 20, 21 ∴ The Probability distribution is given by : The Mean is given by : E 𝑋﷯= 𝑖=1﷮𝑛﷮𝑥𝑖𝑃𝑖﷯ =14 × 2﷮15﷯+15× 1﷮15﷯+16× 2﷮15﷯+17× 3﷮15﷯+18× 1﷮15﷯ +19× 2﷮15﷯+20× 3﷮15﷯+21× 1﷮15﷯ = 28 + 15 + 32 + 51 + 18 + 38 + 60 + 21﷮15﷯ = 263﷮15﷯ = 17.53 Hence the mean is 17.53 The variance of x is given by : Var 𝑋﷯=𝐸 𝑋﷮2﷯﷯− 𝐸 𝑋﷯﷯﷮2﷯ So, finding 𝐸 𝑋﷮2﷯﷯ E 𝑿﷮𝟐﷯﷯= 𝑖 = 1﷮𝑛﷮ 𝑥﷮𝑖﷯﷮2﷯𝑝𝑖﷯ = 14﷮2﷯× 2﷮15﷯+ 15﷮2﷯× 1﷮15﷯+ 16﷮2﷯× 2﷮15﷯+ 17﷮2﷯× 3﷮15﷯ + 18﷮2﷯× 1﷮15﷯+ 19﷮2﷯× 2﷮15﷯+ 20﷮2﷯× 3﷮15﷯+ 21﷮2﷯× 1﷮15﷯ = 196 × 2 + 225 ×1+256 ×2+289 ×3+324 ×1+361 ×2+400 ×3+441 ×1﷮15﷯ = 392 + 225 + 512 + 867 + 324 + 722 + 1200 + 441﷮15﷯ = 𝟒𝟔𝟖𝟑﷮𝟏𝟓﷯ Now, Var 𝑿﷯=𝑬 𝑿﷮𝟐﷯﷯− 𝑬 𝑿﷯﷯﷮𝟐﷯ = 4683﷮15﷯ – 263﷮15﷯﷯﷮2﷯ = 4683﷮15﷯ – 69169﷮225﷯ = 4683 15﷯ − 69169﷮225﷯ = 70245 − 69169﷮225﷯ = 1076﷮225﷯ = 4.78 Standard deviation 𝜎﷮𝑥﷯ is given by 𝜎﷮𝑥﷯ = ﷮𝑉𝑎𝑟(𝑥)﷯ 𝝈﷮𝒙﷯ = ﷮𝟒.𝟕𝟖﷯ = 2.18

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail