SOLUTION: the average monthly sales of 5000 firms are normally distributed ,its mean and standard deviation are 36000 and 10000 respectively,find
a)the number of firm have sales over 40000
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-> SOLUTION: the average monthly sales of 5000 firms are normally distributed ,its mean and standard deviation are 36000 and 10000 respectively,find
a)the number of firm have sales over 40000
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Question 1075070: the average monthly sales of 5000 firms are normally distributed ,its mean and standard deviation are 36000 and 10000 respectively,find
a)the number of firm have sales over 40000
b)the percentage of firms having sales between rs.38500 and 41000
c)the no.of.firm having sales between rs.30000 and rs.40000
the relevant extract of the area table(under normal curve)is,
Z 0.25 0.40 0.5 0.6
area 0.0987 0.1554 0.1915 0.2257 Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! a) Probability(P) ( X > 40000 ) = 1 - P ( X < 40000 )
Z-score = (40000 - 36000) / 10000 = 0.4
consult Z-tables for probability associated with 0.4
P ( X < 40000) = 0.6554
P ( X > 40000) = 1 - 0.6554 = 0.3446
The number of firms with sales > rs40000 is 5000 * 0.3446 = 1723
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b) P ( 38500 < X < 41000 ) = P ( X < 41000 ) - P ( X < 38500 )
Z-score = ( 41000 - 36000 ) / 10000 = 0.5
P ( X < 41000 ) = 0.6915
Z-score = ( 38500 - 36000 ) / 10000 = 0.25
P ( X < 38500 ) = 0.5987
P ( 38500 < X < 41000 ) = 0.6915 - 0.5987 = 0.0928
The number of firms with sales > rs38500 and < rs41000 is 5000 * 0.0928 = 464
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c) P ( 30000 < X < 40000 ) = P ( X < 40000 ) - P ( X < 30000 )
P ( X < 40000 ) = 0.6554 Note we this in a)
Z-score = ( 30000 - 36000 ) / 10000 = -0.6
P ( X < 30000 ) = 0.2743
P ( 30000 < X < 40000 ) = 0.6554 - 0.2743 = 0.3811
The number of firms with sales > rs30000 and < rs40000 is 5000 * 0.3811 = 1905.5 is approx 1906
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