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Cost- sampling decisons
σ x¯ =S/√n
Where is the standard error of he mean is the standard devaluation of the population and n is the sample size. Here the standard error of the mean is expressed in terms of money i, e. ± Rs. 10. If the standard deviation is 100 the factor for 95 percent confidence is 1.96 of standard deviation. Thus the actual size of the sample would be
10/ 1.96 = 100/√n
√n = 196/10
N = 384
The total cost of the budget been reduced from Rs.8000 to Rs. 7680= ( 384x 20)
Another alternative may be to increase the allowable error from a lower level to higher level provided it does not affect the attitude of the respondents. For example if the error is increased to 15, the sample size will be
σ x¯ =s/√n
1.5/1.96= 100√n
√n= 196/ 15
N= 169
Thus the total budget will be reduced from Rs. 8000 to Rs. 3380= ( 196x 20)
Table of contents The tables of contents is an outline of the order of the chapters sections and sub section with their respective pages. If report includes a n
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#Solve the following Linear Programming Problem using Simplex method. Maximize Z= 3x1 + 2X2 Subject to the constraints: X1+ X2 = 4 X1 - X2 = 2 X1, X2 = 0
what is meant by Application Areas of Linear Programming?
1 strategic implication of product and process decisions 2 process planning and design 3 work measurement
A paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y paper i
#questA paper mill produces two grades of paper viz., X and Y. Because of raw material restrictions, it cannot produce more than 400 tons of grade X paper and 300 tons of grade Y p
Standard Deviation The standard deviation of samples (s) can be estimated by the pilot study past studies and the ranges of distribution. A pilot study may be conducted
Deviation Taken from Assumed Mean This methods is assorted when the arithmetic average is a fractional value. Taking deviation from fractional value would be a ver
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