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MHT CET202123 Sep 2021Evening ShiftMathematicsProbabilityActual

For the probability distribution given by following array |l|l|l|l|l|l|l|l| X & 5 & 6 & 7 & 8 & 9 & 10 & 11 P ( X = x ) & 0.07 & 0.2 & 0.3 & k & 0.07 & 0.04 & 0.02 array Var ( X )=

Options

  1. A2.56
  2. B2.85
  3. C1.65
  4. D3.85

Correct answer

C. 1.65

Step-by-step solution

We have 0.07+0.2+0.3+ k +0.07+0.04+0.02=1 k =0.3 array |l|l|l|l| x _ i & p _ i & p _ i x _ i & p _ i x _ i ^2 5 & 0.07 & 0.35 & 1.75 6 & 0.2 & 1.2 & 7.2 7 & 0.3 & 2.1 & 14.7 8 & 0.3 & 2.4 & 19.2 9 & 0.07 & 0.63 & 5.67 10 & 0.04 & 0.4 & 4 11 & 0.02 & 0.22 & 2.42 2 |l| Total & 7.3 & 54.94 array Variance ( x )= p _ i x _ i ^2- ( p _ i x _ i )^2=(54.94)-(7.3)^2=1.65

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