If a cubical die is thrown, then the mean and variance of the random variable X, giving the number on the face that shows up, are respectively

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Accepted Answer

Correct answer: D. (d) X= 1 ,2 ,3 ,4 ,5 ,6 andP (1 )=P (2 ) …P (6 )= 16 Mean= x_i P= 1+2+3+4+5+66= 216= 72 Variance= x_i^2 P- ( Mcan )^2 = 16(1^2+2^2+3^2+4^2+5^2+6^2)- 494 = 16 91- 494= 3512 ⇒ Variance= 3512

$If a cubical die is thrown, then the mean and variance of the random variable X, giving the number on the face that shows up, are respectively$

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