# 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
by gutsy860
have u got it?
Yep, thanks....
Thank u

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)
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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
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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) ]