MHT CET20239 May 2023Evening ShiftMathematicsProbabilityActual
A random variable X assumes values 1,2 , 3, ., n with equal probabilities, if var ( X )= E ( X ) , then n is
Options
- A4
- B5
- C7
- D9
Correct answer
C. 7
Step-by-step solution
aligned & X =1,2,3, n & aligned P ( X ) & = 1 n E ( X ) & = _ i =1 ^ n x_ i p _ i & = (1+2+3+ + n ) n & = n ( n +1) 2 n E ( X ) & = n +1 2 aligned aligned aligned & Var ( X )= _ i =1 ^ n x_ i ^2 p _ i -[ E ( X )]^2 & = 1^2+2^2+3^2+ + n ^2 n - ( n +1 2 )^2 & = n ( n +1)(2 n +1) 6 n - ( n +1 2 )^2 & = ( n +1)(2 n +1) 6 - ( n +1 2 )^2 & Var (X)=E(X) & ...[Given] & ( n +1)(2 n +1) 6 - ( n +1 2 )^2= n +1 2 & 2 n ^2+ n +2 n +1 6 - ( n ^2+2 n +1 4 )= n +1 2 & 4 n^2+6 n+2-3 n^2-6 n-3 12 = n+1 2 & n ^2-1=6( n +1) & n ^2-1=6