SOLUTION: A company specialises in packing of washing powder for a national chain of supermarkets.The machine used to pack the washing powder is calibrated to pack a mean of 510g.However,the

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Question 261695: A company specialises in packing of washing powder for a national chain of supermarkets.The machine used to pack the washing powder is calibrated to pack a mean of 510g.However,the packing machine is 100% reliable and there is some variation in the mean.The variability as measured by the standrd deviation is 2.5g.It is also known that the weight packed by the machine follows a normal distribution.The supermarket chain has recently been complaining because they think that too many washing powder boxes are either underweight or overweight.
a)What is the percentage of the bags filled by the machine will weigh less that 507.5g?
b)What is the probability that a bag filled by the machine weighs less than 512g?
c)What percentage of the bags will weigh between 512 and 515g?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A company specialises in packing of washing powder for a national chain of supermarkets.
The machine used to pack the washing powder is calibrated to pack a mean of 510g.
However,the packing machine is not 100% reliable and there is some variation in the mean.The variability as measured by the standrd deviation is 2.5g.
It is also known that the weight packed by the machine follows a normal distribution.The supermarket chain has recently been complaining because they think that too many washing powder boxes are either underweight or overweight.
a)What is the percentage of the bags filled by the machine will weigh less that 507.5g?
z(507.5) = (507.5-510)/2.5 = -1
P(x<507.5) = P(z<-1) = 0.1587 = 15.87%
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b)What is the probability that a bag filled by the machine weighs less than 512g?
z(512) = (512-510)/2.5 = 0.8
P(x<512) = P(z<0.8) = 0.7881 = 78.81%
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c)What percentage of the bags will weigh between 512 and 515g?
Find the z-scores then find the % of population between those
scores. I get 18.91% using a TI calculator.
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Cheers,
Stan H.