English, asked by kushwaha19, 1 year ago

The weekly wages of 2000 workmen are normally distributed with mean wage of Rs 70 and wage standard deviation of Rs 5. Estimatethe number of workers whose weekly wage are a.between Rs 70 and Rs 71b.between Rs 69 and Rs 73 c.more than Rs 72, and d.less than Rs 65


kvnmurty: subject should probably be maths.

Answers

Answered by kvnmurty
2
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
=================================================
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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