3.
The weekly wages of 2000 workmen are normally distributed with mean wage of Rs
70 and wage standard deviation of Rs 5. Estimate the number of workers whose
weekly wage are

a. between Rs 70 and Rs 71

b. between Rs 69 and Rs 73

c. more than Rs 72, and

d.
less than Rs 65

1
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Answers

2014-09-02T03:13:22+05:30

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The area under the curve of normal distribution gives the probabilities. I hope you have the z tables or cumulative normal distribution tables or error function tables.

z-tables have mean value = 0 and probability at P(z<= 0) = 0.50
ZTable(z < 1.0) gives the probability for z to be between 0 and 1.0 = P(0 < z <1.0)

So, P(z<1.0) = P(z<0) + P(0<z<1.0) = 0.50 + ZTable(1.00)
Also, P(z< -2.0) = P(z<0) - P(-2<Z<0) = 0.5 - P(0<Z<2.0) = 0.5 - ZTable(2.00)
P(3<z<4) = ZTable (4.00) - ZTable(3.00)
========================================
Mu = mean = average = Rs 70, N = 2000,  Sigma = Rs 5 = std deviation

X = wage        Z = normal distribution function variable = [ X - Mean ] / Sigma

1) Number of estimated workers with Wages X between Rs 70 and Rs 71

       X2 = Rs 71    Z2 = (71-70)/5 = 0.2
       X1 = Rs 70    Z1 = (70-70)/5 = 0
       P( 70 < X < 71)  = P(0<Z<0.20) = ZTable (0.20) = 0.0793
       Estimation of Workers : N * Probability = 2000 * 0.0793 = 158.6 or 159

2) Estimation of workers with wages Rs 69 < X < Rs 73
       Z2 = (73-70)/5 = 0.6        Z1 = (69-70)/5 = -0.2
       P(-0.2 < Z < 0.6) = P(-0.2 <Z<0) + P(0<Z<0.6) = P(0<Z<0.2) + P(0<Z<0.6)
                                 = ZTable(0.2)+ZTable(0.6) = 0.0793 + 0.2257 =0.3050
       Estimation of Workers : N * Probability = 2000 * 0.3050 = 610

3)  Estimation of workers with wages more than Rs 72
       Z = (72-70)/5 = 0.4
       P(Z>0.4) = 1 - P(Z<0.4) = 1 - [ P(Z<0) + P(0<Z<0.4) ] = 1 - [ 0.5 + ZTable(0.4)]
                   = 0.5 - ZTable (0.4) = 0.5 - 0.1554 = 0.3446
       Estimation of Workers : N * Probability = 2000 * 0.3446 = 689.2

 4) Estimation of workers with wages less than Rs 65
       Z = (65-70)/5  = -1
       P(Z< -1) = P(Z<0) - P(-1<Z<0) = 0.50 - P(0<Z<1) = 0.50 - ZTable(1.00)
                   = 0.500 - 0.3413 = 0.1587
      Estimation of number of workers :  N * Probability = 317.4
===============================================
If you have error function values, then Multiply the following probabilities with N =2000.

a)  1/2  { erf [ (71-70)/ 5 √2 ] } = 1/2 erf (1/5√2)

b)  1/2 { erf (73 - 70) / 5√2 ] - erf [(69 - 70)/5√2 ]  } = 1/2 [ erf (3/5√2) - erf (-1/5√2) ]
         = 1/2 [ erf (3/5√2) - 1 + erf(1/5√2) ]
c) 1/2 { 1 - erf [ ( 72 - 70)/ 5√2 ] } = 1/2 [ erf (√2 * 1/5 )

d) 1/2  { 1 - erf [ 5 / 5√2] } = 1/2 { 1 - erf (1/√2) ]


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could you answer this please:- A salesman paid 12 visits to his area sales manager and noted that he had to wait for 10,15,20,17,11,25,30,27,36,40,5 and 26 minutes, respectively, before being called in his office. The area sales manager claims that the salesmen wishing to meet him do not have to wait for more than 20 minutes before being called in. Using the sign test, verify at 0.05 level of significance the claim made by the area sales manager.