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2014-10-06T17:59:09+05:30

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N = 2000          Mean wage = μ = Rs 70
Standard deviation of wages σ = Rs 5

X1 = Rs 70       X2 = Rs 71

Z1 = (X1 - μ) / σ = 70-70)/5 = 0             Z2 = (X2 - μ) / σ = (71 - 70)/ 5 = 0.2

Probability of a person having wages between Rs 70 and Rs71 = 
     P(Z1 <= Z  <= Z2)   P(0<= Z <= 0.2) = ZTable (0.2) = 0.0793 

Number of persons with wages between Rs 70 and Rs 71 = 0.0793 * N 
                 = 0.0793 * 2,000 = 158.6

 So answer is 158 persons

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2)  X1 = Rs 69    X2 = Rs73
     Z1 = (X1 - μ)σ  =  -2/5 = -0.4                 Z2 = (X2 - μ) / σ = 3/5 = 0.6

Probability (Rs 69 < X < Rs 73)  =  Probabilty (-0.2 <= Z <= 0.6)
        = P( -0.2 < = Z <= 0.0) + P(0<= Z <= 0.6 )
               = P(0 < Z <0.2) + P( 0 <=Z<=0.6)
          = 0.0793 + 0.2257 = 0.305

Number of persons with wages between Rs 69 &Rs 73 
      = N * 0.305 = 710
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3. X  == wages  > Rs 72

Z = (72 - 70 )/ 5 = 0.4

P (X  < 72 ) = P( Z < 0.4 ) = 0.5 + P (0 <= Z <= 0.4)
                     = 0.5 + 0.155 = 0.655

P(X > 72 ) = 1 - 0.655 = 0.345

Number of persons with wages above  Rs 72 = 0.345 * 2000 = 690

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4.
            X < Rs 65
            Z < (65 - 70 ) / 5 = -1
       P( Z < -1 ) = P(Z > 1 )  = 0.5  -  ZTable (1) = 0.5 - 0.341 = 0.159

        Number o f  persons with wages less than Rs 65 = 
                         0.159 * 2000 =  318
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